Samacheer Kalvi Class 12 Chemistry Solutions Chapter 7 Chemical Kinetics

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Question 20. If the initial concentration of the reactant is doubled, the time for half-reaction is also doubled. Then the order of the reaction is ..................
(a) Zero
(b) one
(c) Fraction
(d) none
Answer: (a) Zero
In simple words: For a zero-order reaction, the time it takes for half of the reactant to be used up (half-life) doubles when the starting amount of reactant also doubles. This shows a direct relationship between half-life and initial concentration for these reactions.

🎯 Exam Tip: Remember that for zero-order reactions, half-life is directly proportional to the initial concentration, while for first-order reactions, it is independent of initial concentration.

Question 21. In a homogeneous reaction A → B + C + D, the initial pressure was P0 and after time t it was P. Expression for rate constant in terms of P0, P and t will be
(a) \( k = \frac{2.303}{t} \log \frac{2P_0}{3P_0 - P} \)
(b) \( k = \frac{2.303}{t} \log \frac{2P_0}{P_0 - P} \)
(c) \( k = \frac{2.303}{t} \log \frac{3P_0 - P}{2P_0} \)
(d) \( k = \frac{2.303}{t} \log \frac{2P_0}{3P_0 - 2P} \)
Answer: (a) \( k = \frac{2.303}{t} \log \frac{2P_0}{3P_0 - P} \)
In simple words: This question asks for the rate constant formula for a specific type of reaction involving gas pressures. The correct formula uses the initial pressure \(P_0\) and the pressure at time \(t\) (\(P\)) in a logarithmic expression. It reflects how pressure changes over time in this reaction.

🎯 Exam Tip: For gas-phase reactions, remember to use partial pressures instead of concentrations in rate law expressions, and always convert time units if necessary to match the rate constant units.

Question 22. If 75% of a first-order reaction was completed in 60 minutes, 50% of the same reaction under the same conditions would be completed in .........
(a) 20 minutes
(b) 30 minutes
(c) 35 minutes
(d) 75 minutes
Answer: (b) 30 minutes
In simple words: For a first-order reaction, the time it takes for 75% of the reaction to finish is twice the time it takes for 50% to finish. So, if 75% takes 60 minutes, then 50% will take half of that, which is 30 minutes. The half-life remains constant for a first-order reaction.

🎯 Exam Tip: Always remember that for a first-order reaction, \( t_{75\%} = 2 \times t_{50\%} \). This relationship is very useful for quick calculations.

Question 23. The half-life period of a radioactive element is 140 days. After 560 days, 1 g of the element will be reduced to
(a) \( \frac{1}{2} \) g
(b) \( \frac{1}{4} \) g
(c) \( \frac{1}{8} \) g
(d) \( \frac{1}{16} \) g
Answer: (d) \( \frac{1}{16} \) g
In simple words: Since the half-life is 140 days, the substance halves every 140 days. After 560 days, which is four half-lives (560 / 140 = 4), the initial 1g will be reduced four times: \( 1 \rightarrow 1/2 \rightarrow 1/4 \rightarrow 1/8 \rightarrow 1/16 \). Radioactive decay follows first-order kinetics.

🎯 Exam Tip: For radioactive decay, calculate the number of half-lives that have passed, then divide the initial amount by 2 for each half-life to find the remaining amount.

Question 24. The correct difference between first and second-order reactions is that ..................
(a) A first-order reaction can be catalysed; a second-order reaction cannot be catalysed.
(b) The half-life of a first-order reaction does not depend on [A0]; the half-life of a second-order reaction does depend on [A0].
(c) The rate of a first-order reaction does not depend on reactant concentrations; the rate of a second-order reaction does depend on reactant concentrations.
(d) The rate of a first-order reaction does depend on reactant concentrations; the rate of a second-order reaction does not depend on reactant concentrations,
Answer: (b) The half-life of a first-order reaction does not depend on [A0]; the half-life of a second-order reaction does depend on [A0].
In simple words: For a first-order reaction, how long it takes for half the substance to react doesn't change based on the starting amount. But for a second-order reaction, this half-life period does change if you start with a different amount. This is a key difference in how these reactions behave.

🎯 Exam Tip: Understanding how half-life depends on initial concentration is crucial for distinguishing reaction orders. First-order half-life is constant, while zero-order and second-order half-lives are concentration-dependent.

Question 25. After 2 hours, a radioactive substance becomes \( \frac{1}{16} \)th of original amount. Then the half life (in mm) is ..................
(a) 120 minutes
(b) 30 minutes
(c) 30 minutes
(d) 15 minutes
Answer: (c) 30 minutes
In simple words: When a substance reduces to \( \frac{1}{16} \) of its original amount, it means four half-lives have passed (since \( (1/2)^4 = 1/16 \)). If this took 2 hours (120 minutes), then each half-life must be 120 minutes divided by 4, which is 30 minutes. This is typical for first-order decay.

🎯 Exam Tip: For radioactive decay (which is always first-order), use the relationship \( \text{Remaining fraction} = (\frac{1}{2})^n \), where \( n \) is the number of half-lives, to find \( n \), then divide total time by \( n \) to find \( t_{1/2} \).

II. Answer the following questions:

Question 1. Define average rate and instantaneous rate.
Answer:
1. Average rate: The average rate of a reaction describes how the concentration of a reactant or product changes over a specific time period. It gives an overall view of the reaction speed during that interval.
2. Instantaneous rate: The instantaneous rate of reaction shows how fast the reaction is happening at one exact moment in time. It is found by looking at the slope of the concentration-time graph at that particular point. This rate is usually determined by the time derivative of concentration.
In simple words: Average rate is the speed of a reaction over a longer time, like measuring how far a car goes in an hour. Instantaneous rate is the speed at one exact moment, like looking at the car's speedometer right now.

🎯 Exam Tip: Clearly state that average rate is over an interval and instantaneous rate is at a specific moment. Mentioning graphical representation (slope) for instantaneous rate can earn extra marks.

Question 2. Define rate law and rate constant.
Answer:
1. Rate law: The rate law is an equation that connects the reaction rate to the concentrations of the reactants. Each concentration term in the rate law is raised to a specific power, which might not be the same as its stoichiometric coefficient in the balanced chemical equation. It tells us how the reaction rate depends on the amounts of starting materials.
2. Rate constant (k): The rate constant is a proportionality constant in the rate law. It shows the reaction rate when the concentration of each reactant is set to one unit. If all reactant concentrations are 1, then the rate constant equals the reaction rate. This value changes with temperature but not with concentration.
In simple words: The rate law is a math rule that shows how fast a reaction goes based on how much stuff you start with. The rate constant (k) is a special number in that rule; it tells you the actual speed when you have a certain amount of starting material.

🎯 Exam Tip: When defining rate law, emphasize that powers are experimentally determined and not necessarily stoichiometric coefficients. For rate constant, highlight its temperature dependence and concentration independence.

Question 3. Derive integrated rate law for a zero-order reaction A → product.
Answer:
A zero-order reaction is one where the rate does not depend on the reactant concentration over a wide range. Such reactions are uncommon. Let's consider a simple zero-order reaction: \( A \rightarrow \text{Product} \).
The rate law for this reaction is: \( \text{Rate} = k[A]^0 \).
Since \( [A]^0 = 1 \), the rate law simplifies to: \( \text{Rate} = k \).
We can also write the rate of reaction as: \( \text{Rate} = -\frac{d[A]}{dt} \).
Combining these, we get: \( -\frac{d[A]}{dt} = k \).
Rearranging the equation: \( -d[A] = k dt \).
Now, we integrate this equation. We integrate from the initial concentration \( [A_0] \) at time \( t = 0 \) to the concentration \( [A] \) at time \( t \):
\[ \int_{[A_0]}^{[A]} -d[A] = \int_{0}^{t} k dt \]
Integrating gives: \( -([A] - [A_0]) = k(t - 0) \)
So, \( -[A] + [A_0] = kt \)

\( \implies [A] = [A_0] - kt \).
This is the integrated rate law for a zero-order reaction. This equation shows that the concentration of reactant decreases linearly with time.
In simple words: For a zero-order reaction, the speed doesn't change, no matter how much chemical you have. The formula we get from this shows that the amount of chemical left simply goes down in a straight line over time.

🎯 Exam Tip: Ensure you show both the differential rate law and the integration steps clearly. The final integrated equation, \( [A] = [A_0] - kt \), is a key result to remember.

Question 4. Define the half-life of a reaction. Show that for a first-order reaction half-life is independent of initial concentration.
Answer:
The half-life (\( t_{1/2} \)) of a reaction is defined as the time needed for the concentration of a reactant to drop to half of its initial value. It's a measure of how quickly a reaction proceeds.

For a first-order reaction, the half-life is constant and does not depend on the initial concentration. The rate constant (\( k \)) for a first-order reaction is given by the integrated rate law:
\[ k = \frac{2.303}{t} \log \frac{[A_0]}{[A]} \]
At half-life (\( t = t_{1/2} \)), the concentration of the reactant \( [A] \) becomes half of the initial concentration, i.e., \( [A] = \frac{[A_0]}{2} \).
Substitute these values into the integrated rate law:
\[ k = \frac{2.303}{t_{1/2}} \log \frac{[A_0]}{[A_0]/2} \]
\[ k = \frac{2.303}{t_{1/2}} \log (2) \]
We know that \( \log(2) \approx 0.3010 \).

\( \implies k = \frac{2.303 \times 0.3010}{t_{1/2}} \)

\( \implies k = \frac{0.693}{t_{1/2}} \)
Rearranging this equation to solve for \( t_{1/2} \):
\[ t_{1/2} = \frac{0.693}{k} \]
This equation shows that for a first-order reaction, the half-life (\( t_{1/2} \)) depends only on the rate constant (\( k \)). Since the rate constant (\( k \)) is a fixed value at a given temperature, the half-life for a first-order reaction is independent of the initial concentration of the reactant. This makes first-order reactions predictable in terms of how long it takes for a certain fraction to react. This is a crucial property for understanding many natural processes, like radioactive decay.
In simple words: Half-life is how long it takes for half of a chemical to disappear. For a first-order reaction, this time is always the same, no matter how much chemical you start with. The formula only uses the rate constant, which is a fixed speed number.

🎯 Exam Tip: Clearly define half-life and then show the derivation from the first-order integrated rate law. The key is to demonstrate that \( [A_0] \) cancels out, leaving \( t_{1/2} \) dependent only on \( k \).

Question 5. What is an elementary reaction? Give the differences between the order and molecularity of a reaction.
Answer:
**Elementary reaction:** An elementary reaction is a single step in a reaction mechanism where reactants directly change into products without any intermediate steps. Each and every individual step in a complex reaction is an elementary reaction.

**Differences between order and molecularity:**
**Order of a reaction:**
1. The order of a reaction is the total sum of the powers of the concentration terms of the reactants in the experimentally determined rate law. It tells us how the rate is influenced by reactant concentrations.
2. The order can be zero, fractional, or an integer value. This means it's not always a whole number.

**Molecularity of a reaction:**
1. The molecularity of a reaction is the number of reactant species (atoms, ions, or molecules) that come together and collide in an elementary step. It's directly from the balanced equation of that single step.
2. Molecularity is always a whole number (1, 2, or 3) and can never be zero or a fraction. Reactions with molecularity greater than three are very rare because it's unlikely for more than three particles to collide at the same time and in the right way. This fundamental difference helps classify reactions.
In simple words: An elementary reaction is one simple step of a bigger chemical process. The 'order' of a reaction tells you how reactant amounts affect speed, which can be any number. 'Molecularity' tells you how many particles bump into each other in that single step, and it's always a whole number like 1, 2, or 3.

🎯 Exam Tip: Remember that order is an experimental quantity from the rate law, while molecularity is a theoretical concept based on the mechanism of an elementary step. Molecularity applies only to elementary steps, not overall reactions.

Question 6. Explain the rate-determining step with an example.
Answer:
The **rate-determining step** (or rate-limiting step) in a chemical reaction mechanism is the slowest step in the sequence of elementary steps. This slowest step limits the overall rate of the entire reaction, much like the slowest person in a relay race determines the team's overall time. All faster steps before it will not speed up the overall reaction beyond the pace of this slowest step.

**Example: Decomposition of hydrogen peroxide (\( H_2O_2 \)) catalyzed by iodide ions (\( I^- \))**
Overall reaction: \( 2H_2O_2(aq) \rightarrow 2H_2O(l) + O_2(g) \)
Experimentally, the reaction is found to be first order with respect to both \( H_2O_2 \) and \( I^- \), which means \( I^- \) is actively involved in the reaction.

The mechanism for this reaction involves two steps:
**Step 1:** \( H_2O_2(aq) + I^-(aq) \rightarrow H_2O(l) + OI^-(aq) \) (Slow)
**Step 2:** \( H_2O_2(aq) + OI^-(aq) \rightarrow H_2O(l) + O_2(g) + I^-(aq) \) (Fast)

In this example, **Step 1 is the rate-determining step** because it is the slowest step. The speed of the overall reaction is controlled by how fast this first step occurs. It involves both \( H_2O_2 \) and \( I^- \), which matches the experimental rate law. The catalyst \( I^- \) is consumed in the slow step but regenerated in the fast step, so it is not used up in the overall reaction. This concept is vital for optimizing chemical processes.
In simple words: The rate-determining step is simply the slowest step in a series of chemical changes. It's like a bottleneck: the overall speed of the whole process can only go as fast as this one slow step. For hydrogen peroxide breakdown, the first step where it reacts with iodide is the slow one.

🎯 Exam Tip: Clearly define the rate-determining step as the slowest step. Provide a simple, well-known example with its mechanism, identifying which step is slow/fast and explaining why the slow step is rate-determining.

Question 7. Describe the graphical representation of first order reaction.
Answer:
The integrated rate law for a first-order reaction is given by:
\( \ln[A] = \ln[A_0] - kt \)
This equation is in the form of a straight line equation, \( y = c + mx \), where:
- \( y = \ln[A] \) (the natural logarithm of the reactant concentration at time \( t \))
- \( c = \ln[A_0] \) (the y-intercept, which is the natural logarithm of the initial concentration)
- \( m = -k \) (the slope of the line, which is the negative of the rate constant)
- \( x = t \) (time)

If we plot \( \ln[A] \) on the y-axis against time (\( t \)) on the x-axis, we will get a straight line with a negative slope. This downward-sloping straight line confirms that the reaction is first-order. From the slope of this line, which is \( -k \), we can easily calculate the rate constant (\( k \)) of the reaction. This graphical method is an important way to verify the order of a reaction and determine its rate constant. The relationship shows that the concentration decreases exponentially with time.

🎯 Exam Tip: When describing graphical representation, always link the plot axes directly to the integrated rate law and state clearly what the slope and y-intercept represent.

Question 8. Write the rate law for the following reactions.
a) A reaction that is 3/2 order in x and zero-order in y.
b) A reaction that is second order in NO and first order in Br2.
Answer:
a) For a reaction that is 3/2 order in x and zero-order in y:
\( \text{Rate} = k[x]^{3/2}[y]^0 \)
Since \( [y]^0 = 1 \), the rate law simplifies to: \( \text{Rate} = k[x]^{3/2} \)
b) For a reaction that is second order in NO and first order in Br2:
\( \text{Rate} = k[NO]^2[Br_2]^1 \)
This is also often written as: \( \text{Rate} = k[NO]^2[Br_2] \). The powers in rate laws indicate how sensitive the reaction rate is to changes in reactant concentrations. A zero-order reactant, like 'y' in the first example, does not affect the rate at all.
In simple words: The rate law tells us how fast a reaction happens based on how much of each chemical is present. If it's "3/2 order in x and zero-order in y", it means the speed depends on x in a specific way, but not at all on y. If it's "second order in NO and first order in Br2", the speed depends on the square of NO's amount and directly on Br2's amount.

🎯 Exam Tip: Remember to include the rate constant \( k \) in all rate law expressions and write concentrations in square brackets. Any reactant that is "zero-order" will have its concentration term raised to the power of zero, which simplifies to 1.

Question 9. Explain a catalyst on reaction rate with an example.
Answer:
A **catalyst** is a substance that changes (usually increases) the rate of a chemical reaction without being used up itself. Catalysts work by providing a new reaction pathway that has a lower activation energy, which is the minimum energy needed for reactants to turn into products. By lowering this energy barrier, more reactant molecules can overcome it, leading to more successful collisions and a faster reaction rate. Catalysts are regenerated at the end of the reaction, so they can be used again.

**How a catalyst affects reaction rate:**
* A catalyst introduces a new mechanism for the reaction with a lower activation energy.
* Because the activation energy is lower, a larger fraction of reactant molecules has enough energy to react.
* This leads to an increased number of effective collisions, which speeds up the reaction.
* The catalyst itself does not undergo any permanent chemical change and can be recovered after the reaction.

**Example:** The decomposition of potassium chlorate (\( KClO_3 \)) to potassium chloride (\( KCl \)) and oxygen gas (\( O_2 \)) is quite slow at lower temperatures.
\( 2KClO_3(s) \rightarrow 2KCl(s) + 3O_2(g) \)
However, if we add manganese dioxide (\( MnO_2 \)) as a catalyst, the reaction proceeds much faster at a lower temperature. \( MnO_2 \) provides an alternative pathway with a lower activation energy, speeding up the oxygen production significantly. This shows how catalysts are essential in many industrial processes to make them more efficient.
In simple words: A catalyst is like a shortcut for a chemical reaction. It helps the reaction happen faster by finding an easier path, so less energy is needed for the chemicals to change. The catalyst itself doesn't get used up. For example, adding manganese dioxide makes potassium chlorate break down much quicker.

🎯 Exam Tip: When explaining catalysts, always mention that they provide an alternative pathway with lower activation energy and are not consumed. A clear example illustrating this effect will enhance your answer.

Question 10. The rate law for a reaction of A, B, and C have been found to be rate = k[A]2 [B] [L]3/2. How would the rate of reaction change when
1. The concentration of [L] is quadrupled
2. The concentration of both [A] and [B] are doubled
3. The concentration of [A] is reduced to (1/2)
4. The concentration of [A] is reduced to (1/3) and the concentration of [L] is quadrupled.
Answer:
The given rate law is: \( \text{Rate} = k[A]^2[B]^1[L]^{3/2} \)

1. **When the concentration of [L] is quadrupled ([L] becomes 4[L]):**
New Rate \( = k[A]^2[B]^1(4[L])^{3/2} \)
New Rate \( = k[A]^2[B]^1 (4^{3/2})[L]^{3/2} \)
Since \( 4^{3/2} = (\sqrt{4})^3 = 2^3 = 8 \),
New Rate \( = 8 \times k[A]^2[B]^1[L]^{3/2} \)
The rate of reaction increases by 8 times.

2. **When the concentration of both [A] and [B] are doubled ([A] becomes 2[A], [B] becomes 2[B]):**
New Rate \( = k(2[A])^2(2[B])^1[L]^{3/2} \)
New Rate \( = k(4[A]^2)(2[B]^1)[L]^{3/2} \)
New Rate \( = 8 \times k[A]^2[B]^1[L]^{3/2} \)
The rate of reaction increases by 8 times.

3. **When the concentration of [A] is reduced to (1/2) ([A] becomes \( \frac{1}{2}[A] \)):**
New Rate \( = k(\frac{1}{2}[A])^2[B]^1[L]^{3/2} \)
New Rate \( = k(\frac{1}{4}[A]^2)[B]^1[L]^{3/2} \)
New Rate \( = \frac{1}{4} \times k[A]^2[B]^1[L]^{3/2} \)
The rate of reaction is reduced to 1/4 times.

4. **When the concentration of [A] is reduced to (1/3) and the concentration of [L] is quadrupled ([A] becomes \( \frac{1}{3}[A] \), [L] becomes 4[L]):**
New Rate \( = k(\frac{1}{3}[A])^2[B]^1(4[L])^{3/2} \)
New Rate \( = k(\frac{1}{9}[A]^2)[B]^1(8[L]^{3/2}) \)
New Rate \( = \frac{8}{9} \times k[A]^2[B]^1[L]^{3/2} \)
The rate of reaction is reduced to 8/9 times. Understanding how each reactant affects the rate helps in controlling and optimizing reactions.
In simple words: This asks how changing the amounts of chemicals A, B, and L will affect the speed of the reaction based on its rule. For example, if you make L four times more, the reaction will go 8 times faster. If you double both A and B, it also goes 8 times faster. If A is halved, it goes 4 times slower. If A is reduced to one-third and L is quadrupled, the speed becomes 8/9 of the original.

🎯 Exam Tip: Always write down the initial rate law first. Then, substitute the new concentrations into the rate law and simplify the expression to determine the factor by which the rate changes. Pay close attention to the powers of each concentration term.

Question 11. The rate of formation of a dimer in a second order reaction is 7.5 x 10-3 mol L-1s-1 at 0.05 mol L-1 monomer concentration. Calculate the rate constant.
Answer:
Let's consider the dimerization of a monomer (M) to form a dimer, which is a second-order reaction.
The reaction can be written as: \( 2M \rightarrow (M)_2 \)
The rate law for a second-order reaction with respect to M is: \( \text{Rate} = k[M]^n \)
Given that the reaction is second order, \( n = 2 \).
So, the rate law is: \( \text{Rate} = k[M]^2 \)

We are given:
Rate \( = 7.5 \times 10^{-3} \text{ mol L}^{-1}\text{s}^{-1} \)
Monomer concentration \( [M] = 0.05 \text{ mol L}^{-1} \)

Now, we can rearrange the rate law to solve for the rate constant \( k \):
\( k = \frac{\text{Rate}}{[M]^2} \)
Substitute the given values:
\( k = \frac{7.5 \times 10^{-3} \text{ mol L}^{-1}\text{s}^{-1}}{(0.05 \text{ mol L}^{-1})^2} \)
\( k = \frac{7.5 \times 10^{-3} \text{ mol L}^{-1}\text{s}^{-1}}{0.0025 \text{ mol}^2\text{L}^{-2}} \)
\( k = \frac{7.5 \times 10^{-3}}{2.5 \times 10^{-3}} \text{ mol}^{-1}\text{L}\text{s}^{-1} \)

\( \implies k = 3 \text{ mol}^{-1}\text{L}\text{s}^{-1} \).
The rate constant for this dimerization reaction is \( 3 \text{ mol}^{-1}\text{L}\text{s}^{-1} \). This value helps compare the speeds of different second-order reactions.
In simple words: We have a reaction where single units (monomers) join to make pairs (dimers), and it's a second-order reaction. We know how fast it's going and how much monomer there is. We use these numbers in the rate law formula to find the rate constant, which tells us the reaction's basic speed, and it comes out to 3 mol\(^{-1}\)L s\(^{-1}\).

🎯 Exam Tip: Always write down the correct rate law based on the given order. Pay close attention to units and ensure they cancel out correctly to give the appropriate units for the rate constant of that order.

Question 12. For a reaction x +y + z → products, the rate law is given by rate \( = k [x]^{3/2} [y]^{1/2} \) what is the overall order of the reaction and what is the order of the reaction with respect to z.
Answer: The overall order of a reaction is found by adding the powers of the concentration terms in its rate law. For this reaction, the rate law is given as \( \text{Rate} = k[x]^{3/2}[y]^{1/2} \). So, the overall order is \( \frac{3}{2} + \frac{1}{2} = \frac{4}{2} = 2 \), which means it's a second-order reaction. Since the concentration of 'z' is not included in the rate law expression, the reaction is zero order with respect to reactant 'z'. This means changing the amount of 'z' will not change how fast the reaction goes.
In simple words: The overall order is found by adding the powers of \( x \) and \( y \), which is 2. The order for \( z \) is zero because its concentration is not in the rate law.

🎯 Exam Tip: Remember that reaction orders are determined experimentally and may not always match the stoichiometric coefficients in the balanced equation.

Question 13. Explain briefly the collision theory of bimolecular reactions.
Answer: Collision theory helps explain how chemical reactions happen, especially for reactions involving two molecules. It was developed by Max Trautz and William Lewis. This theory states that chemical reactions occur when reactant molecules bump into each other, or 'collide'. The more collisions there are, the faster the reaction generally is. For example, if you have two gases, A2 and B2, reacting, the rate of collision depends on how much of A2 and B2 are present. However, not all collisions lead to a reaction. For a collision to be 'effective' and cause a reaction, the molecules must:
1. Have enough energy: They must hit each other with a minimum amount of energy called 'activation energy'. If they don't have this energy, they just bounce off without reacting.
2. Have the right orientation: The molecules must also hit each other in a specific way, with the correct parts facing each other, for the reaction to occur. This is called the 'orientation factor'. Only a small fraction of all collisions actually meet these two conditions and lead to product formation. This is why not all reactions are super fast, even if molecules are colliding very often.
In simple words: Chemical reactions happen when molecules hit each other. For a reaction to happen, the molecules need enough energy and must hit in the right way. Not all collisions are successful.

🎯 Exam Tip: Key terms like 'activation energy', 'effective collisions', and 'orientation factor' are essential when explaining collision theory.

Question 14. Write Arrhenius equation and explains the terms involved.
Answer: The Arrhenius equation describes how the rate constant of a chemical reaction changes with temperature. It is given by: \( k = A \cdot e^{-E_a/(RT)} \). Here, \( k \) is the rate constant, which tells us how fast the reaction happens. \( A \) is the Arrhenius factor, also called the frequency factor, representing how often molecules collide with the correct orientation. \( E_a \) is the activation energy, the minimum energy needed for a reaction to occur. \( R \) is the gas constant, a universal constant. \( T \) is the absolute temperature, measured in Kelvin. The exponential term \( e^{-E_a/(RT)} \) represents the fraction of molecules that have enough energy to overcome the activation barrier at a given temperature. This equation shows that a reaction speeds up when the temperature increases because more molecules have enough energy to react.
In simple words: The Arrhenius equation (\( k = A \cdot e^{-E_a/(RT)} \)) shows how reaction speed changes with temperature. \( k \) is the reaction rate, \( A \) is a collision factor, \( E_a \) is energy needed to start the reaction, \( R \) is a constant, and \( T \) is temperature.

🎯 Exam Tip: Make sure to convert temperature to Kelvin (K) for all Arrhenius equation calculations, as it uses absolute temperature.

Question 15. The decomposition of \( \text{Cl}_2\text{O}_7 \) at 500K in the gas phase to \( \text{Cl}_2 \) and \( \text{O}_2 \) is a first order reaction. After 1 minute at 500K, the pressure of \( \text{Cl}_2\text{O}_7 \) falls from 0.08 to 0.04 atm. Calculate the rate constant in \( \text{s}^{-1} \).
Answer: To find the rate constant for this first-order decomposition, we use the integrated rate law. For a first-order reaction, the rate constant \( k \) can be calculated using the formula:
\( k = \frac{2.303}{t} \log \frac{[A_0]}{[A]} \)
Here, \( t \) is 1 minute, the initial concentration \( [A_0] \) is 0.08 atm, and the final concentration \( [A] \) is 0.04 atm.
Substitute the values into the equation:
\( k = \frac{2.303}{1 \text{ min}} \log \frac{0.08 \text{ atm}}{0.04 \text{ atm}} \)
\( k = 2.303 \log 2 \)
We know that \( \log 2 \) is approximately 0.3010.
So, \( k = 2.303 \times 0.3010 \)
\( k = 0.6932 \text{ min}^{-1} \)
To convert the rate constant from minutes to seconds, we divide by 60:
\( k = \frac{0.6932}{60} \text{ s}^{-1} \)
\( k \approx 0.01155 \text{ s}^{-1} \)
Therefore, the rate constant is approximately \( 1.153 \times 10^{-2} \text{ s}^{-1} \). A first-order reaction's rate depends linearly on the concentration of one reactant, meaning its half-life is constant.
In simple words: We used the first-order reaction formula. We plugged in the time (1 minute) and the starting (0.08 atm) and ending (0.04 atm) pressures. After calculating, we converted the rate constant from minutes to seconds.

🎯 Exam Tip: Always pay attention to the units requested in the question (e.g., \( \text{s}^{-1} \) instead of \( \text{min}^{-1} \)) and perform necessary conversions.

Question 16. Give examples for a zero-order reaction.
Answer: Here are some common examples of reactions that follow zero-order kinetics:
1. **Photochemical Reaction:** The reaction between hydrogen gas (\( \text{H}_2 \)) and chlorine gas (\( \text{Cl}_2 \)) in the presence of light (\( h\nu \)) to form hydrogen chloride (\( \text{HCl} \)).
\( \text{H}_2(\text{g}) + \text{Cl}_2(\text{g}) \xrightarrow{h\nu} 2\text{HCl}(\text{g}) \)
2. **Decomposition on a Surface:** The breakdown of nitrous oxide (\( \text{N}_2\text{O} \)) gas when it is heated on a hot platinum surface. The platinum acts as a catalyst.
\( \text{N}_2\text{O}(\text{g}) \rightleftharpoons \text{N}_2(\text{g}) + \frac{1}{2}\text{O}_2(\text{g}) \)
3. **Iodination of Acetone:** This reaction, when carried out in an acidic solution, is zero-order specifically with respect to iodine (\( \text{I}_2 \)). This means the rate of the reaction does not depend on the concentration of iodine.
\( \text{CH}_3\text{COCH}_3 + \text{I}_2 \xrightarrow{\text{H}^+} \text{ICH}_2\text{COCH}_3 + \text{HI} \)
The rate law for this is often written as \( \text{Rate} = k[\text{CH}_3\text{COCH}_3][\text{H}^+][\text{I}_2]^0 \).
Zero-order reactions are common in heterogeneous catalysis or when one reactant is present in a very large excess.
In simple words: Examples include hydrogen and chlorine reacting with light, nitrous oxide breaking down on platinum, and acetone reacting with iodine in acid. In these, the reaction speed does not depend on the amount of certain reactants.

🎯 Exam Tip: For zero-order reactions, the rate constant's units are usually concentration per unit time, e.g., mol L\(^{-1}\) s\(^{-1}\).

Question 17. Explain pseudo-first-order reaction with an example.
Answer: A pseudo-first-order reaction is originally a higher-order reaction (like second order) that behaves like a first-order reaction. This happens when one of the reactants is present in a much larger amount (in 'large excess') compared to the others. Because its concentration hardly changes during the reaction, it acts like a constant.
A good example is the acid hydrolysis of an ester, such as methyl acetate:
\( \text{CH}_3\text{COOCH}_3(\text{aq}) + \text{H}_2\text{O}(\text{l}) \xrightarrow{\text{H}^+} \text{CH}_3\text{COOH}(\text{aq}) + \text{CH}_3\text{OH}(\text{aq}) \)
The actual rate law for this reaction is \( \text{Rate} = k[\text{CH}_3\text{COOCH}_3][\text{H}_2\text{O}] \). This looks like a second-order reaction.
However, if we do this reaction in a lot of water, the amount of water changes so little that its concentration stays almost constant. We can then combine the original rate constant \( k \) with the constant concentration of water to form a new 'pseudo' rate constant, let's call it \( k' = k[\text{H}_2\text{O}]_{\text{constant}} \).
So, the rate equation becomes \( \text{Rate} = k'[\text{CH}_3\text{COOCH}_3] \).
This new rate law shows that the reaction now depends only on the concentration of the ester, making it appear as a first-order reaction. Many biological reactions in living systems occur in aqueous environments and often exhibit pseudo-first-order kinetics due to the vast excess of water. This simplification helps in studying complex reactions.
In simple words: A pseudo-first-order reaction is when a higher-order reaction acts like a first-order one because one reactant is in huge excess. An example is an ester reacting with a lot of water.

🎯 Exam Tip: Always identify the reactant in large excess when explaining pseudo-first-order reactions, as this is the key factor.

Question 18. Identify the order for the following reactions
i) Rusting of iron
ii) Radioactive disintegration of \( ^{92}\text{U}^{238} \)
iii) \( 2\text{A} + \text{B} \rightarrow \text{products}; \text{rate} = k [\text{A}]^{1/2} [\text{B}]^2 \)
Answer: Let's identify the reaction order for each case:
(i) **Rusting of Iron:** Rusting is a very complex process involving many steps, and it happens very slowly. Because it's so complex and slow, it's very difficult to define a simple rate law or a specific order for the overall rusting process. It is generally not assigned a simple reaction order.
(ii) **Radioactive Disintegration of Uranium (\( ^{92}\text{U}^{238} \)):** All radioactive decay processes, including the disintegration of uranium-238, follow first-order kinetics. This means the rate of decay depends only on the amount of radioactive substance present, and its half-life is constant.
(iii) **Reaction \( 2\text{A} + \text{B} \rightarrow \text{products} \) with rate \( k[\text{A}]^{1/2} [\text{B}]^2 \):** The order of a reaction is the sum of the exponents of the concentration terms in its rate law.
For this reaction, the rate law is given as \( \text{Rate} = k[\text{A}]^{1/2} [\text{B}]^2 \).
The order with respect to \( \text{A} \) is \( \frac{1}{2} \).
The order with respect to \( \text{B} \) is 2.
The overall order of the reaction is \( \frac{1}{2} + 2 = \frac{1}{2} + \frac{4}{2} = \frac{5}{2} = 2.5 \).
So, this is a 2.5 order reaction. Reaction orders can be fractional, as seen here. Reaction order is determined experimentally and does not always match the stoichiometric coefficients of the balanced chemical equation.
In simple words: Rusting has no simple order. Radioactive decay is always first-order. For the third reaction, the total order is 2.5, found by adding the powers of the concentrations.

🎯 Exam Tip: Remember that the order of a reaction with respect to a particular reactant is its exponent in the rate law, and the overall order is the sum of these exponents.

Question 19. A gas phase reaction has energy of activation 200 kJ mol\(^{-1}\). If the frequency factor of the reaction is \( 1.6 \times 10^{13} \text{ s}^{-1} \). Calculate the rate constant at 600 K. (\( e^{-40.09} = 3.8 \times 10^{-18} \))
Answer: To calculate the rate constant \( k \) for this gas phase reaction, we use the Arrhenius equation: \( k = A \cdot e^{-E_a/(RT)} \).
First, list the given values:
Activation energy, \( E_a = 200 \text{ kJ mol}^{-1} = 200 \times 10^3 \text{ J mol}^{-1} \)
Frequency factor, \( A = 1.6 \times 10^{13} \text{ s}^{-1} \)
Temperature, \( T = 600 \text{ K} \)
Gas constant, \( R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1} \)
We are also given that \( e^{-40.09} = 3.8 \times 10^{-18} \).
Now, substitute these values into the Arrhenius equation:
\( k = (1.6 \times 10^{13} \text{ s}^{-1}) \times e^{\frac{-(200 \times 10^3 \text{ J mol}^{-1})}{(8.314 \text{ J K}^{-1} \text{ mol}^{-1}) \times (600 \text{ K})}} \)
First, calculate the exponent \( \frac{-E_a}{RT} \):
\( \frac{-200 \times 10^3}{8.314 \times 600} \approx -40.09 \)
So, the equation becomes:
\( k = (1.6 \times 10^{13} \text{ s}^{-1}) \times e^{-40.09} \)
Using the given value for \( e^{-40.09} \):
\( k = (1.6 \times 10^{13}) \times (3.8 \times 10^{-18}) \)
Multiply the numbers and exponents:
\( k = (1.6 \times 3.8) \times (10^{13} \times 10^{-18}) \)
\( k = 6.08 \times 10^{(13-18)} \)
\( k = 6.08 \times 10^{-5} \text{ s}^{-1} \)
The rate constant for this reaction at 600 K is \( 6.08 \times 10^{-5} \text{ s}^{-1} \). A larger activation energy generally leads to a smaller rate constant, meaning a slower reaction, especially at lower temperatures. This value indicates how fast the reaction proceeds under these conditions.
In simple words: We used the Arrhenius equation with the given activation energy, frequency factor, and temperature. We calculated the exponent, then multiplied by the frequency factor to find the rate constant.

🎯 Exam Tip: Ensure that all units are consistent (e.g., energy in Joules if R is in J/mol K) before performing calculations with the Arrhenius equation.

Question 20. For the reaction \( 2\text{x} + \text{y} \rightarrow \text{L} \) find the rate law from the following data.

🎯 Exam Tip: To find reaction orders, look for experiments where one reactant's concentration changes while others remain constant, then compare the corresponding rates.

Question 21. How do concentrations of the reactant influence the rate of reaction?
Answer: The concentration of reactants has a direct impact on the rate of a chemical reaction. Generally, as the concentration of reactants increases, the reaction rate also increases. This effect can be understood using the collision theory of reaction rates. This theory states that reactions happen when reactant molecules collide with each other. When there are more reactant molecules packed into the same space (higher concentration), there is a greater chance for them to bump into each other. More collisions mean more opportunities for effective collisions that lead to a chemical reaction. Therefore, a higher concentration leads to a faster reaction rate. For most reactions, if you double the concentration of a reactant, the rate will increase, though not always by double, depending on the reaction order.
In simple words: Higher reactant concentration usually means a faster reaction. This is because more molecules are present, leading to more collisions and therefore more successful reactions.

🎯 Exam Tip: Always link concentration changes to the frequency of effective collisions when explaining its effect on reaction rate.

Question 22. How does the nature of the reactant influence rate of reaction?
Answer: The inherent nature of the reactants plays a crucial role in determining how fast a chemical reaction proceeds. Different substances react at different speeds, even under the same conditions. Chemical reactions involve breaking old bonds in reactants and forming new bonds to create products. The amount of energy needed to break these bonds, and the stability of the new bonds formed, varies greatly from one substance to another. This 'net energy change' is unique to each reactant, which means some reactions require more energy input to start, or proceed more easily than others. For example, the redox reaction between ferrous ammonium sulphate and potassium permanganate (\( \text{KMnO}_4 \)) is very quick, and you'll see the pink color of \( \text{KMnO}_4 \) disappear almost instantly. However, another redox reaction, like the one between oxalic acid and \( \text{KMnO}_4 \), is much slower. This reaction needs heat (around 60°C) to happen at a noticeable speed. The pink color disappears only after heating. Reactants with weaker bonds or more reactive functional groups generally react faster because less energy is needed to initiate bond breaking. These examples show that the type of chemicals involved directly affects how quickly a reaction occurs, based on their molecular structure and bonding properties.
In simple words: Different chemicals react at different speeds because their chemical bonds and structures are unique. Some bonds are easier to break, making the reaction faster, while others need more energy or specific conditions.

🎯 Exam Tip: Provide specific contrasting examples of fast and slow reactions (e.g., ionic vs. covalent bond breaking) to illustrate how reactant nature affects rate.

Question 23. The rate constant for a first order reaction is \( 1.54 \times 10^{-3} \text{ s}^{-1} \). Calculate its half life.
Answer: To calculate the half-life (\( t_{1/2} \)) for a first-order reaction, we use the specific formula that relates half-life to the rate constant (\( k \)).
The formula for the half-life of a first-order reaction is:
\( t_{1/2} = \frac{0.693}{k} \)
Given the rate constant \( k = 1.54 \times 10^{-3} \text{ s}^{-1} \).
Substitute the value of \( k \) into the formula:
\( t_{1/2} = \frac{0.693}{1.54 \times 10^{-3} \text{ s}^{-1}} \)
Now, perform the calculation:
\( t_{1/2} \approx 450 \text{ s} \)
So, the half-life for this first-order reaction is 450 seconds. For a first-order reaction, the half-life is constant and does not depend on the initial concentration of the reactant. This means it takes 450 seconds for half of the initial reactant to be consumed.
In simple words: The half-life is found by dividing 0.693 by the rate constant. Here, it is 450 seconds.

🎯 Exam Tip: The formula \( t_{1/2} = \frac{0.693}{k} \) is exclusively for first-order reactions; do not apply it to other reaction orders.

Question 24. The half life of the homogeneous gaseous reaction \( \text{SO}_2\text{Cl}_2 \rightarrow \text{SO}_2 + \text{Cl}_2 \) which obeys first order kinetics Is 8.0 minutes. How long will it take for the concentration of \( \text{SO}_2\text{Cl}_2 \) to be reduced to 1% of the initial value?
Answer: To find the time it takes for the concentration to reduce to 1% for a first-order reaction, we first need to calculate the rate constant \( k \).
1. **Calculate the rate constant (\( k \)):**
Given the half-life (\( t_{1/2} \)) is 8.0 minutes for this first-order reaction.
The formula for \( k \) in a first-order reaction is:
\( k = \frac{0.693}{t_{1/2}} \)
Substitute the given \( t_{1/2} \):
\( k = \frac{0.693}{8.0 \text{ minutes}} \)
\( k = 0.086625 \text{ minutes}^{-1} \approx 0.087 \text{ minutes}^{-1} \)
2. **Calculate the time (\( t \)) for concentration to reduce to 1%:**
We use the integrated rate law for a first-order reaction:
\( k = \frac{2.303}{t} \log \left( \frac{[A_0]}{[A]} \right) \)
We want to find \( t \) when \( [A] \) is 1% of \( [A_0] \). This means \( \frac{[A_0]}{[A]} = \frac{100}{1} = 100 \).
Rearrange the formula to solve for \( t \):
\( t = \frac{2.303}{k} \log \left( \frac{[A_0]}{[A]} \right) \)
Substitute \( k \) and the concentration ratio:
\( t = \frac{2.303}{0.087 \text{ min}^{-1}} \log(100) \)
We know that \( \log(100) = 2 \).
\( t = \frac{2.303}{0.087} \times 2 \)
\( t \approx 26.47 \times 2 \)
\( t \approx 52.94 \text{ minutes} \)
Therefore, it will take approximately 52.94 minutes for the concentration of \( \text{SO}_2\text{Cl}_2 \) to be reduced to 1% of its initial value. Reducing the concentration to 1% requires multiple half-lives, as each half-life reduces the remaining amount by 50%. This highlights how reaction rates change over time.
In simple words: First, we found the rate constant from the half-life. Then, using the first-order formula, we calculated the time needed for the substance to drop to 1% of its original amount.

🎯 Exam Tip: When using the integrated rate law, ensure consistent units for time and rate constant (e.g., both in minutes or both in seconds).

Question 25. The time for half change in the first-order decomposition of a substance A is 60 seconds. Calculate the rate constant. How much of A will be left after 180 seconds?
Answer: This problem involves a first-order decomposition reaction. We need to calculate the rate constant and then the remaining concentration after a certain time.
1. **Calculate the rate constant (\( k \)):**
Given that the half-life (\( t_{1/2} \)) is 60 seconds.
For a first-order reaction, the rate constant is calculated as:
\( k = \frac{0.693}{t_{1/2}} \)
Substitute the value of \( t_{1/2} \):
\( k = \frac{0.693}{60 \text{ s}} \)
\( k \approx 0.01155 \text{ s}^{-1} \)
2. **Calculate the percentage of A left after 180 seconds:**
Use the integrated rate law for a first-order reaction:
\( k = \frac{2.303}{t} \log \left( \frac{[A_0]}{[A]} \right) \)
We want to find \( [A] \) (as a percentage of \( [A_0] \)) after \( t = 180 \text{ s} \).
Rearrange the formula to solve for the ratio:
\( \log \left( \frac{[A_0]}{[A]} \right) = \frac{k \cdot t}{2.303} \)
Substitute the known values:
\( \log \left( \frac{[A_0]}{[A]} \right) = \frac{(0.01155 \text{ s}^{-1}) \times (180 \text{ s})}{2.303} \)
\( \log \left( \frac{[A_0]}{[A]} \right) \approx \frac{2.079}{2.303} \)
\( \log \left( \frac{[A_0]}{[A]} \right) \approx 0.9027 \)
Now, take the antilog (inverse log) of both sides:
\( \frac{[A_0]}{[A]} = 10^{0.9027} \)
\( \frac{[A_0]}{[A]} \approx 8.00 \)
This means \( [A] = \frac{[A_0]}{8} \).
If we assume \( [A_0] = 100\% \), then \( [A] = \frac{100\%}{8} = 12.5\% \).
Alternatively, notice that 180 seconds is exactly 3 half-lives (180 / 60 = 3).
After 1 half-life: \( 100\% \times \frac{1}{2} = 50\% \)
After 2 half-lives: \( 50\% \times \frac{1}{2} = 25\% \)
After 3 half-lives: \( 25\% \times \frac{1}{2} = 12.5\% \)
Therefore, after 180 seconds, 12.5% of substance A will be left over. Understanding half-life helps predict how much reactant will remain over time, especially in fields like nuclear chemistry or pharmacology. This method works well for integer multiples of half-lives.
In simple words: First, we found the rate constant using the half-life. Then, since 180 seconds is three half-lives, we calculated that 12.5% of the substance would remain.

🎯 Exam Tip: For first-order reactions, knowing that a certain time corresponds to a whole number of half-lives can provide a quick mental check or alternative solution method.

Question 26. A zero-order reaction is 20% complete in 20 minutes. Calculate the value of the rate constant. In what time will the reaction be 80% complete?
Answer: This question deals with a zero-order reaction. For such reactions, the rate is constant, and the amount of reactant consumed depends linearly on time.
1. **Calculate the rate constant (\( k \)):**
A zero-order reaction is 20% complete in 20 minutes. This means 20 units of reactant have reacted in 20 minutes.
The rate constant for a zero-order reaction is given by:
\( k = \frac{[A_0] - [A]}{t} = \frac{x}{t} \), where \( x \) is the amount reacted.
Here, \( x = 20\% \) of initial amount (let's assume 20 units) and \( t = 20 \text{ minutes} \).
\( k = \frac{20 \text{ units}}{20 \text{ minutes}} = 1 \text{ unit min}^{-1} \) (If initial concentration is 100%, then \( k = 1\% \text{ min}^{-1} \)).
So, the rate constant for this reaction is 1% per minute.
2. **Calculate the time for the reaction to be 80% complete:**
We need to find the time (\( t \)) when 80% of the reactant has been consumed.
Using the same formula, \( t = \frac{x}{k} \).
Here, \( x = 80\% \) (or 80 units) and \( k = 1\% \text{ min}^{-1} \).
\( t = \frac{80 \text{ units}}{1 \text{ unit min}^{-1}} \)
\( t = 80 \text{ minutes} \)
Therefore, the reaction will be 80% complete in 80 minutes. In a zero-order reaction, the rate constant's unit is typically concentration per unit time, such as mol L\(^{-1}\) s\(^{-1}\) or % min\(^{-1}\). The constant rate in zero-order reactions makes these calculations straightforward.
In simple words: First, we found the rate constant using the given completion percentage and time. Then, we used that constant to calculate the time needed for 80% completion.

🎯 Exam Tip: For zero-order reactions, the amount reacted is directly proportional to time, simplifying calculations involving completion percentages.

Question 27. The activation energy of a reaction is 225 k cal mol\(^{-1}\) and the value of rate constant at 40°C is \( 1.8 \times 10^{-5} \text{ s}^{-1} \). Calculate the frequency factor, A.
Answer: To calculate the frequency factor (\( A \)) for the reaction, we will use the Arrhenius equation in its logarithmic form.
First, gather the given information:
Activation energy, \( E_a = 225 \text{ kcal mol}^{-1} \). Convert to calories: \( E_a = 22500 \text{ cal mol}^{-1} \)
Temperature, \( T = 40^\circ\text{C} \). Convert to Kelvin: \( T = 40 + 273 = 313 \text{ K} \)
Rate constant, \( k = 1.8 \times 10^{-5} \text{ s}^{-1} \)
Gas constant, \( R = 1.987 \text{ cal K}^{-1} \text{ mol}^{-1} \) (used for energy in calories).
The logarithmic form of the Arrhenius equation is:
\( \log k = \log A - \frac{E_a}{2.303RT} \)
Rearrange to solve for \( \log A \):
\( \log A = \log k + \frac{E_a}{2.303RT} \)
Now, substitute the values:
\( \log A = \log(1.8 \times 10^{-5}) + \frac{22500 \text{ cal mol}^{-1}}{2.303 \times (1.987 \text{ cal K}^{-1} \text{ mol}^{-1}) \times (313 \text{ K})} \)
First calculate \( \log(1.8 \times 10^{-5}) \):
\( \log(1.8 \times 10^{-5}) = \log(1.8) - 5 \approx 0.2553 - 5 = -4.7447 \)
Next, calculate the second term:
\( \frac{22500}{2.303 \times 1.987 \times 313} \approx \frac{22500}{1436.43} \approx 15.7089 \)
So, \( \log A = -4.7447 + 15.7089 \)
\( \log A \approx 10.9642 \)
Finally, find \( A \) by taking the antilog:
\( A = 10^{10.9642} \)
\( A \approx 9.208 \times 10^{10} \text{ s}^{-1} \)
The frequency factor \( A \) for this reaction is approximately \( 9.208 \times 10^{10} \text{ s}^{-1} \). This value gives an idea of how many collisions between reactant molecules are successful in forming products, considering their proper orientation.
In simple words: We used the logarithmic form of the Arrhenius equation. We plugged in the rate constant, activation energy, gas constant, and temperature to calculate the frequency factor \( A \).

🎯 Exam Tip: Be careful with unit consistency; if \( E_a \) is in calories, use R in cal/mol K. If \( E_a \) is in Joules, use R in J/mol K.

Question 28. Benzene diazonium chloride in aqueous solution decomposes according to the equation \( \text{C}_6\text{H}_5\text{N}_2\text{Cl} \rightarrow \text{C}_6\text{H}_5\text{Cl} + \text{N}_2 \). Starting with an initial concentration of 10 g L\(^{-1}\) volume of \( \text{N}_2 \) gas obtained at 50°C at different intervals of time was found to be as under:

🎯 Exam Tip: When proving a reaction is first-order using experimental data, ensure that the calculated rate constant \( k \) remains consistent across multiple data points.

III. Evaluate Yourself

Question 1. Write the rate expression for the following reactions, assuming them as elementary reactions.
i) \( 3\text{A} + 5\text{B}_2 \rightarrow 4\text{CD} \)
ii) \( \text{X}_2 + \text{Y}_2 \rightarrow 2\text{XY} \)
Answer: When we assume reactions are elementary, their rate expressions can be written directly from their stoichiometry. The rate is expressed in terms of the change in concentration of reactants (with a negative sign) or products (with a positive sign) over time, divided by their stoichiometric coefficients.
(i) For the reaction: \( 3\text{A} + 5\text{B}_2 \rightarrow 4\text{CD} \)
The rate expression is:
\( \text{Rate} = -\frac{1}{3} \frac{d[A]}{dt} \)
\( \implies = -\frac{1}{5} \frac{d[B_2]}{dt} \)
\( \implies = +\frac{1}{4} \frac{d[CD]}{dt} \)
(ii) For the reaction: \( \text{X}_2 + \text{Y}_2 \rightarrow 2\text{XY} \)
The rate expression is:
\( \text{Rate} = -\frac{d[X_2]}{dt} \)
\( \implies = -\frac{d[Y_2]}{dt} \)
\( \implies = +\frac{1}{2} \frac{d[XY]}{dt} \)
These expressions show how the concentrations of each species change relative to each other during the reaction. The rate of reaction is defined to be positive, so a negative sign is used for reactants because their concentrations decrease over time.
In simple words: For elementary reactions, the rate expression uses the stoichiometric coefficients. Reactant changes have a negative sign, and product changes have a positive sign, both divided by their coefficient.

🎯 Exam Tip: Remember to include the negative sign for reactants and the positive sign for products in the rate expression.

Question 2. Consider the decomposition of \( \text{N}_2\text{O}_5(\text{g}) \) to form \( \text{NO}_2(\text{g}) \) and \( \text{O}_2(\text{g}) \). At a particular instant \( \text{N}_2\text{O}_5 \) disappears at a rate of \( 2.5 \times 10^{-2} \text{ mol dm}^{-3} \text{ s}^{-1} \). At what rates are \( \text{NO}_2 \) and \( \text{O}_2 \) formed? What is the rate of the reaction?
Answer: Let's determine the rates of formation for \( \text{NO}_2 \) and \( \text{O}_2 \) and the overall reaction rate based on the given decomposition of \( \text{N}_2\text{O}_5 \).
The balanced chemical equation for the decomposition is:
\( 2\text{N}_2\text{O}_5(\text{g}) \rightarrow 4\text{NO}_2(\text{g}) + \text{O}_2(\text{g}) \)
The rate of disappearance of \( \text{N}_2\text{O}_5 \) is given as \( 2.5 \times 10^{-2} \text{ mol dm}^{-3} \text{ s}^{-1} \).
From the stoichiometry of the reaction, the general rate expression is:
\( \text{Rate} = -\frac{1}{2} \frac{d[\text{N}_2\text{O}_5]}{dt} = +\frac{1}{4} \frac{d[\text{NO}_2]}{dt} = +\frac{d[\text{O}_2]}{dt} \)
1. **Rate of formation of \( \text{NO}_2 \):**
Using the relationship \( \frac{1}{4} \frac{d[\text{NO}_2]}{dt} = -\frac{1}{2} \frac{d[\text{N}_2\text{O}_5]}{dt} \).
\( \frac{d[\text{NO}_2]}{dt} = \frac{4}{2} \left( -\frac{d[\text{N}_2\text{O}_5]}{dt} \right) \)
\( \frac{d[\text{NO}_2]}{dt} = 2 \times (2.5 \times 10^{-2} \text{ mol dm}^{-3} \text{ s}^{-1}) \)
\( \frac{d[\text{NO}_2]}{dt} = 5.0 \times 10^{-2} \text{ mol dm}^{-3} \text{ s}^{-1} \)
So, \( \text{NO}_2 \) is formed at a rate of \( 5.0 \times 10^{-2} \text{ mol dm}^{-3} \text{ s}^{-1} \).
2. **Rate of formation of \( \text{O}_2 \):**
Using the relationship \( \frac{d[\text{O}_2]}{dt} = -\frac{1}{2} \frac{d[\text{N}_2\text{O}_5]}{dt} \).
\( \frac{d[\text{O}_2]}{dt} = \frac{1}{2} \times (2.5 \times 10^{-2} \text{ mol dm}^{-3} \text{ s}^{-1}) \)
\( \frac{d[\text{O}_2]}{dt} = 1.25 \times 10^{-2} \text{ mol dm}^{-3} \text{ s}^{-1} \)
Thus, \( \text{O}_2 \) is formed at a rate of \( 1.25 \times 10^{-2} \text{ mol dm}^{-3} \text{ s}^{-1} \).
3. **Overall rate of the reaction:**
The overall rate of the reaction is defined by any of the terms in the general rate expression. Let's use the first term:
\( \text{Rate of reaction} = -\frac{1}{2} \frac{d[\text{N}_2\text{O}_5]}{dt} \)
\( \text{Rate of reaction} = \frac{1}{2} \times (2.5 \times 10^{-2} \text{ mol dm}^{-3} \text{ s}^{-1}) \)
\( \text{Rate of reaction} = 1.25 \times 10^{-2} \text{ mol dm}^{-3} \text{ s}^{-1} \)
The overall rate of the reaction is \( 1.25 \times 10^{-2} \text{ mol dm}^{-3} \text{ s}^{-1} \). The stoichiometric coefficients are critical for converting the rate of disappearance/formation of one species into the overall reaction rate. This problem shows how to relate the rate of change of one species to others in a reaction.
In simple words: We used the balanced equation to find how the rates of different chemicals are linked. Then, we calculated the formation rates of \( \text{NO}_2 \) and \( \text{O}_2 \) and the total reaction rate from the given rate of \( \text{N}_2\text{O}_5 \) disappearance.

🎯 Exam Tip: Always write down the balanced chemical equation and the general rate expression based on stoichiometry before solving for individual rates.

Question 3. For a reaction, X + Y → Product quadrupling [x], increases the rate by a factor of 8. Quailrupling both [x] and [y] increases the rate by a factor of 16. Find the order of the reaction with respect to x and y. what is the overall order of the reaction?
Answer:
Let the rate law be: \( \text{Rate} = k[X]^m[Y]^n \)
From the first condition, when \( [X] \) is quadrupled, the rate increases by a factor of 8 (with \( [Y] \) constant):
\( 8 = (4)^m \)
\( 2^3 = (2^2)^m \)
\( 2^3 = 2^{2m} \)
\( \implies 3 = 2m \)
\( \implies m = \frac{3}{2} = 1.5 \)
From the second condition, when both \( [X] \) and \( [Y] \) are quadrupled, the rate increases by a factor of 16:
\( 16 = (4)^m (4)^n \)
Since we know \( m = 1.5 \):
\( 16 = (4)^{1.5} (4)^n \)
\( 16 = (4^{\frac{3}{2}}) (4)^n \)
\( 16 = (\sqrt{4^3}) (4)^n \)
\( 16 = (\sqrt{64}) (4)^n \)
\( 16 = 8 (4)^n \)
\( \implies (4)^n = \frac{16}{8} \)
\( \implies (4)^n = 2 \)
\( (2^2)^n = 2^1 \)
\( 2^{2n} = 2^1 \)
\( \implies 2n = 1 \)
\( \implies n = \frac{1}{2} = 0.5 \)
So, the order with respect to \( X \) is 1.5, and the order with respect to \( Y \) is 0.5.
The overall order of the reaction is \( m + n = 1.5 + 0.5 = 2 \). This means the reaction is second order overall.
In simple words: We find how much the rate changes when reactant X is changed, and then when both X and Y are changed. This helps us find the power for each reactant in the rate equation. Adding these powers gives the total order of the reaction.

🎯 Exam Tip: When solving rate law problems, always write down the general rate law first. Substitute the given concentration changes and rate changes into the equation to find the individual orders (m and n).

Question 4. Find the individual and overall order of the following reaction using the given data. 2NO(g) + Cl2 (g) → 2NOCl(g)

🎯 Exam Tip: When using tabular data to find reaction orders, choose experiments where the concentration of only one reactant changes while others remain constant. This simplifies the calculations significantly.

Question 6. The rate constant for a first order reaction is 2.3 x 10^-4 s^-1. If the initial concentration of the reactant is 0.01 M. what concentration will remain after 1 hour?
Answer:
Given:
Rate constant, \( k = 2.3 \times 10^{-4} \text{ s}^{-1} \)
Initial concentration, \( [A_0] = 0.01 \text{ M} \)
Time, \( t = 1 \text{ hour} = 60 \text{ minutes} = 60 \times 60 \text{ s} = 3600 \text{ s} \)

For a first-order reaction, the integrated rate law is:
\( k = \frac{2.303}{t} \log \frac{[A_0]}{[A]} \)
We need to find the concentration remaining, \( [A] \).
Rearrange the equation:
\( \log \frac{[A_0]}{[A]} = \frac{k \times t}{2.303} \)
Substitute the given values:
\( \log \frac{0.01}{[A]} = \frac{(2.3 \times 10^{-4} \text{ s}^{-1}) \times (3600 \text{ s})}{2.303} \)
\( \log \frac{0.01}{[A]} = \frac{0.828}{2.303} \)
\( \log \frac{0.01}{[A]} = 0.3595 \)

To find \( \frac{0.01}{[A]} \), take the antilog of 0.3595:
\( \frac{0.01}{[A]} = \text{antilog}(0.3595) \)
\( \frac{0.01}{[A]} = 2.288 \)

Now, calculate \( [A] \):
\( [A] = \frac{0.01}{2.288} \)
\( [A] = 0.00437 \text{ M} \)
After 1 hour, the concentration of the reactant remaining will be 0.00437 M. This shows how the concentration decreases over time in a first-order reaction.
In simple words: We use a special formula for reactions where the speed depends on just one reactant. By putting in the starting amount, the reaction speed number, and the time, we can figure out how much of the reactant is left.

🎯 Exam Tip: Always ensure that the units of time for the rate constant and the given time are consistent. Convert all time values to seconds or minutes as required by the rate constant's unit before calculation.

Question 7. Hydrolysis of an ester in an aqueous solution was studied by titrating the of the ester at different time intervals are given below. Show that, the reaction follows first order kinetics.

🎯 Exam Tip: To prove a reaction is first-order, calculate the rate constant at multiple time points. If these values are consistent, it confirms the reaction order. Small variations are acceptable due to experimental error.

Question 8. For a first order reaction the rate constant at 500K is 8 x 10^-4 s^-1. Calculate the frequency factor, if the energy of activation for the reaction is 190 kJ mol^-1.
Answer:
Given:
Rate constant, \( k = 8 \times 10^{-4} \text{ s}^{-1} \)
Temperature, \( T = 500 \text{ K} \)
Energy of activation, \( E_a = 190 \text{ kJ mol}^{-1} = 190 \times 10^3 \text{ J mol}^{-1} \)
Gas constant, \( R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1} \)

The Arrhenius equation is:
\( k = A e^{\frac{-E_a}{RT}} \)
To find the frequency factor \( A \), we can rearrange this equation into its logarithmic form:
\( \ln k = \ln A - \frac{E_a}{RT} \)
Or, using \( \log_{10} \):
\( \log k = \log A - \frac{E_a}{2.303RT} \)

Rearranging to solve for \( \log A \):
\( \log A = \log k + \frac{E_a}{2.303RT} \)

Substitute the given values into the equation:
\( \log A = \log(8 \times 10^{-4}) + \frac{190 \times 10^3 \text{ J mol}^{-1}}{2.303 \times 8.314 \text{ J K}^{-1} \text{ mol}^{-1} \times 500 \text{ K}} \)
\( \log A = \log(8) + \log(10^{-4}) + \frac{190000}{9576.89} \)
\( \log A = 0.9031 - 4 + 19.839 \)
\( \log A = 16.7421 \)

To find \( A \), take the antilog of 16.7421:
\( A = \text{antilog}(16.7421) \)
\( A = 5.522 \times 10^{16} \text{ s}^{-1} \)
The frequency factor, also known as the pre-exponential factor, is \( 5.522 \times 10^{16} \text{ s}^{-1} \). This value represents how often molecules collide with the correct orientation at infinite temperature.
In simple words: The frequency factor (A) tells us how often particles bump into each other in the right way to react, before considering how much energy they have. We used a special formula called the Arrhenius equation to work it out from the reaction speed and energy needed.

🎯 Exam Tip: Remember the logarithmic form of the Arrhenius equation (\( \log k = \log A - \frac{E_a}{2.303RT} \)) as it's frequently used. Pay close attention to unit conversions, especially for activation energy (kJ to J).

12th Chemistry Guide Chemical Kinetics Elements Additional Questions and Answers

I. Match The Following

Question 1. Rate constant Unit i. Zero order reaction a. mol^-2L^-2s^-1 ii. First order reaction b. molL^-1s^-1 iii. Second order reaction c. s^-1 iv. Third order reaction d. mol^-1Ls^-1
Answer:
i. Zero order reaction - b. molL^-1s^-1
ii. First order reaction - c. s^-1
iii. Second order reaction - d. mol^-1Ls^-1
iv. Third order reaction - a. mol^-2L^-2s^-1
In simple words: The unit for a reaction's speed constant changes depending on the reaction's order. Each order has its own unique unit, which helps us identify the reaction type.

🎯 Exam Tip: Memorize the units for rate constants of different reaction orders. A common formula for the unit of a rate constant is \( (\text{mol L}^{-1})^{1-n} \text{ s}^{-1} \), where \( n \) is the order of the reaction.

Question 2. I. Catalyst II. Order III. Molecularity IV. Half life of first order reaction
(a) can be negative
(b) independent of initial concentration
(c) lowers the activation energy
(d) can not be fraction
Answer: (I) c, (II) a, (III) d, (IV) b
I. Catalyst - (c) lowers the activation energy
II. Order - (a) can be negative
III. Molecularity - (d) can not be fraction
IV. Half life of first order reaction - (b) independent of initial concentration
In simple words: Catalysts speed up reactions by lowering the energy needed. Reaction order can sometimes be negative, but molecularity is always a whole number. For a first-order reaction, how long it takes for half of the reactant to be used up doesn't depend on how much you started with.

🎯 Exam Tip: Understand the key definitions and characteristics of each concept. Pay attention to the specific properties of first-order reactions, especially regarding half-life and its independence from initial concentration.

II. Assertion And Reason

Question 1. Assertion (A): Rate of a reaction can be \( -\frac{d[\text{Reactant}]}{dt} \) Reason (R): As time increases concentration of reactant decreases.
Answer: (i) Both A and R are true and R explains A.
In simple words: The assertion is true because reaction rate is measured by how quickly reactants are used up, shown by the negative sign. The reason is also true because reactants get used up over time. The reason correctly explains why we use a negative sign in the assertion.

🎯 Exam Tip: Remember that the rate of disappearance of a reactant is always expressed with a negative sign, while the rate of formation of a product has a positive sign. The rate is always a positive value.

Question 2. Assertion (A) : For a gas phase reaction the unit of reaction rate is atm s^-1 Reason (R) : For a gas phase reaction, the concentration of the gaseous species is expressed in terms of mole per litre.
Answer: (ii) A is correct but R is wrong.
Correct Reason: For a gas phase reaction, the concentration of the gaseous species is expressed in atm.
In simple words: The first statement is correct because for gas reactions, we measure rate using pressure units like atmospheres per second. The second statement is wrong because for gases, we usually talk about pressure (like atm) instead of moles per liter to describe their amounts.

🎯 Exam Tip: Always be mindful of the physical state of the reactants. For gas-phase reactions, concentration can be expressed in terms of partial pressures, and the rate unit reflects this (e.g., atm s^-1 or Pa s^-1).

Question 3. Assertion (A) : For a general reaction xA + yB → products, rate law is Rate = k[A]^m[B]ⁿ Reason (R) : The values of m and n can be deduced from the stoichiometry of the reaction.
Answer: (ii) A is correct but R is wrong.
Correct Reason: The values of m and n can be deduced from experiments.
In simple words: The first statement is true: the rate law shows how the reaction speed depends on reactant amounts, with powers 'm' and 'n'. But the second statement is false; you cannot simply look at the balanced equation (stoichiometry) to find 'm' and 'n'. These values must be found through actual experiments.

🎯 Exam Tip: A common misconception is that reaction orders (m, n) are the same as stoichiometric coefficients (x, y). Emphasize that orders must be determined experimentally, not by balancing the equation.

Question 4. Assertion (A) : Acid hydrolysis of an ester is a pseudo first order reaction. Reason (R) : The rate of the reaction does not depend on the concentration of one of the reactants water.
Answer: (i) Both A and R are correct and R explains A.
In simple words: The assertion is true because acid hydrolysis of an ester often acts like a first-order reaction, even though two things are reacting. The reason is also true: if water is present in very large amounts, its concentration hardly changes, so the reaction rate doesn't depend on it. This explains why it behaves like a first-order reaction.

🎯 Exam Tip: Recognize pseudo-first-order reactions when one reactant is in vast excess. Its concentration effectively remains constant, simplifying the rate law to appear first order with respect to the other reactant.

III. Choose The Correct Statements

Question 1.
(i) During the reaction the concentration of the reactant increases.
(ii) Unit of rate = Unit of concentration
(iii) For a gas phase reaction the unit of rate is atm s^-1
(a) (i) & (ii)
(b) (i) & (ii)
(c) (ii) &(iii)
(d) (iii) only
Answer: (d) (iii) only
Correct statement for (i): During the reaction, the concentration of the reactant decreases.
Correct statement for (ii): Unit of rate is (concentration unit) / (time unit), e.g., mol L^-1 s^-1 or atm s^-1, not just concentration.
In simple words: Only statement (iii) is correct. As a reaction happens, the amount of reactants goes down, not up. The unit for reaction speed is amount per time, not just amount. So, only the unit for gas reactions using atmosphere per second is right.

🎯 Exam Tip: Always remember that reactant concentrations decrease over time, and the unit of reaction rate combines concentration (or pressure for gases) with a time unit. Don't confuse the unit of rate with the unit of concentration itself.

Question 2.
(i) The rate of a reaction decreases with time as the reaction proceeds.
(ii) The average rate can be used to predict the rate of the reaction at any instant.
(iii) The rate of the reaction at a particular instant during the reaction is called the instantaneous rate.
(a) (i) & (ii)
(b) (i) & (iii)
(c) (ii) & (iii)
(d) (i), (ii) & (iii)
Answer: (b) (i) & (iii)
Correct statement for (ii): The average rate cannot be used to predict the rate of the reaction at any instant.
In simple words: Statements (i) and (iii) are correct. Reaction speeds usually slow down over time because reactants get used up. The speed at any exact moment is called the instantaneous rate. But, the average speed over a long time cannot tell us the exact speed at one particular moment.

🎯 Exam Tip: Distinguish clearly between average rate and instantaneous rate. Instantaneous rate is more precise and useful for understanding reaction mechanisms at specific points in time. Average rate gives a general idea over a duration.

Question 3.
(i) Rate constant is a proportionality constant
(ii) Rate of a reaction depends on the initial concentration of the reactants.
(iii) Rate constant does not depend on the initial concentration of the reactants.
(a) (i) & (ii)
(b) (i) & (iii)
(c) (ii) & (iii)
(d) (i), (ii) & (iii)
Answer: (b) (i) & (iii)
In simple words: Statements (i) and (iii) are correct. The rate constant is a fixed number that links reactant amounts to reaction speed. This constant does not change if you just change the starting amount of reactants. However, the overall speed of the reaction itself generally depends on the initial concentrations of reactants.

🎯 Exam Tip: Remember that the rate constant (\( k \)) is independent of reactant concentrations and only changes with temperature or catalyst. The rate of reaction, however, depends on both \( k \) and reactant concentrations.

Question 4.
(i) Order is assigned for each elementary step of mechanism.
(ii) Order can be fractional.
(iii) Molecularity can not be fractional.
(a) (i) & (ii)
(b) (i) & (iii)
(c) (ii) & (iii)
(d) (i), (ii) & (iii)
Answer: (c) (ii) & (iii)
Correct statement for (i): Molecularity is assigned for each elementary step of mechanism, not order. Order is for the overall reaction.
In simple words: Statements (ii) and (iii) are correct. The order of a reaction can be a fraction, but molecularity (how many particles collide in one step) must always be a whole number. Reaction order is about the total reaction, while molecularity is about each small step.

🎯 Exam Tip: Clearly differentiate between order and molecularity. Molecularity is theoretical and applies to individual elementary steps, always a whole number. Order is an experimentally determined value for the overall reaction and can be zero, integer, or fractional.

IV. Choose The Incorrect Statements

Question 1.
(i) All radio active reactions are first order reaction.
(ii) Half life of first order reaction depends on the initial concentration of the reactants.
(iii) For a first order reaction a plot of In[A] against t gives a straight line with positive slope.
(a) (i) & (ii)
(b) (i) & (iii)
(c) (ii) & (iii)
(d) (i), (ii) & (iii)
Answer: (c) (ii) & (iii)
Correct statement for (ii): Half-life of a first-order reaction does not depend on the initial concentration of the reactants.
Correct statement for (iii): For a first-order reaction, a plot of \( \ln[A] \) against \( t \) gives a straight line with a negative slope.
In simple words: Statements (ii) and (iii) are incorrect. For first-order reactions, the time it takes for half the reactant to disappear never changes, no matter how much you start with. Also, when you graph the natural log of reactant amount versus time for a first-order reaction, the line goes downwards (negative slope). Statement (i) is actually correct.

🎯 Exam Tip: Remember the two key characteristics of first-order reactions: constant half-life (independent of initial concentration) and a linear \( \ln[A] \) vs. \( t \) plot with a negative slope. These are critical for identifying and solving first-order problems.

Question 2.
(i) Acid hydrolysis of an ester is a second order reaction.
(ii) Isomerisation of cyclo propane to propene is a zero order reaction.
(iii) In acid hydrolysis of ester the concentration of ester remains constant.
(a) (i) & (ii)
(b) (i) & (iii)
(c) (ii) & (iii)
(d) (i), (ii) & (iii)
Answer: (d) (i), (ii) & (iii)
Correct statement for (i): Acid hydrolysis of an ester is a pseudo-first-order reaction (not second order).
Correct statement for (ii): Isomerisation of cyclopropane to propene is a first-order reaction (not zero order).
Correct statement for (iii): In acid hydrolysis of ester, the concentration of water remains constant (not ester), as water is in large excess.
In simple words: All three statements are incorrect. Acid hydrolysis of ester acts like a first-order reaction, not second-order. The change from cyclopropane to propene is a first-order reaction, not zero-order. Also, in ester hydrolysis, it's the water that stays constant in amount, not the ester.

🎯 Exam Tip: Be precise with reaction orders for common reactions. Pseudo-first-order reactions are often mistaken for their actual higher order. Remember standard examples for zero, first, and second-order reactions.

Question 3.
(i) A zero order reaction is independent of the concentration of the reactants.
(ii) Iodination of acetone in acid medium is zero order with respect to acetone.
(iii) Zero order reactions are quite common.
(a) (i) & (ii)
(b) (i) & (iii)
(c) (ii) & (iii)
(d) (i), (ii) & (iii)
Answer: (c) (ii) & (iii)
Correct statement for (ii): Iodination of acetone in acid medium is zero order with respect to iodine (not acetone).
Correct statement for (iii): Zero-order reactions are rare (not quite common).
In simple words: Statements (ii) and (iii) are incorrect. For the iodination of acetone, the reaction speed does not depend on iodine's amount, not acetone's. Also, zero-order reactions are not very common; they usually happen in special conditions like on a surface. Statement (i) is correct.

🎯 Exam Tip: For zero-order reactions, the rate is constant and does not depend on reactant concentration. Familiarize yourself with classic examples of zero-order reactions, such as enzymatic reactions or reactions on metal surfaces, and remember they are generally rare.

Question 4.
(i) Rate of a reaction is directly proportional to the number of collision per second.
(ii) Collision rate in gases can be calculated from kinetic theory of gases.
(iii) Number of collisions is inversely proportional to the concentration of the reactants.
(a) (i) & (ii)
(b) (i) & (iii)
(c) (ii) & (iii)
(d) (i), (ii) & (iii)
Answer: (b) (i) & (iii)
Reason for (i) being incorrect: Only *effective* collisions (those with sufficient energy and correct orientation) are directly proportional to the reaction rate, not all collisions.
Correct statement for (iii): The number of collisions is directly proportional to the concentration of the reactants.
In simple words: Statements (i) and (iii) are incorrect. The speed of a reaction depends only on the number of successful bumps between molecules, not all bumps. Also, more reactant means more bumps, so the number of collisions goes up with concentration, not down. Statement (ii) is correct.

🎯 Exam Tip: Distinguish between total collisions and effective collisions. Reaction rate is governed by effective collisions, which require both proper orientation and activation energy. The total collision frequency is directly proportional to concentration.

V. Choose The Best Answer

Question 1. As the reaction proceeds the concentration of the reactant
(a) increases
(b) decreases
(c) remains the same
(d) tends to be maximum
Answer: (b) decreases
In simple words: As a chemical reaction moves forward, the starting materials (reactants) get used up to make new substances. This means their amount goes down over time.

🎯 Exam Tip: This is a fundamental concept in chemical kinetics. Reactants are consumed, and products are formed. Always relate the change in concentration to the progress of the reaction.

Question 2. For reaction xA + yB → lC + mD regarding its rate which is incorrect
(a) \( -\frac{1}{x} \frac{d[A]}{dt} \)
(b) \( +\frac{1}{y} \frac{d[B]}{dt} \)
(c) \( +\frac{1}{l} \frac{d[C]}{dt} \)
(d) \( +\frac{1}{m} \frac{d[D]}{dt} \)
Answer: (b) \( +\frac{1}{y} \frac{d[B]}{dt} \)
In simple words: The incorrect statement is (b). For reactants like B, their amount goes down during the reaction, so the rate should be shown with a minus sign. The plus sign in option (b) makes it wrong for a reactant.

🎯 Exam Tip: Remember the convention for expressing reaction rates: negative signs are used for reactants (indicating disappearance) and positive signs for products (indicating appearance). The stoichiometric coefficient divides the rate of change of concentration.

Question 3. The rate constant of a reaction is 2.3 x 10^-2 lit mol^-1 S^-1 The order of the reaction is
(a) Zero order
(b) First order
(c) Second order
(d) Third order
Answer: (c) Second order
In simple words: The unit for the rate constant (lit mol^-1 s^-1) tells us the reaction's order. A unit of mol^-1 L s^-1 (which is the same as lit mol^-1 s^-1) always means it's a second-order reaction.

🎯 Exam Tip: The units of the rate constant are crucial for determining the order of a reaction. For a general reaction of order 'n', the unit of the rate constant is typically \( (\text{concentration})^{1-n} \text{ time}^{-1} \). For second-order reactions, this becomes \( (\text{mol L}^{-1})^{1-2} \text{ s}^{-1} = (\text{mol L}^{-1})^{-1} \text{ s}^{-1} = \text{mol}^{-1} \text{ L s}^{-1} \).

Question 4. The sum of exponential terms in the rate law is called as
(a) Molecularity
(b) Rate constant
(c) Order
(d) Rate
Answer: (c) Order
In simple words: In a reaction's rate law, the small numbers (exponents) that show how much each reactant affects the speed are added together. This sum is called the overall order of the reaction.

🎯 Exam Tip: Distinguish order from molecularity. Order is determined experimentally as the sum of exponents in the rate law, while molecularity is the number of reacting species in an elementary step.

Question 5. The value of order of a reaction can be
(a) Zero
(b) Fractional
(c) whole number
(d) All of the options
Answer: (d) All of the options
In simple words: The order of a reaction, which tells us how reaction speed depends on reactant amounts, can be a whole number (like 0, 1, 2), or even a fraction. It's not limited to just one type of value.

🎯 Exam Tip: Remember that reaction order is an experimentally determined value and can take various forms: zero, positive integer, or even fractional, and occasionally negative. This broad range makes it distinct from molecularity, which is always a positive integer.

Question 6. The value of molecularity of a reaction can be
(a) Zero
(b) Fractional
(c) whole number
(d) all the above
Answer: (c) whole number
In simple words: Molecularity is always a whole number because it refers to the actual number of molecules or atoms that collide in a single step of a reaction. This count must be a complete, indivisible number.

🎯 Exam Tip: Remember that molecularity specifically describes the number of reacting species in an elementary step, which must be an integer, unlike the reaction order which can be zero or fractional.

Question 7. During the decomposition of H2O2 (H2O2 → 2H2O + O2). 48g of O2 is formed per Minute at particular instant. The rate of formation of water at this instant in mol min-1 is
(a) 1
(b) 2
(c) 3
(d) 4
Answer: (c) 3
In simple words: The decomposition of hydrogen peroxide creates water and oxygen. If 48 grams of oxygen are made each minute, this is equivalent to 1.5 moles of oxygen. Based on the balanced reaction, water is formed at twice the rate of oxygen, so 3 moles of water are formed per minute.

🎯 Exam Tip: Always balance the chemical equation first and use the stoichiometric coefficients to relate the rates of different reactants and products accurately.

Question 8. If the rate law of a reaction is Rate = k[A]3/2[B]1 the order of the reaction with respect to A is
(a) 1.5
(b) 1
(c) 2.5
(d) 3
Answer: (a) 1.5
In simple words: The order of a reaction with respect to a specific reactant is the exponent of its concentration term in the rate law. In the given rate law, the concentration of A has an exponent of 3/2, which means the order for A is 1.5.

🎯 Exam Tip: Remember that the order of a reaction with respect to a particular reactant is the power to which its concentration term is raised in the experimentally determined rate law.

Question 9. If the rate law of a reaction is Rate = k[A]3/2[B]1 the order of the reaction with respect to B is
(a) 1.5
(b) 1
(c) 2.5
(d) 3
Answer: (b) 1
In simple words: The order of a reaction for a specific reactant is the power of its concentration term in the rate law. In this rate law, the concentration of B has an exponent of 1. Therefore, the reaction order for B is 1.

🎯 Exam Tip: Be careful to identify the exponent for each specific reactant when determining its individual order in the rate law.

Question 10. If the rate law of a reaction is Rate = k[A]3/2[B]¹ the overall order of the reaction is
(a) 1.5
(b) 1
(c) 2.5
(d) 3
Answer: (c) 2.5
In simple words: The overall order of a reaction is found by adding up all the individual orders for each reactant in the rate law. Here, the order for A is 3/2 (1.5) and for B is 1, so the total order is 1.5 + 1 = 2.5.

🎯 Exam Tip: The overall order is the sum of the exponents of the concentration terms in the rate law, representing the total dependence of the reaction rate on the concentrations of all reactants.

Question 11. If the rate law for a reaction is Rate = k[A]2 [B]°, the unit of rate constant k for the
(a) molL-1V
(b) mol-1LS-1
(c) mol-2L-2S-1
(d) S-1
Answer: (b) mol-1LS-1
In simple words: The unit of the rate constant (k) depends on the overall order of the reaction. For a reaction with the rate law Rate = k[A]^2[B]^0, the overall order is 2. The unit for the rate constant of a second-order reaction is mol to the power of minus one, liter to the power of one, and second to the power of minus one.

🎯 Exam Tip: Always derive the units of the rate constant using the general formula: Unit of k = (Unit of concentration)^(1-n) * (Unit of time)^-1, where 'n' is the overall order of the reaction.

Question 12. Which among the following is the unit of k of a zero order reaction ?
(a) mol-1L-1S-1
(b) mol-1LS-1
(c) mol-2L-2S-1
(d) S-1
Answer: (a) mol-1L-1S-1
In simple words: For a zero-order reaction, the rate of the reaction does not depend on the concentration of reactants. This means the rate constant (k) has the same units as the reaction rate itself, which are mol per liter per second.

🎯 Exam Tip: For a zero-order reaction, the rate constant's unit directly reflects the rate of concentration change over time, as the rate is independent of reactant concentrations.

Question 13. The rate law of a reaction is Rate = k[A] [B], if the concentration of B is taken in large excess, the order of the reaction is
(a) zero order
(b) first order
(c) pseudo first order
(d) second order
Answer: (c) pseudo first order
In simple words: When one reactant is present in a very large amount, its concentration changes so little during the reaction that it can be considered constant. This makes the reaction appear to be of a lower order than it truly is, making it a "pseudo first order" reaction.

🎯 Exam Tip: A pseudo-order reaction occurs when one or more reactants are in such large excess that their concentrations remain essentially constant throughout the reaction, simplifying the observed reaction order.

Question 14. The rate determining step of a reaction is .......... step.
(a) fast
(b) slow
(c) equilibrium
(d) intermediate
Answer: (b) slow
In simple words: In a multi-step reaction, the overall rate of the reaction is limited by its slowest step. This slowest step is called the rate-determining step, much like the slowest person in a relay race determines the team's overall time.

🎯 Exam Tip: To identify the rate-determining step, always look for the elementary step in a reaction mechanism that has the highest activation energy or the slowest intrinsic rate.

Question 15. From the following which is a second order rate constant?
(a) k = 5.47 x 10-4s-1
(b) k = 3.9 x 10-3mol lit s-1
(c) k = 3.94 x 10-4 lit mol -1 s-1
(d) k = 3.98 x 10-5 lit mol-2s-1
Answer: (c) k = 3.94 x 10-4 lit mol -1 s-1
In simple words: The unit of the rate constant (k) tells us the order of the reaction. For a second-order reaction, the correct unit for 'k' is liters per mole per second. Option (c) matches this unit exactly, showing it is a second-order rate constant.

🎯 Exam Tip: Memorize or be able to quickly derive the units of the rate constant for zero, first, second, and third-order reactions to quickly identify reaction orders from given rate constant units.

Question 16. Total number of reactant species that are involved in an elementary step is called
(a) order
(b) molecularity
(c) rate
(d) rate constant
Answer: (b) molecularity
In simple words: "Molecularity" refers to the exact number of reactant particles (like atoms or molecules) that come together and collide in one single step of a chemical reaction. It counts these particles directly.

🎯 Exam Tip: Molecularity is always an integer and applies only to elementary reaction steps, while reaction order is an experimentally determined value that can be zero, fractional, or an integer and applies to overall reactions.

Question 17. For an elementary reaction 2A + B → C + D the molecularity is
(a) zero
(b) one
(c) two
(d) three
Answer: (d) three
In simple words: For a simple reaction like this, the molecularity is found by adding up the numbers in front of each reactant. Here, two A molecules and one B molecule give a total of three.

🎯 Exam Tip: For elementary reactions, the molecularity is always equal to the sum of the stoichiometric coefficients of the reactants.

Question 18. Which is correct for first order reactions,
(a) t 1/2 \( \propto \) (concn)-1
(b) t 1/2 \( \propto \) (concn)
(c) t 1/2 \( \propto \) (concn)0
(d) t 14/2 \( \propto \) (concn)1/2
Answer: (c) t 1/2 \( \propto \) (concn)0
In simple words: For a first-order reaction, the half-life period (the time it takes for half the reactant to be used up) is always the same, no matter how much reactant you start with. So, it does not depend on the initial concentration.

🎯 Exam Tip: This independence of half-life from initial concentration is a defining feature of first-order reactions and helps distinguish them from zero-order and second-order reactions.

Question 19. Time required for the reactant concentration to reach one half of its initial value is called
(a) Half life period
(b) First order
(c) zero order
(d) Second order
Answer: (a) Half life period
In simple words: The "half-life period" is a chemistry term for the time it takes for the amount of a reactant to drop to exactly half of what it started with.

🎯 Exam Tip: Understand that the half-life is a characteristic time for a reaction, and its behavior (constant or concentration-dependent) varies with the order of the reaction.

Question 20. Molecularity can be determined by
(a) Stoichiometry
(b) experiment
(c) mechanism
(d) none of the above
Answer: (c) mechanism
In simple words: Molecularity tells us how many particles crash into each other in one step of a reaction. You figure this out by looking at the step-by-step "mechanism" of the reaction.

🎯 Exam Tip: Molecularity is a theoretical concept derived from the reaction mechanism, whereas the reaction order is an experimentally determined quantity.

Question 21. Rate constant of a reaction is equal to the rate of the reaction when the concentration of the reactants is
(a) zero
(b) unity
(c) infinity
(d) fractional
Answer: (b) unity
In simple words: The rate constant (k) is a special number in the reaction speed rule. Its value becomes equal to the total speed of the reaction when the concentration of each reactant is exactly one, or "unity."

🎯 Exam Tip: This definition highlights the intrinsic nature of the rate constant, which reflects the reaction's inherent speed independent of reactant concentrations when normalized to unity.

Question 22. For a first order reaction a plot of ln[A] against t yields a straight line with a slope which is
(a) positive
(b) negative
(c) zero
(d) at infinitv
Answer: (b) negative
In simple words: For a first-order reaction, if you graph the natural logarithm of the reactant's amount against time, you get a straight line that points downwards. This means the line has a negative slope because the concentration decreases over time.

🎯 Exam Tip: Remember that the slope of a plot of \( \text{ln}[A] \) vs. time for a first-order reaction is \( -k \), while for a zero-order reaction, a plot of \( [A] \) vs. time has a slope of \( -k \).

Question 23. Which of the following is not a first order reaction ?
(i) Decomposition of dinitrogen pentoxide.
(ii) Decomposition of N2O on hot platinum surface.
(iii) Decomposition of thionyl chloride.
(a) (i) only
(b) (i) & (ii)
(c) (ii) only
(d) (ii) & (iii)
Answer: (c) (ii) only
In simple words: The breaking down of N2O on a hot platinum surface is not a first-order reaction. Instead, it is a zero-order reaction because the hot surface quickly gets covered, making the speed independent of how much N2O is around.

🎯 Exam Tip: Remember that heterogeneous catalytic reactions can often exhibit zero-order kinetics due to surface saturation effects, where the active sites on the catalyst become fully occupied.

Question 24. Which of the following are zero order reaction ?
(i) Isomerisation of cyclo propane to propene.
(ii) Iodination of acetone in acid medium with respect to Iodine.
(iii) Photo chemical reaction between H2 and Cl2
(a) (i) only
(b) (i) & (ii)
(c) (ii) only
(d) (ii) & (iii)
Answer: (d) (ii) & (iii)
In simple words: The iodination of acetone in acid (for iodine) and the light-driven reaction of hydrogen and chlorine are examples of zero-order reactions. In these cases, the speed of the reaction does not change with the amount of certain reactants.

🎯 Exam Tip: Zero-order reactions are relatively rare but commonly found in enzyme-catalyzed reactions (at high substrate concentrations), surface-catalyzed reactions (when the surface is saturated), and some photochemical reactions.

Question 25. Acid hydrolysis of an ester is an example for ............ order reaction.
(a) zero
(b) first
(c) pseudo first
(d) second
Answer: (c) pseudo first
In simple words: The acid breakdown of an ester is a "pseudo first-order" reaction. This is because water, a reactant, is present in such a huge amount that its concentration barely changes, making the reaction only seem to depend on the ester's concentration.

🎯 Exam Tip: Recognize pseudo-first-order reactions by identifying a reaction where one reactant is in vast excess, allowing its concentration to be treated as constant, thus simplifying the observed reaction order.

Question 26. Base hydrolysis of an ester is an example for ............ order reaction
(a) zero
(b) first
(c) pseudo first
Answer: (c) pseudo first
In simple words: While base hydrolysis of an ester is usually a second-order reaction, if the base is in a very large amount, its concentration stays almost the same. In this case, the reaction appears to be a "pseudo first-order" reaction.

🎯 Exam Tip: While base hydrolysis of an ester is fundamentally a second-order reaction, it can be simplified to pseudo-first-order kinetics if one of the reactants (typically the base) is in significant excess.

Question 27. The half life period of a first order reaction is
(a) t1/2 = \( \frac{2.303}{k} \log 2 \)
(b) t1/2 = \( \frac{2.303}{k} \times 0.3010 \)
(c) t1/2 = \( \frac{0.6932}{k} \)
(d) all the above
Answer: (d) all the above
In simple words: The half-life period for a first-order reaction can be written in many ways, but they all mean the same thing. All the given options are correct mathematical forms of how to calculate the half-life.

🎯 Exam Tip: Recognize the various mathematical expressions for the half-life of a first-order reaction; they are all derived from the same integrated rate law.

Question 28. If the initial concentration of the reactant is doubled, the half life period of a first order reaction is
(a) doubled
(b) tripled
(c) quadrupled
(d) remains the same
Answer: (d) remains the same
In simple words: A special thing about first-order reactions is that their half-life period does not depend on how much reactant you start with. So, if you double the initial amount, the half-life stays exactly the same.

🎯 Exam Tip: This independence of half-life from initial concentration is a defining feature of first-order reactions and helps distinguish them from zero-order (half-life increases with concentration) and second-order reactions (half-life decreases with concentration).

Question 29. The half life period of a zero order reaction is
(a) t1/2 = \( \frac{[A]_0}{[k]} \)
(b) t1/2 = \( \frac{2[A]_0}{[k]} \)
(c) t1/2 = \( \frac{[A]_0}{2k} \)
(d) t1/2 = \( \frac{k}{2[A]_0} \)
Answer: (c) t1/2 = \( \frac{[A]_0}{2k} \)
In simple words: For a zero-order reaction, the half-life period is directly related to the starting amount of the reactant. It is calculated by dividing the initial concentration by two times the rate constant.

🎯 Exam Tip: Remember that for a zero-order reaction, the half-life is directly proportional to the initial concentration, meaning it changes if you start with a different amount of reactant.

Question 30. If the initial concentration of the reactant is doubled, the half life period of a zero order reaction is
(a) doubled
(b) tripled
(c) quadrupled
(d) remains the same
Answer: (a) doubled
In simple words: For a zero-order reaction, if you double the initial amount of reactant, the half-life period will also double. This is because the reaction rate is constant and does not speed up with more reactant.

🎯 Exam Tip: Understand the relationship between half-life and initial concentration for different reaction orders: directly proportional for zero-order, independent for first-order, and inversely proportional for second-order reactions.

Question 31. The half life period of a second order reaction is
(a) \( \frac{[A]_0}{2k} \)
(b) \( \frac{1}{k[A]_0} \)
(c) \( \frac{1}{k[A]_0} \)
(d) \( \frac{2[A]_0}{k} \)
Answer: (c) \( \frac{1}{k[A]_0} \)
In simple words: For a second-order reaction, the half-life period is found by dividing 1 by the rate constant multiplied by the starting amount of reactant. This means the half-life gets shorter if you start with more reactant.

🎯 Exam Tip: Note the inverse relationship between half-life and initial concentration for second-order reactions, which distinguishes it from zero-order (direct relationship) and first-order (no relationship).

Question 32. The half life period of a first order reaction is 5 minutes, the time required for 99.9% completion is nearly equal to
(a) 99.9 minu tes
(b) 49.95 minutes
(c) 50 minutes
Answer: (c) 50 minutes
In simple words: For a first-order reaction, the time needed for 99.9% of the reaction to finish is approximately ten times its half-life. Since the half-life is 5 minutes, 99.9% completion will take about 50 minutes.

🎯 Exam Tip: For first-order reactions, a useful approximation is that 99.9% completion takes about 10 half-lives; this saves time on calculations in exams.

Question 33. The time required for 99.9% completion of a first-order reaction is equal to
(a) \( 2t_{1/2} \)
(b) \( 5t_{1/2} \)
(c) \( 10t_{1/2} \)
(d) \( 100t_{1/2} \)
Answer: (c) \( 10t_{1/2} \)
In simple words: For a first-order reaction, the time it takes for 99.9% of the reactant to be used up is approximately ten times the half-life period of that reaction.

🎯 Exam Tip: This rule of thumb (\( t_{99.9\%} \approx 10 t_{1/2} \)) is specific to first-order reactions and is derived from the logarithmic nature of their integrated rate law.

Question 34. Collision theory was proposed independently by
(a) MaxTrautz
(b) William Lewis
(c) both (a) & (b)
(d) none of the options
Answer: (c) both (a) & (b)
In simple words: Max Trautz and William Lewis both came up with the ideas of collision theory on their own. This theory helps explain how molecules must hit each other to react.

🎯 Exam Tip: When discussing collision theory, remember to credit both Max Trautz and William Lewis for their independent contributions to its development.

Question 35. Collision theory is based on
(a) Arrhenius theory
(b) Kinetic theory of gases
(c) Ostwald theory
(d) Gas laws
Answer: (b) Kinetic theory of gases
In simple words: Collision theory, which explains that molecules must collide with enough energy to react, is built on the ideas of the kinetic theory of gases. This theory describes how gas particles move and crash into each other.

🎯 Exam Tip: Understand that the kinetic theory of gases provides the framework for collision theory by describing the motion and collision frequency of molecules, which are essential for chemical reactions.

Question 36. In order to react the colliding molecules must possess a minimum energy called
(a) Kinetic energy
(b) activation energy
(c) potential energy
(d) bond energy
Answer: (b) activation energy
In simple words: For molecules to react successfully when they collide, they need to have a minimum amount of energy. This necessary energy is called "activation energy," and it helps them overcome the initial energy barrier.

🎯 Exam Tip: Activation energy is the minimum energy required for a reaction to occur; it determines the temperature sensitivity of the reaction rate.

Question 37. The rate of a reaction ............ with increasing temperature
(a) increases
(b) decreases
(c) becomes zero
(d) remains the same
Answer: (a) increases
In simple words: When a reaction gets hotter, the molecules move faster, leading to more frequent and stronger collisions. This means more molecules have enough energy to react, so the reaction speeds up.

🎯 Exam Tip: This direct relationship between temperature and reaction rate is a fundamental principle, explained by collision theory and the Arrhenius equation.

Question 38. For many reactions near room temperature, reaction rate tends to ............ when the temperature is increased by 10 C.
(a) decreases
(b) double
(c) triple
(d) quaruple
Answer: (b) double
In simple words: For many chemical reactions, especially around room temperature, the speed of the reaction generally doubles for every 10-degree Celsius increase in temperature. This happens because molecules get more energy and collide more often.

🎯 Exam Tip: This "rule of thumb" (rate doubling for every 10°C rise) is a useful generalization, though the exact increase can vary depending on the activation energy of the specific reaction.

Question 39. The slope of the straight line obtained from the plot of lnk Vs 1/T is T
(a) \( - \frac{E_a}{RT} \)
(b) \( - \frac{E_a}{R} \)
(c) \( \frac{E_a}{RT} \)
(d) \( \frac{E_a}{R} \)
Answer: (b) \( - \frac{E_a}{R} \)
In simple words: When you graph the natural logarithm of the rate constant (lnk) against 1 divided by the temperature (1/T), you get a straight line. The steepness of this line, called the slope, is equal to the negative of the activation energy divided by the gas constant.

🎯 Exam Tip: The Arrhenius plot (ln k vs 1/T) is a powerful tool to experimentally determine the activation energy of a reaction from the slope.

Question 40. The y intercept of the straight line obtained from the plot of lnk vs 1/T is
(a) ln k
(b) \( - \frac{E_a}{R} \)
(c) ln A
(d) \( - \frac{E_a}{RT} \)
Answer: (c) ln A
In simple words: In the Arrhenius graph where lnk is plotted against 1/T, the point where the line crosses the y-axis (the y-intercept) tells you the natural logarithm of the frequency factor 'A'. This 'A' factor shows how often molecules collide with the right energy and direction.

🎯 Exam Tip: The y-intercept of an Arrhenius plot provides the value of ln A, where A is the pre-exponential factor or frequency factor, which relates to the frequency of collisions and proper orientation.

Question 41. Rate of a reaction ............ with increase in concentration of the reactants
(a) decreases
(b) increases
(c) remains the same
(d) is slower
Answer: (b) increases
In simple words: For most chemical reactions, if you increase the amount of reactants, the molecules bump into each other more often. More collisions mean more chances for them to react, so the reaction speeds up.

🎯 Exam Tip: This direct relationship between concentration and reaction rate is a fundamental principle, a direct consequence of collision theory.

Question 42. As concentration of the reactant increases the number of collisions between the molecules
(a) decreases
(b) increases
(c) no change
(d) is slower
Answer: (b) increases
In simple words: When you put more reactant molecules into the same space, they are packed closer together. This makes them hit each other more frequently, leading to a higher number of collisions.

🎯 Exam Tip: This relationship between concentration and collision frequency is a fundamental aspect of collision theory, explaining why reaction rates typically increase with concentration.

Question 43. Increase in surface area of the reactant............the rate of the reaction
(a) increases
(b) decreases
(c) no change
(d) is slower
Answer: (a) increases
In simple words: For reactions that involve solid reactants, making the solid into smaller pieces (increasing surface area) means more of its parts can touch other reactants. This extra contact makes the reaction happen faster.

🎯 Exam Tip: This factor is particularly relevant for heterogeneous reactions involving solids, where the reaction often occurs at the interface between phases.

Question 44. Increase in surface area of the reactant .......... the number of collisions between the molecules.
(a) increases
(b) decreases
(c) no change
(d) is slower
Answer: (a) increases
In simple words: When the surface area of a reactant is made larger, there are more places for other molecules to bump into it. This directly causes more collisions to happen, which helps speed up the reaction.

🎯 Exam Tip: This is a physical factor influencing reaction rate, primarily affecting heterogeneous reactions by providing more sites for contact and interaction.

Question 45. Generally addition of a catalyst ............ the rate of a reaction
(a) increases
(b) decreases
(c) inhibits
Answer: (a) increases
In simple words: When you add a catalyst to a reaction, it almost always makes the reaction happen faster. Catalysts do this by finding an easier path for the reaction that needs less energy to start.

🎯 Exam Tip: Remember that catalysts increase reaction rates without being consumed themselves and do not change the equilibrium position of a reversible reaction, only the speed at which equilibrium is reached.

Question 46. A catalyst ............ the activation energy of a reaction.
(a) increases
(b) decreases
(c) has no change
(d) none of the options
Answer: (b) decreases
In simple words: A catalyst helps a reaction go faster by lowering the "activation energy," which is the minimum energy needed for the reaction to start. It provides an easier path for the molecules to react.

🎯 Exam Tip: Catalysts lower the activation energy by changing the reaction mechanism, but they do not affect the energy of the reactants or products.

Question 47. What is the rate law of the reaction 2A + 2B → C + 2D. If the concentration of A is doubled at constant [B] the rate of the reaction increases by factor 4. If the concentration of B is doubled at constant [A], the rate is doubled.
(a) Rate = k [A] [B]2
(b) Rate = k [A] [B]
(c) Rate = k [A] 1/2 [B]2
(d) Rate = k [A]2[B]
Answer: (d) Rate = k [A]2[B]
In simple words: To find the rate law, we look at how the speed of the reaction changes. If doubling reactant A makes the rate 4 times faster, it means A has a power of 2. If doubling reactant B makes the rate 2 times faster, B has a power of 1. So, the rate law is k times A squared times B.

🎯 Exam Tip: Determine the order of reaction with respect to each reactant by observing how the rate changes when only one reactant's concentration is altered, keeping others constant.

Question 48. The concentration of a reactant decreases from 0.5 M to 0.3 M in 10 minutes. The rate of the reaction is
(a) 0.01mol L-1 S-1
(b) 0.02 mol L-1 S-1
(c) 0.08mol L-1 S-1
Answer: (b) 0.02 mol L-1 S-1
In simple words: The reaction rate is found by how much the concentration changes over a period of time. Here, the concentration decreases by 0.2 M (from 0.5 M to 0.3 M) in 10 minutes. Dividing the change in concentration by the time gives a rate of 0.02 M per minute.

🎯 Exam Tip: Pay close attention to the units of time (seconds, minutes, hours) specified in the problem and the answer options to ensure consistency in your calculations.

Question 49. The rate of a reaction is doubled for every 10°C rise in temperature. The increase in reaction rate as a result of temperature rise from 10 C to 50°C is
(a) 4
(b) 8
(c) 16
(d) 32
Answer: (c) 16
In simple words: When the temperature goes up by 10°C, the reaction speed doubles. If the temperature increases from 10°C to 50°C, that's a 40°C rise (4 intervals of 10°C). So, the rate doubles four times, which means it becomes 16 times faster.

🎯 Exam Tip: Remember that a 10°C rise doubles the rate; for multiple 10°C increments, multiply the rate by 2 for each increment.

Question 50. For the reaction \( 2N_2O_5 \rightarrow 4NO_2 + O_2 \). If \( \frac{-d[N_2O_5]}{dt} = k_1[N_2O_5] \), \( \frac{d[NO_2]}{dt} = k_2[N_2O_5] \) and \( \frac{d[O_2]}{dt} = k_3[N_2O_5] \), then the relation between \( k_1, k_2 \) and \( k_3 \) is
(a) \( k_1 = k_2 = k_3 \)
(b) \( k_1 = 2k_2 = 4k_3 \)
(c) \( 2k_1 = 4k_2 = k_3 \)
(d) \( 2k_1 = k_2 = 4k_3 \)
Answer: (d) \( 2k_1 = k_2 = 4k_3 \)
In simple words: To find the relationship between the rate constants, we compare the rates of disappearance and formation of each chemical using their coefficients in the balanced equation. This comparison helps establish how the individual rate constants relate to each other.

🎯 Exam Tip: Always relate the rates of reactants and products using their stoichiometric coefficients to find the correct relationship between rate constants.

Question 51. Rate law of a reaction \( A + B \rightarrow C \) is Rate \( = k[A][B]^2 \). If the concentration A and B are doubled at constant volume then the rate of the reaction will be increased by
(a) two times
(b) four times
(c) eight times
(d) sixteen times
Answer: (c) eight times
In simple words: If you double the concentration of A (rate changes by \( 2^1 \)) and double the concentration of B (rate changes by \( 2^2 \)), the total rate increases by \( 2 \times 4 = 8 \) times. The rate law shows how reactant concentrations affect reaction speed.

🎯 Exam Tip: For rate law problems, calculate the change in rate by raising each concentration factor to its respective order and then multiplying these changes.

Question 52. If 60% of a first order reaction is complete in 60 minutes, its half life period is approximately (log4 = 0.6)
(a) 50 minutes
(b) 45 minutes
(c) 25 minutes
(d) 40 minutes
Answer: (b) 45 minutes
In simple words: For a first-order reaction, the half-life is the time it takes for half of the reactant to be used up. We can calculate the rate constant from the given data and then use it to find the half-life.

🎯 Exam Tip: For first-order reactions, the half-life is constant and can be found using the formula \( t_{1/2} = \frac{0.693}{k} \).

Question 53. Half life period of a first order reaction is 1386 seconds. The rate constant of the reaction is
(a) \( 5.0 \times 10^{-3} s^{-1} \)
(b) \( 0.5 \times 10^{-2} s^{-1} \)
(c) \( 0.5 \times 10^{-3} s^{-1} \)
(d) \( 5.0 \times 10^{-2} s^{-1} \)
Answer: (c) \( 0.5 \times 10^{-3} s^{-1} \)
In simple words: The half-life for a first-order reaction is related to its rate constant. If you know one, you can easily find the other using a simple formula.

🎯 Exam Tip: The formula \( k = \frac{0.693}{t_{1/2}} \) is essential for first-order reactions and allows quick calculation of the rate constant from the half-life.

Question 54. For the reaction \( 2A + B \rightarrow 3C + D \). Which of the following does not express the reaction rate?
(a) \( \frac{d[D]}{dt} \)
(b) \( \frac{1}{2} \frac{d[A]}{dt} \)
(c) \( -\frac{1}{3} \frac{d[C]}{dt} \)
(d) \( \frac{d[B]}{dt} \)
Answer: (c) \( -\frac{1}{3} \frac{d[C]}{dt} \)
In simple words: When writing the rate of a reaction, the coefficient for reactants is positive and for products is negative. Also, we always use a positive value for the rate, so a negative sign is added to reactants if they are disappearing. Products are always forming, so their rate term is positive. The rate is divided by its stoichiometric coefficient. Here, C is a product, so its term should be positive, not negative.

🎯 Exam Tip: Remember to use negative signs for reactants (disappearance) and positive signs for products (formation) when expressing reaction rates. Also, divide by the stoichiometric coefficient.

Question 55. The half life period of a first order reaction is 10 minutes. If the initial concentration is 0.08 mol L\(^{-1}\) at what time the concentration will become 0.01 mol L\(^{-1}\)
(a) 10 minutes
(b) 20 minutes
(c) 30 minutes
(d) 40 minutes
Answer: (c) 30 minutes
In simple words: For a first-order reaction, the half-life means the time it takes for the concentration to halve. So, to go from 0.08 M to 0.01 M, the concentration must halve three times. This takes three half-lives.

🎯 Exam Tip: For first-order reactions, each half-life reduces the concentration by half, independent of the initial amount. You can count the number of half-lives needed.

Question 56. For a reaction the rate law is Rate \( = k [A]^{3/2}[B]^{-1/2} \), the over all order of reaction is
(a) 2
(b) 1
(d) 3/2
Answer: (b) 1
In simple words: The overall order of a reaction is found by adding up all the powers of the concentration terms in the rate law. Here, \( 3/2 + (-1/2) \) gives us 1. The reaction depends on both A and B, but the total influence combined is a first-order dependency.

🎯 Exam Tip: To determine the overall reaction order, always sum the exponents of the concentration terms in the rate law expression.

Question 57. A first order reaction is half completed in 45 minutes. How long does it need 99.9% of the reaction to be completed?
(a) 5 hours
(b) 7.5 hours
(c) 10 hours
(d) 20 hours
Answer: (b) 7.5 hours
In simple words: For a first-order reaction, reaching 99.9% completion takes about ten times its half-life. Since the half-life is 45 minutes, multiplying that by 10 gives 450 minutes, which is 7.5 hours.

🎯 Exam Tip: For first-order reactions, \( t_{99.9\%} \) is approximately \( 10 \times t_{1/2} \), which is a useful shortcut for quick calculations.

Question 58. If the activation energy for a reaction at TK is 2.303 RT J mol\(^{-1}\), the ratio of rate constant to frequency factor is
(a) \( 2 \times 10^{-3} \)
(b) \( 2 \times 10^{-2} \)
(c) \( 10^{-1} \)
(d) \( 10^{-2} \)
Answer: (c) \( 10^{-1} \)
In simple words: The Arrhenius equation shows how the rate constant is linked to activation energy and temperature. If the activation energy is given as 2.303 RT, then after putting this value in the Arrhenius equation, the ratio of the rate constant (k) to the frequency factor (A) simplifies to \( 10^{-1} \). This calculation comes from the exponential part of the equation.

🎯 Exam Tip: The Arrhenius equation \( k = Ae^{-E_a/RT} \) is key; substitute the given activation energy directly into the exponent to find the ratio \( k/A \).

Question 59. Chemical reactions with very high Ea values are generally
(a) very fast
(b) very slow
(c) moderately fast
(d) spontaneous
Answer: (b) very slow
In simple words: Activation energy (Ea) is like a barrier that reactants need to cross to become products. If this barrier is very high, it means that very few reactant molecules will have enough energy to react, making the reaction proceed very slowly.

🎯 Exam Tip: A higher activation energy means more energy is needed to start the reaction, leading to a slower reaction rate at a given temperature.

Question 60. If the activation energy of a reaction is zero, the rate constant of this reaction
(a) increases with increase in temperature
(b) decreases with increase in temperature
(c) decreases with decrease in temperature
(d) is independent of temperature
Answer: (d) is independent of temperature
In simple words: The Arrhenius equation shows that if the activation energy is zero, the rate constant becomes equal to the frequency factor. This means the rate constant does not change with temperature.

🎯 Exam Tip: For a zero activation energy, all collisions are effective, and the rate becomes solely dependent on the frequency of collisions, which is not affected by temperature in the Arrhenius equation.

VI. Two mark Questions

Question 1. Define rate of a reaction.
Answer: The rate of a chemical reaction is defined as the change in concentration of any reacting species (either reactant or product) per unit of time. This shows how quickly reactants are consumed or products are formed. For instance, if you have a lot of reactants changing into products very fast, the rate of reaction is high.
In simple words: Reaction rate tells us how fast the amounts of chemicals change in a reaction over time.

🎯 Exam Tip: Clearly state that the rate is measured as a change in concentration (or pressure for gases) over time, and specify that it can be for either reactants (decreasing) or products (increasing).

Question 2. Write the rate expression of the following reaction. \( N_2 + 3H_2 \rightarrow 2NH_3 \)
Answer: The rate expression for the reaction \( N_2 + 3H_2 \rightarrow 2NH_3 \) can be written by considering the change in concentration of each reactant and product over time, divided by their stoichiometric coefficients. For reactants, a negative sign is used because their concentrations decrease. For products, a positive sign is used because their concentrations increase.
Rate \( = -\frac{d[N_2]}{dt} = -\frac{1}{3}\frac{d[H_2]}{dt} = +\frac{1}{2}\frac{d[NH_3]}{dt} \)
In simple words: The rate is how fast nitrogen and hydrogen are used up, and how fast ammonia is made, adjusted by their numbers in the chemical equation.

🎯 Exam Tip: Always include the negative sign for reactants and a positive sign for products, and divide by the stoichiometric coefficient to correctly express the rate of reaction.

Question 3. Why a negative sign is introduced in the rate expression ?
Answer: A negative sign is introduced in the rate expression for reactants because the concentration of reactants decreases as the reaction progresses. By convention, the rate of a reaction must always be a positive value. Since the change in concentration of a reactant (\( \Delta[\text{reactant}] \)) would be negative (final concentration - initial concentration), adding a negative sign makes the overall rate positive. This ensures consistency in expressing reaction speeds.
In simple words: We put a minus sign for reactants because their amount goes down, and we want the reaction rate number to always be positive.

🎯 Exam Tip: Emphasize that the negative sign is a convention to make the reaction rate positive, as reactant concentration change is inherently negative.

Question 4. Define Order of a reaction.
Answer: The order of a reaction is defined as the sum of the powers of the concentration terms of the reactants in the experimentally determined rate law. It shows how the rate of reaction depends on the concentration of each reactant. For example, if doubling a reactant's concentration quadruples the rate, the reaction is second order with respect to that reactant.
In simple words: The order of a reaction is the total power of reactant concentrations in the rate equation, which shows how strongly concentration affects the reaction speed.

🎯 Exam Tip: Stress that reaction order is an experimentally determined value, not always derived from the stoichiometric coefficients of the balanced equation.

Question 5. Define molecularity of a reaction.
Answer: Molecularity of a reaction is the total number of reactant species (atoms, ions, or molecules) that participate in the slowest or elementary step of a chemical reaction. This number represents the count of particles that must collide at the same time to cause the reaction. For example, if two molecules collide and react in one step, the molecularity is two.
In simple words: Molecularity is the count of molecules that bump into each other in a single step of a chemical reaction.

🎯 Exam Tip: Highlight that molecularity is always a whole number (1, 2, or 3, rarely higher) and refers only to elementary steps, not overall reactions.

Question 6. Mention the factors that affect the rate of the reaction.
Answer: The main factors that affect the rate (speed) of a chemical reaction are:
1. **Nature and state of the reactant:** Different chemicals react at different speeds. Solids, liquids, and gases also react at different rates.
2. **Concentration of the reactant:** Higher concentration means more particles, so more collisions and a faster reaction.
3. **Surface area of the reactant:** For solids, a larger surface area (like powder instead of a lump) means more places for reaction, speeding it up.
4. **Temperature of the reaction:** Generally, higher temperatures make reactions faster because particles move more quickly and collide with more energy.
5. **Presence of a catalyst:** A catalyst is a substance that speeds up a reaction without being used up itself. It provides an easier path for the reaction to happen.
In simple words: How fast a reaction goes depends on what chemicals are reacting, how much of them there is, how much surface they have (for solids), the heat, and if a helper chemical (catalyst) is present.

🎯 Exam Tip: When listing factors, provide a brief explanation for each to show understanding of why it influences the reaction rate.

VII. Three mark Questions

Question 1. Write the differences between rate and rate constant of a reaction.
Answer: Here are the differences between the rate of a reaction and the rate constant of a reaction:
**Rate of a reaction:**
1. It describes how fast reactants turn into products. It is essentially the speed of the reaction at any moment.
2. It is measured as the decrease in reactant concentration or the increase in product concentration over time.
3. It changes as the concentrations of reactants change. For example, the rate usually slows down as reactants are used up.

**Rate constant (k) of a reaction:**
1. It is a specific constant value that links the rate of reaction to the concentrations of reactants at a given temperature. It is a proportionality constant.
2. It is equal to the rate of reaction when the concentration of each reactant is exactly one unit (e.g., 1 mol/L).
3. It does not depend on the initial concentrations of reactants. However, it is strongly affected by temperature and the presence of a catalyst. The rate constant is a fundamental property of a specific reaction under specific conditions.
In simple words: The reaction rate is how fast chemicals are changing right now, and it changes all the time. The rate constant is a fixed number for that reaction at a certain temperature, showing its natural speed, and it doesn't change with how much chemical you start with.

🎯 Exam Tip: Distinguish clearly that the rate changes during a reaction, while the rate constant (k) is fixed for a given temperature and describes the intrinsic speed of the reaction.

Question 2. Give three examples for first order reaction.
Answer: Here are three examples of first-order reactions:
(i) **Decomposition of dinitrogen pentoxide:** This is a common example where one molecule of \( N_2O_5 \) breaks down. The rate of this reaction depends only on the concentration of \( N_2O_5 \).
\[ N_2O_5(g) \rightarrow 2NO_2(g) + \frac{1}{2}O_2(g) \]
(ii) **Decomposition of sulphuryl chloride:** In this reaction, \( SO_2Cl_2 \) breaks down into \( SO_2 \) and \( Cl_2 \). Its rate is directly proportional to the concentration of \( SO_2Cl_2 \).
\[ SO_2Cl_2(l) \rightarrow SO_2(g) + Cl_2(g) \]
(iii) **Decomposition of hydrogen peroxide in aqueous solution:** Hydrogen peroxide breaks down into water and oxygen. This process is often catalyzed, but in an aqueous solution, it follows first-order kinetics with respect to hydrogen peroxide.
\[ H_2O_2(aq) \rightarrow H_2O(l) + \frac{1}{2}O_2(g) \]
In simple words: First-order reactions are those where the speed depends only on how much of one specific chemical is present. Examples include the breakdown of \( N_2O_5 \), \( SO_2Cl_2 \), and \( H_2O_2 \).

🎯 Exam Tip: When providing examples, always include the balanced chemical equation to clearly illustrate the reaction.

Question 3. Calculate the half -life period of zero order reaction.
Answer: To calculate the half-life period (\( t_{1/2} \)) for a zero-order reaction, we start with its integrated rate law, which is \( [A] = [A_0] - kt \). Half-life is the time when the concentration of reactant \( [A] \) becomes half of its initial concentration, so \( [A] = \frac{[A_0]}{2} \).
Substitute this into the integrated rate law:
\( \frac{[A_0]}{2} = [A_0] - kt_{1/2} \)

\( kt_{1/2} = [A_0] - \frac{[A_0]}{2} \)

\( kt_{1/2} = \frac{[A_0]}{2} \)

Therefore, the half-life period for a zero-order reaction is:
\( t_{1/2} = \frac{[A_0]}{2k} \)
This means that for a zero-order reaction, the half-life depends on the initial concentration of the reactant, unlike first-order reactions. As the initial concentration increases, the half-life also increases.
In simple words: For a zero-order reaction, half-life is the time when half of the starting chemical is gone. You find it by dividing the starting amount by twice the rate constant.

🎯 Exam Tip: Remember that for zero-order reactions, the half-life is directly proportional to the initial concentration, a key difference from first-order reactions.

Question 4. For the general reaction \( A \rightarrow B \), Plot of concentration of A Vs time is given in the graph below. Answer the following questions on the basis of this graph. (i) What is the order of reaction? (ii) What is the slope of the curve? (iii) What is the unit of rate constant?
Answer: The graph shows a straight line with a negative slope when concentration of A ([A]) is plotted against time (t). This specific linear relationship is characteristic of a zero-order reaction.
(i) **Order of reaction:** It is a zero-order reaction. This is because the plot of reactant concentration versus time is a straight line, which is how zero-order reactions behave.
(ii) **Slope of the curve:** The slope of the curve is the negative of the rate constant, which is denoted as \( -k \). In other words, \( \text{Slope} = -k \). This value tells us how fast the concentration decreases over time.
(iii) **Unit of rate constant:** The unit of the rate constant for a zero-order reaction is mol \( L^{-1} s^{-1} \). This unit indicates the change in concentration per unit time.
In simple words: This graph shows a zero-order reaction because concentration goes down in a straight line over time. The slope of this line is the negative of the rate constant, and the unit for that rate constant is mol per liter per second.

🎯 Exam Tip: Recognize the characteristic linear plot of concentration versus time for a zero-order reaction. Its negative slope is a direct measure of the rate constant.

Question 1. Derive the integrated rate law for a first order reaction.
Answer: Let's derive the integrated rate law for a first-order reaction, which is a reaction where the rate depends only on the concentration of one reactant, raised to the power of one.
Consider a first-order reaction:
\( A \rightarrow \text{products} \)
The differential rate law for this reaction is:
Rate \( = -\frac{d[A]}{dt} = k[A]^1 \)
Now, we rearrange the equation to separate the variables for integration:
\( -\frac{d[A]}{[A]} = k dt \)
Next, we integrate this equation from the initial concentration \( [A_0] \) at time \( t=0 \) to the concentration \( [A] \) at time \( t=t \):
\[ \int_{[A_0]}^{[A]} -\frac{d[A]}{[A]} = \int_{0}^{t} k dt \]
\[ -[\ln[A]]_{[A_0]}^{[A]} = k[t]_{0}^{t} \]

\( -(\ln[A] - \ln[A_0]) = k(t - 0) \)

\( -\ln[A] + \ln[A_0] = kt \)

\( \ln[A_0] - \ln[A] = kt \)
This can be written using logarithm properties as:
\( \ln \frac{[A_0]}{[A]} = kt \)
To convert from natural logarithm (ln) to common logarithm (log base 10), we use the relation \( \ln x = 2.303 \log x \):
\( 2.303 \log \frac{[A_0]}{[A]} = kt \)
Finally, the integrated rate law for a first-order reaction is:
\( k = \frac{2.303}{t} \log \frac{[A_0]}{[A]} \)
This equation allows us to calculate the rate constant or the concentration of reactant at any given time for a first-order reaction. The rate law shows a logarithmic decrease in reactant concentration over time.
In simple words: For a reaction where the speed depends only on one chemical's amount, we can write an equation that connects how much chemical is left, the starting amount, and the time, using logarithms.

🎯 Exam Tip: Clearly show each step of the integration process, from the differential rate law to the final integrated form, and explain the conversion from natural to common logarithms.

Question 2. Derive Arrhenius equation to calculate activation energy from the rate constant \( k_1 \) and \( k_2 \) temperature \( T_1 \) and \( T_2 \) respectively.
Answer: The Arrhenius equation is a formula that describes how the rate constant of a chemical reaction changes with temperature. It is a fundamental concept in chemical kinetics that helps explain why reactions speed up when heated.
Arrhenius proposed that the rate constant (\( k \)) of a reaction is directly proportional to \( e^{-E_a/RT} \), which represents the fraction of molecules with enough energy to react. The base form of the Arrhenius equation is:
\( k = Ae^{-E_a/RT} \)
Where:
* \( k \) = Rate constant
* \( A \) = Arrhenius factor (or frequency factor), representing the frequency of collisions with correct orientation.
* \( E_a \) = Activation energy, the minimum energy required for a reaction to occur.
* \( R \) = Gas constant
* \( T \) = Absolute temperature (in Kelvin)
The frequency factor \( A \) does not change much with temperature, so we can treat it as a constant. Taking the natural logarithm of both sides of the Arrhenius equation gives:
\( \ln k = \ln A - \frac{E_a}{RT} \)
This equation shows that a plot of \( \ln k \) versus \( \frac{1}{T} \) should give a straight line with a negative slope.
Now, let's consider two different temperatures, \( T_1 \) and \( T_2 \), with their corresponding rate constants, \( k_1 \) and \( k_2 \):
For temperature \( T_1 \):
\( \ln k_1 = \ln A - \frac{E_a}{RT_1} \quad \ldots(2) \)
For temperature \( T_2 \):
\( \ln k_2 = \ln A - \frac{E_a}{RT_2} \quad \ldots(3) \)
To find the activation energy (\( E_a \)), we subtract equation (2) from equation (3):
\( \ln k_2 - \ln k_1 = \left( \ln A - \frac{E_a}{RT_2} \right) - \left( \ln A - \frac{E_a}{RT_1} \right) \)

\( \ln \frac{k_2}{k_1} = -\frac{E_a}{RT_2} + \frac{E_a}{RT_1} \)

\( \ln \frac{k_2}{k_1} = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \)

\( \ln \frac{k_2}{k_1} = \frac{E_a}{R} \left( \frac{T_2 - T_1}{T_1T_2} \right) \)
Converting the natural logarithm (\( \ln \)) to common logarithm (\( \log_{10} \)) by multiplying by 2.303:
\( 2.303 \log \frac{k_2}{k_1} = \frac{E_a}{R} \left( \frac{T_2 - T_1}{T_1T_2} \right) \)
Finally, the equation to calculate activation energy from two rate constants at two different temperatures is:
\( \log \frac{k_2}{k_1} = \frac{E_a}{2.303R} \left( \frac{T_2 - T_1}{T_1T_2} \right) \)
This equation is very useful for determining the activation energy of a reaction when experimental data for rate constants at different temperatures are available. It helps us understand the energy requirements for a reaction to proceed.
In simple words: The Arrhenius equation helps us find the energy needed to start a reaction (activation energy) by comparing how fast the reaction happens at two different temperatures.

🎯 Exam Tip: Clearly define all terms in the Arrhenius equation and show the logarithmic transformation and the derivation for two temperatures step-by-step.

Question 3. Explain the factors affecting reaction rate.
Answer: The speed at which a chemical reaction proceeds, known as its rate, is influenced by several key factors. Understanding these factors helps in controlling and predicting reaction outcomes.

**i) Nature and state of the reactant:**
* Different chemical substances have different reactivities, meaning some react faster than others naturally. For example, highly reactive metals like sodium react very quickly with water, while less reactive metals react much slower. This difference in reactivity stems from their electron configuration and bond strengths. A reaction involving the breaking of stronger bonds will typically be slower. For instance, redox reactions often show different rates depending on the specific reactants involved. Also, the physical state matters: gas-phase reactions are generally faster than liquid or solid-phase reactions because molecules in a gas can move and collide more freely.

**ii) Concentration of the reactants:**
* The rate of a reaction generally increases with an increase in the concentration of reactants. This is explained by the collision theory, which states that chemical reactions happen when reactant molecules collide with each other. A higher concentration means there are more molecules in the same space, leading to more frequent collisions. This increased number of collisions raises the probability that effective collisions (those with enough energy and correct orientation) will occur, thus speeding up the reaction.

**iii) Effect of surface area of the reactant:**
* For reactions involving solid reactants, increasing the surface area significantly increases the reaction rate. When a solid is broken into smaller pieces, its total exposed surface area becomes larger. This provides more sites for reactant molecules to interact and collide. For example, powdered sugar dissolves much faster than a sugar cube because the powder has a greater surface area exposed to the water, allowing more rapid interaction. Therefore, a larger surface area leads to more frequent collisions per unit volume per second, accelerating the reaction.

**iv) Effect of Temperature:**
* Increasing the temperature almost always increases the rate of a reaction. According to collision theory, higher temperatures mean reactant molecules have more kinetic energy. This increased energy leads to two main effects: (a) molecules move faster, increasing the frequency of collisions, and (b) more importantly, a larger fraction of these collisions will have energy equal to or greater than the activation energy. This means more effective collisions occur, which dramatically speeds up the reaction. A general rule of thumb is that the rate of many reactions doubles for every 10°C rise in temperature. For instance, magnesium reacts more vigorously with hot water than with cold water.

**v) Effect of Catalyst:**
* A catalyst is a substance that increases the rate of a chemical reaction without being consumed in the process. Catalysts work by providing an alternative reaction pathway that has a lower activation energy. This means that at a given temperature, a larger proportion of reactant molecules will have sufficient energy to react. By lowering the energy barrier, the catalyst enables more molecules to cross it, leading to a faster reaction rate. Catalysts participate in the reaction but are regenerated at the end. For example, the decomposition of hydrogen peroxide is greatly sped up by adding manganese dioxide (MnO\( _2 \)).
In simple words: Reaction speed changes based on the type of chemicals, how much of them there is, how big their exposed surface is (for solids), the temperature, and if a catalyst (a helper chemical) is added. All these factors change how often and how well molecules bump into each other.

🎯 Exam Tip: For each factor, not only state how it affects the rate but also provide a brief, clear explanation (often using collision theory) and an example if possible.

v) Effect of Catalyst

A catalyst is a special substance that changes how fast a chemical reaction happens, but it does not get used up itself. It helps the reaction find a new, easier path to follow, which has lower energy. Because of this lower energy path, more reactant molecules can turn into products, making the reaction go faster. For example, adding manganese dioxide (\(MnO_2\)) helps potassium chlorate decompose quicker.

🎯 Exam Tip: Remember that a catalyst never gets used up in a reaction and always lowers the activation energy, making the reaction faster without changing the final products.

IX. Problems Based On Rate Constant

Question 1. From the following data on \(N_2O_5\) decomposition in \(CCl_4\) at 298K show that the reaction is first order. Also evaluate the rate constant of the reaction.

Answer: For a first order reaction, the rate constant \(k\) can be calculated using the formula:
\[ k = \frac{2.303}{t} \log \frac{V_\infty}{(V_\infty - V_t)} \] Here, \(V_\infty\) is the volume of \(O_2\) at infinite time (34.75 ml), and \(V_t\) is the volume of \(O_2\) at time \(t\). Let's calculate \(k\) for different time points:
At \(t = 10\) min:
\( k = \frac{2.303}{10} \log \frac{34.75}{(34.75 - 6.3)} \)
\( k = \frac{2.303}{10} \log \frac{34.75}{28.45} \)
\( k = 0.2303 \times \log(1.2218) \)
\( k = 0.2303 \times 0.0869 \)
\( k = 0.0200 \) min\(^{-1}\)
At \(t = 15\) min:
\( k = \frac{2.303}{15} \log \frac{34.75}{(34.75 - 8.95)} \)
\( k = \frac{2.303}{15} \log \frac{34.75}{25.8} \)
\( k = 0.1535 \times \log(1.3469) \)
\( k = 0.1535 \times 0.1293 \)
\( k = 0.0198 \) min\(^{-1}\)
At \(t = 20\) min:
\( k = \frac{2.303}{20} \log \frac{34.75}{(34.75 - 11.4)} \)
\( k = \frac{2.303}{20} \log \frac{34.75}{23.35} \)
\( k = 0.11515 \times \log(1.4882) \)
\( k = 0.11515 \times 0.1726 \)
\( k = 0.0198 \) min\(^{-1}\)
The mean value of \(k\) is approximately \(0.01986\) min\(^{-1}\). Since the calculated values of \(k\) are very close to each other, it confirms that the reaction follows first order kinetics. This value of \(k\) represents the rate constant for the reaction.
In simple words: We used a special formula to check the reaction speed at different times. Since the speed constant (k) stayed almost the same each time we calculated it, we know it's a "first order" reaction. The average speed constant is about \(0.01986\) per minute.

🎯 Exam Tip: When proving a reaction is first order, calculate the rate constant \(k\) at multiple time points. If the values are consistent (within experimental error), the reaction is confirmed to be first order.

Question 2. From the following data, show that decomposition of \(H_2O_2\) in aqueous solution follows first order reaction, what is the value of rate constant ?

Answer: For the decomposition of hydrogen peroxide, we use the first order rate equation where initial concentration (at \(t=0\)) is \(V_0\) and concentration at time \(t\) is \(V_t\). The volume of \(KMnO_4\) needed tells us the amount of \(H_2O_2\) remaining.
The formula for a first order reaction is:
\[ k = \frac{2.303}{t} \log \frac{V_0}{V_t} \] Let's calculate \(k\) for different time points:
At \(t = 10\) min:
\( k = \frac{2.303}{10} \log \frac{25}{20} \)
\( k = 0.2303 \times \log(1.25) \)
\( k = 0.2303 \times 0.0969 \)
\( k = 0.0223 \) min\(^{-1}\)
At \(t = 20\) min:
\( k = \frac{2.303}{20} \log \frac{25}{15.6} \)
\( k = 0.11515 \times \log(1.6026) \)
\( k = 0.11515 \times 0.2048 \)
\( k = 0.0236 \) min\(^{-1}\)
At \(t = 30\) min:
\( k = \frac{2.303}{30} \log \frac{25}{12.7} \)
\( k = 0.07677 \times \log(1.9685) \)
\( k = 0.07677 \times 0.2941 \)
\( k = 0.0226 \) min\(^{-1}\)
At \(t = 40\) min:
\( k = \frac{2.303}{40} \log \frac{25}{9.4} \)
\( k = 0.057575 \times \log(2.6595) \)
\( k = 0.057575 \times 0.4248 \)
\( k = 0.0245 \) min\(^{-1}\)
The mean value of \(k\) is approximately \(0.0233\) min\(^{-1}\). Since the rate constant \(k\) is nearly the same at different time intervals, the decomposition of \(H_2O_2\) is a first order reaction.
In simple words: We calculated the reaction speed constant (k) for hydrogen peroxide breaking down at different times. Because k stayed almost the same for each calculation, it means this reaction is a "first order" type. The average speed constant is about \(0.0233\) per minute.

🎯 Exam Tip: When using titration data to determine reaction order, ensure that the volume of titrant directly relates to the concentration of the reactant or product at that time.

Question 3. In a first order reaction, it takes the reactant 40.5 minutes to be 25% decomposed. Calculate the rate constant of the reaction.
Answer: For a first order reaction, the integrated rate law is given by:
\[ k = \frac{2.303}{t} \log \frac{[A]_0}{[A]} \] We are given:
Time \(t = 40.5\) minutes
The reactant is 25% decomposed. This means 25% has reacted, so 75% remains.
If we assume the initial concentration \([A]_0 = 100\%\), then the concentration at time \(t\) is \([A] = 100\% - 25\% = 75\%\).
Now, substitute these values into the formula:
\[ k = \frac{2.303}{40.5} \log \frac{100}{75} \]
\[ k = \frac{2.303}{40.5} \log(1.3333) \]
\[ k = \frac{2.303}{40.5} \times 0.1249 \]
\[ k = 0.05686 \times 0.1249 \]
\[ k = 7.10 \times 10^{-3} \text{ min}^{-1} \]
The rate constant for the reaction is \(7.10 \times 10^{-3} \text{ min}^{-1}\). This constant shows how quickly the reaction proceeds, with a larger value indicating a faster reaction.
In simple words: If 25% of a reactant breaks down in 40.5 minutes, we can use a formula to find its speed constant. This constant, which is about \(7.10 \times 10^{-3}\) per minute, tells us how fast the reaction is going.

🎯 Exam Tip: Always pay attention to whether the question states "decomposed" (reacted) or "remains." This affects how you set up \([A]_0\) and \([A]\) in the rate law equation.

Question 4. The rate of formation of a dimer in a second order reaction is \(7.5 \times 10^{-3} \text{ mol L}^{-1}\text{s}^{-1}\) at \(0.05 \text{ mol L}^{-1}\) monomer concentration. Calculate the rate constant.
Answer: Let's consider the dimerization of a monomer \(M\) as a second order reaction:
\( 2M \rightarrow (M)_2 \)
For a second order reaction, the rate law is given by:
\( \text{Rate} = k[M]^n \)
Since it's a second order reaction, \(n=2\).
So, \( \text{Rate} = k[M]^2 \)
We are given:
Rate of formation \( = 7.5 \times 10^{-3} \text{ mol L}^{-1}\text{s}^{-1} \)
Monomer concentration \([M] = 0.05 \text{ mol L}^{-1} \)
We need to calculate the rate constant \(k\). Rearranging the rate law:
\[ k = \frac{\text{Rate}}{[M]^2} \]
Substitute the given values:
\[ k = \frac{7.5 \times 10^{-3} \text{ mol L}^{-1}\text{s}^{-1}}{(0.05 \text{ mol L}^{-1})^2} \]
\[ k = \frac{7.5 \times 10^{-3}}{0.0025} \]
\[ k = 3 \text{ mol}^{-1}\text{ L s}^{-1} \]
The rate constant for this dimerization reaction is \(3 \text{ mol}^{-1}\text{ L s}^{-1}\). This value helps predict the rate of the reaction at different monomer concentrations.
In simple words: We know how fast a substance (monomer) turns into a pair (dimer) in a "second order" reaction, and we know how much monomer there is. Using these numbers, we calculated the reaction's unique speed constant, which is 3.

🎯 Exam Tip: Always ensure the units of the rate constant match the order of the reaction. For a second-order reaction, the units are typically \(\text{mol}^{-1}\text{ L s}^{-1}\).

X. Problems Based On Half Life Period

Question 1. At 25°C the rate constant of a first order reaction is \(0.45 \text{ s}^{-1}\). What is its half life? Calculate the time required for 12.5% of the reactant to remain.
Answer: For a first order reaction, the half-life (\(t_{1/2}\)) is given by:
\[ t_{1/2} = \frac{0.693}{k} \] We are given the rate constant \(k = 0.45 \text{ s}^{-1}\).
Substitute the value of \(k\):
\[ t_{1/2} = \frac{0.693}{0.45 \text{ s}^{-1}} \]
\[ t_{1/2} = 1.54 \text{ s} \]
Next, we need to find the time required for 12.5% of the reactant to remain.
If 12.5% remains, it means the concentration has been reduced multiple times by half.
Initial: 100%
After 1 half-life: \(100\% \times \frac{1}{2} = 50\%\)
After 2 half-lives: \(50\% \times \frac{1}{2} = 25\%\)
After 3 half-lives: \(25\% \times \frac{1}{2} = 12.5\%\)
So, 3 half-lives are required for 12.5% of the reactant to remain.
Total time \( = 3 \times t_{1/2} \)
Total time \( = 3 \times 1.54 \text{ s} \)
Total time \( = 4.62 \text{ s} \)
So, the half-life is 1.54 seconds, and it takes 4.62 seconds for only 12.5% of the reactant to be left.
In simple words: For this reaction, half of the reactant is used up in 1.54 seconds. To get down to only 12.5% of the original amount, we need to wait for three of these "half-life" periods, which means 4.62 seconds in total.

🎯 Exam Tip: For first-order reactions, the fraction remaining after 'n' half-lives is always \((1/2)^n\). Use this shortcut to quickly determine the number of half-lives for questions like this.

Question 2. The rate constant of a first order reaction is \(1.54 \times 10^{-3} \text{ s}^{-1}\). What is its half life?
Answer: For a first order reaction, the half-life (\(t_{1/2}\)) is calculated using the formula:
\[ t_{1/2} = \frac{0.693}{k} \] We are given the rate constant \(k = 1.54 \times 10^{-3} \text{ s}^{-1}\).
Substitute the value of \(k\) into the formula:
\[ t_{1/2} = \frac{0.693}{1.54 \times 10^{-3} \text{ s}^{-1}} \]
\[ t_{1/2} = \frac{0.693}{0.00154} \text{ s} \]
\[ t_{1/2} = 450 \text{ s} \]
The half-life of this first order reaction is 450 seconds. This is the time it takes for half of the initial reactant to be consumed.
In simple words: For this type of reaction, we can find out how long it takes for half of the original substance to disappear. With a speed constant of \(1.54 \times 10^{-3}\) per second, it takes 450 seconds for half of it to be used up.

🎯 Exam Tip: Remember that for first-order reactions, the half-life is constant and does not depend on the initial concentration of the reactant.

Question 3. The half life period of a first order reaction is 10 mins, what percentage of the reactant will remain after one hour?
Answer: We are given the half-life (\(t_{1/2}\)) of a first order reaction is 10 minutes.
We need to find out what percentage of the reactant will remain after one hour.
First, convert one hour to minutes: 1 hour \( = 60 \) minutes.
Now, calculate how many half-lives have passed in 60 minutes:
Number of half-lives (\(n\)) \( = \frac{\text{Total time}}{\text{Half-life}} \)
\( n = \frac{60 \text{ min}}{10 \text{ min}} = 6 \)
For a first order reaction, the fraction of reactant remaining after \(n\) half-lives is given by the formula:
\[ \text{Fraction remaining} = \left(\frac{1}{2}\right)^n \]
Substitute \(n=6\):
\[ \text{Fraction remaining} = \left(\frac{1}{2}\right)^6 \]
\[ \text{Fraction remaining} = \frac{1}{64} \]
To express this as a percentage, multiply by 100:
\[ \text{Percentage remaining} = \frac{1}{64} \times 100\% \]
\[ \text{Percentage remaining} = 1.5625\% \]
Therefore, after one hour, 1.5625% of the reactant will remain. This shows how quickly the reactant reduces over time.
In simple words: This reaction uses up half of its starting material every 10 minutes. Since one hour has 6 such 10-minute periods, after an hour, only a tiny bit, 1.5625%, of the original reactant will be left.

🎯 Exam Tip: When dealing with percentage remaining, visualize the decay over multiple half-lives or use the formula \([A] = [A]_0 (1/2)^n\), where \(n\) is the number of half-lives.

Question 5. 75% of a first order reaction is completed in 48 minutes. What is its half life?
Answer: For a first order reaction, the integrated rate law is:
\[ k = \frac{2.303}{t} \log \frac{[A]_0}{[A]} \] We are given:
Time \(t = 48\) minutes
75% of the reaction is completed. This means 75% has reacted, so 25% remains.
If we assume the initial concentration \([A]_0 = 100\%\), then the concentration at time \(t\) is \([A] = 100\% - 75\% = 25\%\).
First, calculate the rate constant \(k\):
\[ k = \frac{2.303}{48 \text{ min}} \log \frac{100}{25} \]
\[ k = \frac{2.303}{48} \log(4) \]
\[ k = \frac{2.303}{48} \times 0.6021 \]
\[ k = 0.04798 \times 0.6021 \]
\[ k = 0.02888 \text{ min}^{-1} \]
Now that we have \(k\), we can calculate the half-life (\(t_{1/2}\)) for a first order reaction:
\[ t_{1/2} = \frac{0.693}{k} \]
\[ t_{1/2} = \frac{0.693}{0.02888 \text{ min}^{-1}} \]
\[ t_{1/2} = 24.0 \text{ min} \]
The half-life of the reaction is 24.0 minutes. This indicates that half of the reactant would be consumed in 24 minutes.
Alternatively, since 75% completion means 25% remaining, this is equivalent to two half-lives.
After 1 half-life, 50% remains.
After 2 half-lives, 25% remains (75% completed).
So, \( 2 \times t_{1/2} = 48 \text{ min} \)
Therefore, \( t_{1/2} = \frac{48}{2} = 24 \text{ min} \).
In simple words: This reaction uses up 75% of its material in 48 minutes. Since using up 75% means you've gone through two "half-life" periods, each half-life must be 24 minutes long.

🎯 Exam Tip: For first-order reactions, 75% completion always corresponds to exactly two half-lives. You can use this as a quick check for your calculations or a shortcut to find the half-life.

Question 6. Show that for a first order reaction the time required for 99% completion of the reaction is twice the time required for 90% completion of the reaction.
Answer: For a first order reaction, the integrated rate law is:
\[ t = \frac{2.303}{k} \log \frac{[A]_0}{[A]} \] Let \([A]_0\) be the initial concentration.
For 99% completion:
The reaction is 99% completed, so 1% of the reactant remains.
If \([A]_0 = 100\), then \([A] = 1\).
\[ t_{99\%} = \frac{2.303}{k} \log \frac{100}{1} \]
\[ t_{99\%} = \frac{2.303}{k} \log(100) \]
\[ t_{99\%} = \frac{2.303}{k} \times 2 \]
\[ t_{99\%} = \frac{4.606}{k} \quad \ldots(1) \]
For 90% completion:
The reaction is 90% completed, so 10% of the reactant remains.
If \([A]_0 = 100\), then \([A] = 10\).
\[ t_{90\%} = \frac{2.303}{k} \log \frac{100}{10} \]
\[ t_{90\%} = \frac{2.303}{k} \log(10) \]
\[ t_{90\%} = \frac{2.303}{k} \times 1 \]
\[ t_{90\%} = \frac{2.303}{k} \quad \ldots(2) \]
Now, let's compare \(t_{99\%}\) and \(t_{90\%}\) by dividing equation (1) by equation (2):
\[ \frac{t_{99\%}}{t_{90\%}} = \frac{\frac{4.606}{k}}{\frac{2.303}{k}} \]
\[ \frac{t_{99\%}}{t_{90\%}} = \frac{4.606}{2.303} \]
\[ \frac{t_{99\%}}{t_{90\%}} = 2 \]
\[ t_{99\%} = 2 \times t_{90\%} \]
This shows that for a first order reaction, the time required for 99% completion is indeed twice the time required for 90% completion. This is a characteristic feature of first-order reactions.
In simple words: For a "first order" reaction, it takes twice as long to finish 99% of the reaction as it does to finish 90%. We proved this using the reaction speed formula, which showed that the time for 99% completion is exactly double the time for 90% completion.

🎯 Exam Tip: Remembering key relationships like this for common completion percentages (e.g., 50%, 75%, 90%, 99%) can save time and help verify answers during exams.

XI. Problems Based On Activation Energy

Question 1. For a first order reaction if the rate constant at 25°C is \(3.46 \times 10^{-5} \text{ s}^{-1}\) and the rate constant at 35°C is \(13.50 \times 10^{-5} \text{ s}^{-1}\), calculate the activation energy \(E_a\) and frequency factor \(A\).
Answer: We use the Arrhenius equation to solve this problem. The two-point form of the Arrhenius equation is:
\[ \log \frac{k_2}{k_1} = \frac{E_a}{2.303R} \left( \frac{T_2 - T_1}{T_1 T_2} \right) \] Given values:
\(k_1 = 3.46 \times 10^{-5} \text{ s}^{-1}\) at \(T_1 = 25^\circ C = 25 + 273 = 298 \text{ K}\)
\(k_2 = 13.50 \times 10^{-5} \text{ s}^{-1}\) at \(T_2 = 35^\circ C = 35 + 273 = 308 \text{ K}\)
Gas constant \(R = 8.314 \text{ J K}^{-1}\text{ mol}^{-1}\)
Substitute the values to calculate \(E_a\):
\[ \log \frac{13.50 \times 10^{-5}}{3.46 \times 10^{-5}} = \frac{E_a}{2.303 \times 8.314} \left( \frac{308 - 298}{298 \times 308} \right) \]
\[ \log(3.9017) = \frac{E_a}{19.147} \left( \frac{10}{91784} \right) \]
\[ 0.59126 = \frac{E_a}{19.147} \times 0.0001089 \]
\[ E_a = \frac{0.59126 \times 19.147}{0.0001089} \]
\[ E_a = \frac{11.3204}{0.0001089} \]
\[ E_a = 103952 \text{ J mol}^{-1} = 103.952 \text{ kJ mol}^{-1} \]
Now, we calculate the frequency factor \(A\) using the Arrhenius equation:
\[ \log k = \log A - \frac{E_a}{2.303RT} \] Let's use \(k_1\) and \(T_1\):
\[ \log (3.46 \times 10^{-5}) = \log A - \frac{103952 \text{ J mol}^{-1}}{2.303 \times 8.314 \text{ J K}^{-1}\text{ mol}^{-1} \times 298 \text{ K}} \]
\[ \log(3.46) + \log(10^{-5}) = \log A - \frac{103952}{5705.8} \]
\[ 0.5391 - 5 = \log A - 18.219 \]
\[ -4.4609 = \log A - 18.219 \]
\[ \log A = 18.219 - 4.4609 \]
\[ \log A = 13.7581 \]
\[ A = \text{antilog}(13.7581) \]
\[ A = 5.73 \times 10^{13} \text{ s}^{-1} \]
The activation energy is approximately \(103.95 \text{ kJ mol}^{-1}\) and the frequency factor is \(5.73 \times 10^{13} \text{ s}^{-1}\). These values are important for understanding the energy requirements and collision frequency of the reaction.
In simple words: We used a special formula to find two important numbers for this reaction: the "activation energy" (how much energy is needed to start it) and the "frequency factor" (how often molecules bump into each other in the right way). The activation energy is about \(103.95 \text{ kJ per mole}\), and the frequency factor is about \(5.73 \times 10^{13}\) per second.

🎯 Exam Tip: Ensure consistent units throughout your calculations. Convert temperature to Kelvin and activation energy to Joules if \(R\) is given in \(\text{J K}^{-1}\text{ mol}^{-1}\).

Question 2. The activation energy of a certain reaction is \(100 \text{ kJ mol}^{-1}\). What is the change in the rate constant if the temperature is changed from 25°C to 35°C?
Answer: We need to find the ratio of the rate constants (\(k_2/k_1\)) when the temperature changes. We use the two-point form of the Arrhenius equation:
\[ \log \frac{k_2}{k_1} = \frac{E_a}{2.303R} \left( \frac{T_2 - T_1}{T_1 T_2} \right) \] Given values:
Activation energy \(E_a = 100 \text{ kJ mol}^{-1} = 100 \times 10^3 \text{ J mol}^{-1}\)
Temperature \(T_1 = 25^\circ C = 25 + 273 = 298 \text{ K}\)
Temperature \(T_2 = 35^\circ C = 35 + 273 = 308 \text{ K}\)
Gas constant \(R = 8.314 \text{ J K}^{-1}\text{ mol}^{-1}\)
Substitute the values into the equation:
\[ \log \frac{k_2}{k_1} = \frac{100 \times 10^3 \text{ J mol}^{-1}}{2.303 \times 8.314 \text{ J K}^{-1}\text{ mol}^{-1}} \left( \frac{308 \text{ K} - 298 \text{ K}}{298 \text{ K} \times 308 \text{ K}} \right) \]
\[ \log \frac{k_2}{k_1} = \frac{100000}{19.147} \left( \frac{10}{91784} \right) \]
\[ \log \frac{k_2}{k_1} = 5222.28 \times 0.0001089 \]
\[ \log \frac{k_2}{k_1} = 0.5687 \]
To find the ratio \(k_2/k_1\), take the antilog:
\[ \frac{k_2}{k_1} = \text{antilog}(0.5687) \]
\[ \frac{k_2}{k_1} = 3.70 \]
So, \(k_2 = 3.70 \times k_1\).
This means that when the temperature is increased from 25°C to 35°C, the rate constant increases by a factor of 3.70. A small temperature change can significantly impact reaction rates.
In simple words: If a reaction needs 100 kJ of energy to start, and we warm it up by 10 degrees (from 25°C to 35°C), its speed constant will become 3.70 times faster. This means the reaction will happen much quicker.

🎯 Exam Tip: A good rule of thumb is that for many reactions, the rate roughly doubles for every 10°C rise in temperature. This question shows a specific calculation for how much it actually increases based on activation energy.

Question 3. Decomposition of ethyl bromide and propyl bromide follow first order kinetics and have the same frequency factor 'A'. The rate constant for decomposition of ethyl bromide at 390°C is same as that for propyl bromide at 320'C. If Ea of ethyl bromide reaction is \(230 \text{ kJ mol}^{-1}\). What is the Ea of propyl bromide reaction?
Answer: We are given that both reactions are first order, have the same frequency factor \(A\), and their rate constants \(k\) are equal at different temperatures.
The Arrhenius equation is:
\[ k = A \mathrm{e}^{\frac{-E_a}{RT}} \] Since \(k\) and \(A\) are the same for both reactions at their respective temperatures, we can write:
\[ k_{\text{ethyl}} = k_{\text{propyl}} \]
\[ A_{\text{ethyl}} \mathrm{e}^{\frac{-E_{a, \text{ethyl}}}{RT_{\text{ethyl}}}} = A_{\text{propyl}} \mathrm{e}^{\frac{-E_{a, \text{propyl}}}{RT_{\text{propyl}}}} \]
Since \(A_{\text{ethyl}} = A_{\text{propyl}}\), we can cancel \(A\) from both sides:
\[ \mathrm{e}^{\frac{-E_{a, \text{ethyl}}}{RT_{\text{ethyl}}}} = \mathrm{e}^{\frac{-E_{a, \text{propyl}}}{RT_{\text{propyl}}}} \]
Taking the natural logarithm of both sides:
\[ \frac{-E_{a, \text{ethyl}}}{RT_{\text{ethyl}}} = \frac{-E_{a, \text{propyl}}}{RT_{\text{propyl}}} \]
We can cancel \(R\) from both sides and multiply by -1:
\[ \frac{E_{a, \text{ethyl}}}{T_{\text{ethyl}}} = \frac{E_{a, \text{propyl}}}{T_{\text{propyl}}} \]
Given values:
\(E_{a, \text{ethyl}} = 230 \text{ kJ mol}^{-1} = 230 \times 10^3 \text{ J mol}^{-1}\)
\(T_{\text{ethyl}} = 390^\circ C = 390 + 273 = 663 \text{ K}\)
\(T_{\text{propyl}} = 320^\circ C = 320 + 273 = 593 \text{ K}\)
We need to find \(E_{a, \text{propyl}}\). Rearranging the equation:
\[ E_{a, \text{propyl}} = \frac{E_{a, \text{ethyl}}}{T_{\text{ethyl}}} \times T_{\text{propyl}} \]
\[ E_{a, \text{propyl}} = \frac{230 \times 10^3 \text{ J mol}^{-1}}{663 \text{ K}} \times 593 \text{ K} \]
\[ E_{a, \text{propyl}} = 346.908 \times 593 \]
\[ E_{a, \text{propyl}} = 205662 \text{ J mol}^{-1} \]
\[ E_{a, \text{propyl}} = 205.662 \text{ kJ mol}^{-1} \]
The activation energy for the decomposition of propyl bromide is approximately \(205.66 \text{ kJ mol}^{-1}\). This shows how the activation energy changes with temperature when rate constants are equal.
In simple words: Two different chemicals break down at the same speed if ethyl bromide is at 390°C and propyl bromide is at 320°C. Since we know the energy needed for ethyl bromide to break down, we can figure out that propyl bromide needs about \(205.66 \text{ kJ per mole}\) of energy to start its reaction.

🎯 Exam Tip: When given that rate constants or frequency factors are equal for different reactions or conditions, simplify the Arrhenius equation to relate the remaining variables directly, often leading to a much easier calculation.

Question 4. The rate constant \(k\) values of a reaction were determined at several temperatures. A plot of \(\ln k\) vs \(1/T\) gave a straight line with the slope \(– 2.6 \times 10^4 \text{ K}\). What is the activation energy \(E_a\) of the reaction?
Answer: The Arrhenius equation in its linear form is:
\[ \ln k = \ln A - \frac{E_a}{R} \left( \frac{1}{T} \right) \] This equation is in the form of a straight line \( y = c + mx \), where:
\( y = \ln k \)
\( x = \frac{1}{T} \)
The slope of the line \(m = -\frac{E_a}{R} \).
We are given:
Slope \( = -2.6 \times 10^4 \text{ K} \)
Gas constant \(R = 8.314 \text{ J K}^{-1}\text{ mol}^{-1}\)
Substitute these values into the slope equation:
\[ -2.6 \times 10^4 \text{ K} = -\frac{E_a}{8.314 \text{ J K}^{-1}\text{ mol}^{-1}} \]
Multiply both sides by -1:
\[ 2.6 \times 10^4 \text{ K} = \frac{E_a}{8.314 \text{ J K}^{-1}\text{ mol}^{-1}} \]
Solve for \(E_a\):
\[ E_a = 2.6 \times 10^4 \text{ K} \times 8.314 \text{ J K}^{-1}\text{ mol}^{-1} \]
\[ E_a = 216164 \text{ J mol}^{-1} \]
\[ E_a = 216.164 \text{ kJ mol}^{-1} \]
The activation energy of the reaction is approximately \(216.16 \text{ kJ mol}^{-1}\). The slope of the \(\ln k\) vs \(1/T\) plot provides a direct way to determine activation energy.
In simple words: When we graph how fast a reaction happens at different temperatures, the steepness of the line tells us the "activation energy" – how much energy is needed to start the reaction. With a slope of \(– 2.6 \times 10^4 \text{ K}\), the reaction needs about \(216.16 \text{ kJ per mole}\) of energy to get going.

🎯 Exam Tip: Understand the relationship between the Arrhenius plot slope and activation energy. A negative slope indicates that as temperature increases (\(1/T\) decreases), the rate constant increases.

Question 5. The rate constant for a first order reaction at 45°C is twice that at 35°C. Find the activation energy of the reaction.
Answer: We can use the two-point form of the Arrhenius equation to find the activation energy (\(E_a\)):
\[ \log \frac{k_2}{k_1} = \frac{E_a}{2.303R} \left( \frac{T_2 - T_1}{T_1 T_2} \right) \] Given values:
Temperature \(T_1 = 35^\circ C = 35 + 273 = 308 \text{ K}\)
Temperature \(T_2 = 45^\circ C = 45 + 273 = 318 \text{ K}\)
The rate constant at 45°C is twice that at 35°C, so \(k_2 = 2k_1\).
This means the ratio \(k_2/k_1 = 2\).
Gas constant \(R = 8.314 \text{ J K}^{-1}\text{ mol}^{-1}\)
Substitute the values into the equation:
\[ \log(2) = \frac{E_a}{2.303 \times 8.314} \left( \frac{318 - 308}{308 \times 318} \right) \]
\[ 0.3010 = \frac{E_a}{19.147} \left( \frac{10}{97944} \right) \]
\[ 0.3010 = \frac{E_a}{19.147} \times 0.0001021 \]
\[ E_a = \frac{0.3010 \times 19.147}{0.0001021} \]
\[ E_a = \frac{5.7636}{0.0001021} \]
\[ E_a = 56450.5 \text{ J mol}^{-1} \]
\[ E_a = 56.45 \text{ kJ mol}^{-1} \]
The activation energy of the reaction is approximately \(56.45 \text{ kJ mol}^{-1}\). This value reflects the minimum energy needed for the reaction to occur.
In simple words: When a reaction's speed doubles just by increasing the temperature by 10 degrees (from 35°C to 45°C), it tells us how much energy is needed to start that reaction. In this case, that starting energy, or "activation energy," is about \(56.45 \text{ kJ per mole}\).

🎯 Exam Tip: Questions relating rate constant changes to temperature often involve the two-point Arrhenius equation. Always ensure temperatures are converted to Kelvin before calculation.