CBSE Class 12 Chemistry Chemical Kinetics Notes and Questions

CHEMICAL KINETICS

CONCEPT

Thermodynamics helps us to predict the feasibility of a chemical reaction by using \(\Delta G\) as a parameter, but it cannot tell everything about the rate of reaction. The rate of a chemical reaction is studied in another branch of chemistry called Chemical Kinetics.

Chemical kinetics: The branch of physical chemistry which deals with the study of the rate of reactions and their mechanisms is called chemical kinetics.

Rate of chemical reaction: The change in concentration of any reactant or product per unit time is called the rate of reaction.

TYPES OF RATE OF REACTION:

1. Average rate of reaction: The rate of reaction measured over a long time interval is called the average rate of reaction.
\(\text{Avg rate } \frac{\Delta x}{\Delta t} = -\frac{\Delta [R]}{\Delta t} = +\frac{\Delta [P]}{\Delta t}\)

2. Instantaneous rate of reaction: The rate of reaction measured at a particular instant of time is called the instantaneous rate of reaction.
\(\text{Instantaneous rate } \frac{dx}{dt} = -\frac{d[R]}{dt} = +\frac{d[P]}{dt}\)

FACTORS AFFECTING RATE OF REACTION:

1. Concentration of reactant
2. Surface area
3. Temperature
4. Nature of reactant
5. Presence of catalyst
6. Radiation

RATE CONSTANT (\(k\)): It is equal to the rate of reaction when the concentration of each reactant is unity.

RATE LAW: The rate of reaction is directly proportional to the product of the concentrations of reactants, where each concentration term is raised to some power which may or may not be equal to its stoichiometric coefficient determined experimentally.

For a general reaction:
\(aA + bB \rightarrow cC + dD\)
Rate law \(= k[A]^p[B]^q\)
Where powers \(p\) and \(q\) are determined experimentally.

MOLECULARITY: The total number of reacting species taking part in an elementary chemical reaction is called molecularity.

ORDER OF REACTION: The sum of the powers to which the concentration terms are raised in a rate law expression is called the order of reaction. For the above case, order \(= p + q\); orders of reactions are determined experimentally.

HALF-LIFE PERIOD: The time during which the concentration of the reactant is reduced to half of its initial concentration is called the half-life period.

ACTIVATION ENERGY: The minimum extra amount of energy absorbed by reactant molecules so that their energy becomes equal to the threshold energy is called activation energy.
Activation energy = Threshold energy - Kinetic energy

TEMPERATURE COEFFICIENT: The ratio of the rate constant at two temperatures having a difference of \(10^\circ\text{C}\) is called the temperature coefficient.
Temperature coefficient \(= \frac{\text{Rate constant at } (T + 10)^\circ\text{C}}{\text{Rate constant at } T^\circ\text{C}}\)

Arrhenius Equation:
\(k = A e^{-E_a / RT}\)
Where:
\(k\) = Rate constant
\(A\) = Arrhenius factor (Frequency factor)
\(E_a\) = Activation energy
\(R\) = Gas constant
\(T\) = Temperature

\(\log k = \log A - \frac{E_a}{2.303 RT}\)

Energy of activation can be evaluated as:
\(\log \left(\frac{k_2}{k_1}\right) = \frac{E_a}{2.303 R} \left(\frac{T_2 - T_1}{T_1 T_2}\right)\)
\(\log \left(\frac{k_2}{k_1}\right) = \frac{E_a}{19.15} \left(\frac{T_2 - T_1}{T_1 T_2}\right)\)

1 MARKS QUESTION

Question 1. The gas phase decomposition of acetaldehyde \(\text{CH}_3\text{CHO} \rightarrow \text{CH}_4 + \text{CO}\) follows the rate law. What are the units of its rate constant?
Answer: Rate \(= k[\text{CH}_3\text{CHO}]^{3/2}\)
Order \(= n = \frac{3}{2}\)
Units of rate constant \(= (\text{mol L}^{-1})^{1-n} \text{s}^{-1} = (\text{mol L}^{-1})^{1 - 3/2} \text{s}^{-1} = \text{atm}^{-1/2} \text{s}^{-1}\) (for gaseous reaction in terms of pressure, \(\text{atm}^{-1/2}\text{sec}^{-1}\)).
In simple words: The unit depends on the fractional order of the reaction.

Exam Tip: Always use the general formula \((\text{concentration})^{1-n} \text{time}^{-1}\) to find units of rate constant.

Question 2. State the order with respect to each reactant and overall reaction for \(\text{H}_2\text{O}_2 + 3\text{I}^- + 2\text{H}^+ \rightarrow 2\text{H}_2\text{O} + \text{I}_3^-\), given Rate \(= k[\text{H}_2\text{O}_2]^1[\text{I}^-]^1\).
Answer: Order with respect to \(\text{H}_2\text{O}_2\) is 1.
Order with respect to \(\text{I}^-\) is 1.
Order with respect to \(\text{H}^+\) is 0.
Overall order of reaction = \(1 + 1 + 0 = 2\).
In simple words: Add the exponents of the concentration terms in the rate law to get the overall order.

Exam Tip: Stoichiometric coefficients do not decide the order; only the rate law expression does.

Question 3. Give one example of pseudo first order reaction.
Answer: Hydrolysis of an ester:
\(\text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O} \xrightarrow{\text{H}^+} \text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH}\)
In simple words: A reaction that is bimolecular but follows first-order kinetics because one reactant is present in large excess.

Exam Tip: Mention the excess reactant (like water in aqueous solution) when defining pseudo-order reactions.

Question 4. The conversion of molecules X to Y follows second order kinetics. If concentration of X is increased 3 times, how will it affect the rate of formation of Y?
Answer: Rate \(= k[X]^2\)
When concentration is increased 3 times, new rate \(= k[3X]^2 = k \cdot 9[X]^2 = 9 \times (\text{Initial Rate})\)
The rate of formation will become 9 times.
In simple words: Since it is second order, squaring the concentration factor (\(3^2 = 9\)) gives the increase in rate.

Exam Tip: Rate is proportional to \([\text{Reactant}]^{\text{order}}\).

Question 5. The rate law for a reaction is Rate \(= k[\text{A}][\text{B}]^{3/2}\). Can the reaction be an elementary process? Explain.
Answer: No, the reaction cannot be an elementary process. An elementary process would have a rate law with orders equal to its molecularities, and molecularity must always be a whole number, whereas the order here is fractional (\(3/2\)).
In simple words: Elementary steps cannot have fractional orders because molecules cannot react in fractions.

Exam Tip: Fractional orders always indicate complex, multi-step reactions.

Question 6. What do you understand by 'rate of reaction'?
Answer: The rate of a reaction is defined as the change in concentration of any reactant or product per unit time.
In simple words: It measures how fast reactants turn into products.

Question 7. Name the factors on which the rate of a particular reaction depends.
Answer: 1. Concentration of reactants
2. Temperature
3. Presence of a catalyst
4. Nature of reactants
5. Surface area of reactants
6. Exposure to radiation
In simple words: Conditions like heat, concentration, and catalysts alter how fast a reaction proceeds.

Exam Tip: Memorize at least four key factors for board exams.

Question 8. Why does the rate of reaction not remain constant throughout?
Answer: The rate of reaction does not remain constant throughout because as the reaction proceeds, the concentration of reactants continuously decreases, leading to a decrease in the frequency of effective collisions.
In simple words: As reactants get used up, there are fewer molecules left to react, slowing down the reaction.

Exam Tip: Connect reaction rate directly to reactant concentration.

Question 9. Define specific reaction rate or rate constant.
Answer: The rate constant (\(k\)) is defined as the rate of the reaction when the concentration of each reactant is taken as unity (\(1\text{ mol L}^{-1}\)).
In simple words: It is the rate of a reaction when all reactants are at 1 molar concentration.

Exam Tip: Rate constant is a measure of the intrinsic speed of a reaction at a given temperature.

Question 10. What is half-life period of a reaction?
Answer: The time taken for the concentration of a reactant to be reduced to half of its initial value is called the half-life period (\(t_{1/2}\)).
In simple words: It is the time required for half the reactant to be consumed.

Exam Tip: For a first-order reaction, half-life is independent of initial concentration.

2 MARKS QUESTION

Question 1. The rate of a particular reaction quadruples when the temperature changes from 293K to 313K. Calculate activation energy.
Answer: Given:\br /> \(T_1 = 293\text{ K}, \quad T_2 = 313\text{ K}\)
\(\frac{k_2}{k_1} = 4\)
Using Arrhenius equation:
\(\log\left(\frac{k_2}{k_1}\right) = \frac{E_a}{2.303 R} \left(\frac{T_2 - T_1}{T_1 T_2}\right)\)
\(\log(4) = \frac{E_a}{19.15} \left(\frac{313 - 293}{293 \times 313}\right)\)
\(0.6021 = \frac{E_a}{19.15} \times \frac{20}{91709}\)
\(E_a = \frac{0.6021 \times 19.15 \times 91709}{20} = 52.86\text{ kJ mol}^{-1}\)
In simple words: Substitute the temperature and rate constant ratio into the logarithmic Arrhenius formula to solve for activation energy.

Exam Tip: Always state units (\(\text{kJ mol}^{-1}\)) in the final answer.

Question 2. If the decomposition of nitrogen oxide \(2\text{N}_2\text{O}_5 \rightarrow 4\text{NO}_2 + \text{O}_2\) follows first order kinetics: (i) Calculate the rate constant for a 0.05 M solution if the instantaneous rate is \(1.5 \times 10^{-6}\text{ mol L}^{-1}\text{s}^{-1}\)? (ii) What concentration of \(\text{N}_2\text{O}_5\) would give a rate of \(2.45 \times 10^{-5}\text{ mol L}^{-1}\text{s}^{-1}\)?
Answer: (i) Rate \(= k[\text{N}_2\text{O}_5]\)
\(k = \frac{\text{Rate}}{[\text{N}_2\text{O}_5]} = \frac{1.5 \times 10^{-6}}{0.05} = 3.0 \times 10^{-5}\text{ s}^{-1}\)
(ii) Given Rate \(= 2.45 \times 10^{-5}\text{ mol L}^{-1}\text{s}^{-1}\)
\([\text{N}_2\text{O}_5] = \frac{\text{Rate}}{k} = \frac{2.45 \times 10^{-5}}{3.0 \times 10^{-5}} = 0.82\text{ M}\)
In simple words: Use the first-order rate law equation \(\text{Rate} = k[\text{Reactant}]\) to find either \(k\) or concentration by rearranging.

Exam Tip: Check that first-order rate constant units are \(\text{s}^{-1}\).

Question 3. Write the difference between order and molecularity of a reaction.
Answer:

In simple words: Order is measured from experiments and can be a fraction, whereas molecularity counts the reacting molecules in a step and is a whole integer.

Exam Tip: Present differences in a tabular format for maximum marks.

Question 4. Define Threshold energy and activation energy. How are they related?
Answer: Threshold Energy: The minimum amount of energy which reactant molecules must possess for an effective collision to form products.
Activation Energy: The excess energy required by reactants to undergo a chemical reaction.
Relation: Activation energy = Threshold energy - Average kinetic energy of reacting molecules.
In simple words: Activation energy is the extra push needed to reach the threshold energy barrier.

Exam Tip: Clear definitions paired with the formula fetch full credit.

Question 5(a). Draw a schematic graph showing how the rate of a first-order reaction changes with concentration of reactants. (b) The rate of reaction is given by the equation Rate = \(k[\text{A}]^2[\text{B}]\). What are the units of rate constant for this reaction?
Answer:
(a) Graph description: A straight line passing through the origin representing direct proportionality between Rate and Concentration.

(b) Given Rate \(= k[\text{A}]^2[\text{B}]\)
Overall order \(= 2 + 1 = 3\)
\(k = \frac{\text{Rate}}{[\text{A}]^2[\text{B}]} = \frac{\text{mol L}^{-1}\text{s}^{-1}}{(\text{mol L}^{-1})^2(\text{mol L}^{-1})} = \frac{\text{mol L}^{-1}\text{s}^{-1}}{(\text{mol L}^{-1})^3} = \text{mol}^{-2}\text{L}^2\text{s}^{-1}\)
In simple words: Substitute units into the rearranged rate constant expression.

Exam Tip: For third-order reactions, units of \(k\) are \(\text{mol}^{-2}\text{L}^2\text{s}^{-1}\).

Question 6. List the factors affecting the rate of reaction.
Answer: 1. Concentration of reactants
2. Temperature
3. Catalyst
4. Surface area
5. Nature of reactants
In simple words: These external and internal conditions speed up or slow down a chemical change.

Exam Tip: Listing bullet points cleanly is sufficient for 2 marks.

Question 7. Explain with a suitable example how the molecularity of a reaction is different from the order of a reaction.
Answer: Consider the hydrolysis of methyl acetate:
\(\text{CH}_3\text{COOCH}_3 + \text{H}_2\text{O} \rightarrow \text{CH}_3\text{COOH} + \text{CH}_3\text{OH}\)
Here, two molecules are taking part, so its molecularity is 2 (bimolecular). However, because water is taken in large excess, its concentration remains nearly constant, making the rate depend only on the ester. Thus, its order is 1.
In simple words: A reaction can involve two molecules (molecularity = 2), but behave as a first-order reaction due to excess reagents.

Exam Tip: Pseudo-first-order reactions are the best examples to show the difference between order and molecularity.

Question 8. Define the term 'rate constant' or 'specific reaction rate'.
Answer: Rate constant is defined as the rate of reaction when the concentration of each reactant is unity.
In simple words: It is the reaction rate when reactant concentrations are all set to 1 molar.

Exam Tip: Mention unity concentration to secure full marks.

Question 9. What are Pseudo unimolecular reactions? Explain with the help of a suitable example.
Answer: Reactions that are expected to be of higher order based on their stoichiometry, but actually follow first-order kinetics because one reactant is present in large excess, are called pseudo-first-order (or pseudo-unimolecular) reactions.
Example: Inversion of cane sugar:
\(\text{C}_{12}\text{H}_{22}\text{O}_{11} + \text{H}_2\text{O} \rightarrow \text{C}_6\text{H}_{12}\text{O}_6 + \text{C}_6\text{H}_{12}\text{O}_6\)
In simple words: Bimolecular reactions that act like first-order reactions because water or another reactant is in excess.

Exam Tip: Give both definition and equation for complete credit.

Question 10. What is half life period? Derive an expression for half-life period in case of a first order reaction.
Answer: The time in which the concentration of a reactant is reduced to half of its initial concentration is called half-life (\(t_{1/2}\)).
Derivation:
For a first-order reaction:
\(k = \frac{2.303}{t} \log\left(\frac{[R]_0}{[R]}\right)\)
At \(t = t_{1/2}\), \([R] = \frac{[R]_0}{2}\)
\(k = \frac{2.303}{t_{1/2}} \log\left(\frac{[R]_0}{[R]_0 / 2}\right)\)
\(k = \frac{2.303}{t_{1/2}} \log(2) = \frac{2.303 \times 0.3010}{t_{1/2}} = \frac{0.693}{t_{1/2}}\)
\(t_{1/2} = \frac{0.693}{k}\)
In simple words: By substituting half concentration into the integrated rate law, we find that half-life is independent of initial concentration.

Exam Tip: Memorize the final formula \(t_{1/2} = \frac{0.693}{k}\) for numerical problems.

3 MARKS QUESTION

Question 1. The rate constant for a first order reaction is \(60\text{ s}^{-1}\). How much time will it take to reduce the concentration of the reaction to \(1/10\) of its initial value?
Answer: Given:\br /> \(k = 60\text{ s}^{-1}\)
\([R] = \frac{[R]_0}{10}\)
Using first order rate equation:
\(t = \frac{2.303}{k} \log\left(\frac{[R]_0}{[R]}\right)\)
\(t = \frac{2.303}{60} \log\left(\frac{[R]_0}{[R]_0 / 10}\right)\)
\(t = \frac{2.303}{60} \log(10) = \frac{2.303}{60} \times 1 = 3.84 \times 10^{-2}\text{ s}\)
In simple words: Apply the logarithmic formula relating time, rate constant, and concentration ratio.

Exam Tip: \(\log(10) = 1\), which simplifies calculations significantly.

Question 2. The rate of most of reactions double when their temperature is raised from 298 K to 308 K. Calculate the activation energy of such a reaction.
Answer: Given:\br /> \(T_1 = 298\text{ K}, \quad T_2 = 308\text{ K}\)
\(\frac{k_2}{k_1} = 2\)
\(\log\left(\frac{k_2}{k_1}\right) = \frac{E_a}{2.303 R} \left(\frac{T_2 - T_1}{T_1 T_2}\right)\)
\(\log(2) = \frac{E_a}{2.303 \times 8.314} \left(\frac{308 - 298}{298 \times 308}\right)\)
\(0.3010 = \frac{E_a}{19.147} \times \frac{10}{91784}\)
\(E_a = \frac{0.3010 \times 19.147 \times 91784}{10} = 52.89\text{ kJ mol}^{-1}\)
In simple words: Substitute doubling of rate (\(k_2/k_1 = 2\)) into the Arrhenius temperature equation to solve for \(E_a\).

Exam Tip: Standard value of \(R = 8.314\text{ J K}^{-1}\text{mol}^{-1}\).

Question 3. A first order reaction takes 69.3 min for 50% completion. Set up an equation for determining the time needed for 80% completion.
Answer: For 50% completion (\(t_{1/2}\) = 69.3 min):
\(k = \frac{0.693}{t_{1/2}} = \frac{0.693}{69.3\text{ min}} = 10^{-2}\text{ min}^{-1}\)
For 80% completion, remaining concentration \([R] = [R]_0 - 0.80[R]_0 = 0.20[R]_0 = \frac{[R]_0}{5}\)
\(t = \frac{2.303}{k} \log\left(\frac{[R]_0}{[R]}\right)\)
\(t = \frac{2.303}{10^{-2}} \log(5) = 230.3 \times 0.6990 = 160.9\text{ min}\)
In simple words: First find \(k\) from the half-life, then use it to find the time required for 80% reaction.

Exam Tip: Remember that 80% complete means 20% remains.

Question 4. Following reaction takes place in one step: \(2\text{NO} + \text{O}_2 \rightarrow 2\text{NO}_2\). How will the rate of the reaction change if the volume of reaction vessel is diminished to \(1/3\) of its original volume? Will there be any change in the order of reaction with reduced volume?
Answer: Rate law for the single-step reaction: Rate \(= k[\text{NO}]^2[\text{O}_2]^1\)
When the volume is diminished to \(1/3\), the concentration of each reactant becomes 3 times the original concentration.
New rate \(= k[3\text{NO}]^2[3\text{O}_2]^1 = k \cdot 9[\text{NO}]^2 \cdot 3[\text{O}_2] = 27 \times (\text{Initial rate})\)
The rate of reaction increases 27 times.
There will be no change in the order of the reaction with the reduced volume, as order is an intrinsic characteristic property.
In simple words: Reducing volume triples concentration, and since order is 3, rate increases by \(3^3 = 27\) times. Order remains unchanged.

Exam Tip: Concentration is inversely proportional to volume.

Question 5. The decomposition of \(\text{NH}_3\) on a platinum surface is a zero-order reaction. What are the rate of production of \(\text{N}_2\) and \(\text{H}_2\) if \(k = 2.5 \times 10^{-4}\text{ mol L}^{-1}\text{s}^{-1}\)?
Answer: Reaction: \(2\text{NH}_3 \rightarrow \text{N}_2 + 3\text{H}_2\)
Rate \(= -\frac{1}{2} \frac{d[\text{NH}_3]}{dt} = \frac{d[\text{N}_2]}{dt} = \frac{1}{3} \frac{d[\text{H}_2]}{dt} = k[\text{NH}_3]^0 = k = 2.5 \times 10^{-4}\text{ mol L}^{-1}\text{s}^{-1}\)
Rate of production of \(\text{N}_2\):
\(\frac{d[\text{N}_2]}{dt} = k = 2.5 \times 10^{-4}\text{ mol L}^{-1}\text{s}^{-1}\)
Rate of production of \(\text{H}_2\):
\(\frac{d[\text{H}_2]}{dt} = 3 \times k = 3 \times 2.5 \times 10^{-4} = 7.5 \times 10^{-4}\text{ mol L}^{-1}\text{s}^{-1}\)
In simple words: For zero-order reactions, the rate equals \(k\). Multiply by stoichiometric coefficients to find individual rates of production.

Exam Tip: Stoichiometric coefficients divide rate expressions but multiply rates of production/consumption.

Question 6. How is the rapid change in concentration of reactants/products monitored for fast reactions?
Answer: Fast reactions are monitored using specialized physical methods such as spectroscopic methods (stopped-flow technique), relaxation methods, electrochemical methods, or nuclear magnetic resonance (NMR).
In simple words: Specialized instruments that measure rapid changes in light absorption or electrical properties are used.

Exam Tip: Mention "stopped-flow technique" or "spectroscopic methods" as key buzzwords.

Question 7. What are photochemical reactions? Give two examples.
Answer: Chemical reactions that proceed by the absorption of light (ultraviolet or visible) are called photochemical reactions.
Examples:
1. Photosynthesis: \(6\text{CO}_2 + 6\text{H}_2\text{O} \xrightarrow{\text{light}} \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2\)
2. Combination of hydrogen and chlorine: \(\text{H}_2 + \text{Cl}_2 \xrightarrow{\text{light}} 2\text{HCl}\)
In simple words: Reactions triggered or driven by light energy.

Exam Tip: Always include light or \(h\nu\) in photochemical equations.

Question 8. What is the effect of temperature on the rate of reaction? Explain giving reasons.
Answer: The rate of reaction generally increases with an increase in temperature (often doubling for every \(10^\circ\text{C}\) rise). This is because higher temperature increases the kinetic energy of molecules, leading to a greater fraction of molecules possessing energy equal to or greater than the threshold energy, thereby increasing the frequency of effective collisions.
In simple words: Heating gives molecules more speed and energy, causing more successful collisions per second.

Exam Tip: Mention both kinetic energy increase and successful collision frequency.

Question 9. Comment on free energy change of 'photochemical reactions'.
Answer: For photochemical reactions, the Gibbs free energy change (\(\Delta G\)) is negative (\(\Delta G < 0\)), making them thermodynamically feasible because they absorb photon energy to overcome activation energy barriers.
In simple words: They are spontaneous because light energy assists them in overcoming the energy barrier.

Question 10. State the role of activated complex in a reaction and state its relation with activation energy.
Answer: An activated complex is an unstable, high-energy intermediate state formed during the transition from reactants to products at the peak of the potential energy barrier. The energy required to form this activated complex from the reactants is called the activation energy.
In simple words: It is the unstable transition state at the top of the energy hill, and activation energy is the height of that hill.

Exam Tip: Mention the potential energy peak to define the activated complex clearly.

QUESTIONS CARRYING 5 MARKS

Question 1. What do you understand by the rate of a reaction? How is it expressed? How is the rate of reaction determined?
Answer:
- Definition: The change in concentration of any reactant or product per unit time.
- Expression: For \(\text{A} \rightarrow \text{B}\), Rate \(= -\frac{\Delta[\text{A}]}{\Delta t} = +\frac{\Delta[\text{B}]}{\Delta t}\). Units are \(\text{mol L}^{-1}\text{s}^{-1}\).
- Determination: It is determined by measuring the concentration of reactants or products at various regular time intervals using titration, spectrophotometry, or pressure measurements, and plotting a concentration-versus-time graph.
In simple words: Rate measures how fast reactants decrease or products increase, found by tracking concentration changes over time.

Exam Tip: Include mathematical expressions, units, and experimental methods to earn all 5 marks.

Question 2. What do you understand by order of a reaction? How does rate law differ from law of mass action? Give two examples of each of the reactions of (i) zero order (ii) first order (iii) second order.
Answer:
- Order: The sum of powers of concentration terms in the rate law.
- Rate Law vs. Law of Mass Action: Rate law is determined experimentally and uses actual powers from the rate-determining step, whereas the law of mass action is theoretical and uses stoichiometric coefficients.
- Examples:
(i) Zero order: \(2\text{NH}_3 \xrightarrow{\text{Pt}} \text{N}_2 + 3\text{H}_2\); Photochemical reaction \(\text{H}_2 + \text{Cl}_2 \xrightarrow{h\nu} 2\text{HCl}\)
(ii) First order: Decomposition of \(\text{N}_2\text{O}_5\); Decomposition of \(\text{H}_2\text{O}_2\)
(iii) Second order: \(2\text{HI} \rightarrow \text{H}_2 + \text{I}_2\); \(\text{H}_2 + \text{I}_2 \rightarrow 2\text{HI}\)
In simple words: Order comes from experiments (unlike mass action, which uses equation coefficients). Examples span zero, first, and second-order types.

Exam Tip: Clearly separate the comparison and the examples to ensure complete grading.

Question 3. Derive the equation for the rate constant for a first-order reaction. What would be the units of the first order rate constant if the concentration is expressed in mole per litre and time in seconds?
Answer:
Derivation:
Consider a first-order reaction \(R \rightarrow P\).
Rate \(= -\frac{d[R]}{dt} = k[R]\)
Rearranging: \(\frac{d[R]}{[R]} = -k \, dt\)
Integrating both sides from \(t = 0\) to \(t\):
\(\int_{[R]_0}^{[R]} \frac{d[R]}{[R]} = -k \int_0^t dt\)
\(\ln\left(\frac{[R]}{[R]_0}\right) = -kt\)
\(k = \frac{1}{t} \ln\left(\frac{[R]_0}{[R]}\right) = \frac{2.303}{t} \log\left(\frac{[R]_0}{[R]}\right)\)
Units: \(k = \text{s}^{-1}\).
In simple words: Integrate the differential rate equation using natural logarithms to get the integrated rate expression for first-order reactions.

Question 4. Explain why the rate of reaction increases with increase in temperature.
Answer: An increase in temperature increases the kinetic energy of reacting molecules. This results in an increase in both the collision frequency and the fraction of molecules possessing energy equal to or greater than the activation energy (\(E_a\)). According to collision theory, this leads to a significantly higher number of effective collisions per second, thereby increasing the rate of the reaction.
In simple words: Higher temperatures make molecules move faster and collide harder, allowing more of them to react successfully.

Exam Tip: Connect temperature rise to both kinetic energy and effective collision frequency.

Question 5. Briefly explain the effect of temperature on the rate constant of a reaction.
Answer: The effect of temperature on the rate constant is quantitatively expressed by the Arrhenius equation:
\(k = A e^{-E_a / RT}\)
As temperature increases, the term \(e^{-E_a / RT}\) increases, which causes the rate constant \(k\) to increase. Typically, for every \(10^\circ\text{C}\) rise in temperature, the rate constant approximately doubles.
In simple words: The rate constant grows exponentially with temperature as shown by the Arrhenius equation.

Question 1. The half-life period of two samples are 0.1 and 0.4 seconds. Their initial concentrations are 200 and \(50\text{ mol L}^{-1}\) respectively. What is the order of reaction?
Answer: Given:
\(t_{1/2}^{(1)} = 0.1\text{ s}, \quad [R]_1 = 200\text{ mol L}^{-1}\)
\(t_{1/2}^{(2)} = 0.4\text{ s}, \quad [R]_2 = 50\text{ mol L}^{-1}\)
General half-life formula: \(t_{1/2} \propto \frac{1}{[R]_0^{n-1}}\)
\(\frac{t_{1/2}^{(1)}}{t_{1/2}^{(2)}} = \left(\frac{[R]_2}{[R]_1}\right)^{n-1}\)
\(\frac{0.1}{0.4} = \left(\frac{50}{200}\right)^{n-1}\)
\(\frac{1}{4} = \left(\frac{1}{4}\right)^{n-1}\)
Comparing powers: \(n - 1 = 1 \implies n = 2\)
The order of the reaction is 2.
In simple words: Use the proportionality between half-life and initial concentration for general-order reactions to solve for \(n\).

Exam Tip: Memorize the half-life dependence formula \(t_{1/2} \propto [R]_0^{1-n}\).

Question 2. What is the ratio of \(t_{3/4} : t_{1/2}\) for a first-order reaction?
Answer:
For \(t_{1/2}\) (50% complete):
\(t_{1/2} = \frac{2.303}{k} \log\left(\frac{100}{50}\right) = \frac{2.303}{k} \log(2)\)
For \(t_{3/4}\) (75% complete, 25% remaining):
\(t_{3/4} = \frac{2.303}{k} \log\left(\frac{100}{25}\right) = \frac{2.303}{k} \log(4) = \frac{2.303}{k} \times 2\log(2) = 2 \times t_{1/2}\)
Ratio \(t_{3/4} : t_{1/2} = 2 : 1\)
In simple words: Taking 75% completion takes twice as long as the half-life for first-order reactions.

Question 3. Higher molecularity reactions (viz. molecularity 4 and above) are very rare. Why?
Answer: Higher molecularity reactions are rare because the probability of simultaneous effective collisions among four or more reacting molecules with correct orientation and sufficient energy is extremely low.
In simple words: It is nearly impossible for multiple molecules to collide at the exact same time and place with proper alignment.

Exam Tip: Mention the low probability of simultaneous multi-particle collisions.

Question 4. Consider the reaction \(2\text{A} + \text{B} \rightarrow \text{Products}\). When concentration of B alone was doubled, half-life time does not change. When concentration of A alone is doubled, the rate increases by two times. What is the unit of \(k\) and what is the order of the reaction?
Answer:
- When concentration of B is doubled, half-life does not change, meaning the order with respect to B is 1 (\(n_B = 1\)).
- When concentration of A is doubled, rate doubles, meaning the order with respect to A is 1 (\(n_A = 1\)).
Overall order \(= 1 + 1 = 2\) (Second order).
Unit of rate constant for a second-order reaction: \(\text{mol}^{-1}\text{L s}^{-1}\).
In simple words: Find individual orders from concentration effects, add them for overall order, and write the corresponding units.

Question 5. For a reaction, the energy of activation is \(75\text{ kJ mol}^{-1}\). When the energy of activation of a catalyst is lowered to \(20\text{ kJ mol}^{-1}\), what is the effect of catalyst on the rate of reaction at \(20^\circ\text{C}\)?
Answer:
Difference in activation energy \(\Delta E_a = 75 - 20 = 55\text{ kJ mol}^{-1} = 55000\text{ J mol}^{-1}\)
Using Arrhenius relation \(\frac{k_{\text{cat}}}{k_{\text{uncat}}} = e^{\Delta E_a / RT}\)
\(T = 20^\circ\text{C} = 293\text{ K}\)
\(\log\left(\frac{k_{\text{cat}}}{k_{\text{uncat}}}\right) = \frac{55000}{2.303 \times 8.314 \times 293} = 9.81\)
\(\frac{k_{\text{cat}}}{k_{\text{uncat}}} = 10^{9.81} \approx 6.4 \times 10^9\) times.
The rate of reaction increases enormously by a factor of approximately \(6.4 \times 10^9\).
In simple words: Lowering activation energy with a catalyst exponentially speeds up the reaction.

Exam Tip: Catalysts speed up reactions by lowering activation energy.

Question 6. The gas phase decomposition of \(\text{CH}_3\text{OCH}_3\) follows first order of kinetics: \(\text{CH}_3\text{OCH}_3 \rightarrow \text{CH}_4(\text{g}) + \text{H}_2(\text{g}) + \text{CO}(\text{g})\). The reaction is carried out at a constant volume of the container at \(500^\circ\text{C}\) and has \(t_{1/2} = 14.5\text{ min}\). Initially only dimethyl ether is present at a pressure of \(0.40\text{ atm}\). What is the total pressure of the system after 12 min? Assume ideal behavior.
Answer:
\(k = \frac{0.693}{t_{1/2}} = \frac{0.693}{14.5} = 0.0478\text{ min}^{-1}\)
Reaction table (in terms of pressure):
Initial (\(t = 0\)): \(P_{\text{CH}_3\text{OCH}_3} = 0.40\text{ atm}, \quad P_{\text{CH}_4} = 0, \quad P_{\text{H}_2} = 0, \quad P_{\text{CO}} = 0\)
After \(t = 12\text{ min}\): \(P_{\text{CH}_3\text{OCH}_3} = 0.40 - x, \quad P_{\text{CH}_4} = x, \quad P_{\text{H}_2} = x, \quad P_{\text{CO}} = x\)
Total pressure \(P_t = (0.40 - x) + x + x + x = 0.40 + 2x\)
Also, using first-order rate law for pressure:
\(x = P_0 (1 - e^{-kt}) = 0.40(1 - e^{-0.0478 \times 12}) = 0.40(1 - e^{-0.5736}) = 0.40(1 - 0.5635) = 0.40 \times 0.4365 = 0.1746\text{ atm}\)
Total pressure \(P_t = 0.40 + 2(0.1746) = 0.40 + 0.3492 = 0.7492\text{ atm}\)
In simple words: Calculate pressure changes using gas stoichiometry and the first-order rate equation to find total system pressure.

Exam Tip: Set up pressure tables carefully for gaseous first-order reactions.

Question 7. See the graph and answer the following questions: (1) What is the order of reaction? (2) What is the value of a and b? (Graph: Log [R] vs t, downward sloping line)
Answer:
(1) Order of reaction = 1 (First order, since \(\log[R]\) vs \(t\) is a straight line).
(2) (a) Intercept on y-axis = \(\log[R]_0\)
(b) Slope \(= -\frac{k}{2.303}\)
In simple words: A linear plot of log concentration versus time indicates first-order kinetics with a negative slope.

Exam Tip: Remember slope for \(\log[R]\) vs \(t\) is \(-\frac{k}{2.303}\).

Question 8. (1) What is the order of reaction? (2) What is the value of slope and intercept? (Graph: [R] vs t, downward sloping straight line)
Answer:
(1) Order of reaction = 0 (Zero order).
(2) Intercept = Initial concentration \([R]_0\)
Slope = \(-k\)
In simple words: A straight line in a direct concentration versus time graph represents zero-order kinetics.

Exam Tip: Zero-order graphs have a constant negative slope equal to \(-k\).

Question 9. (1) What is the value of slope? (Graph: [R] vs \(\log A\) or similar logarithmic concentration plot versus time)
Answer:
Slope \(= -\frac{k}{2.303}\)
In simple words: Logarithmic concentration plots always incorporate the conversion factor \(2.303\) in their slopes.

Exam Tip: Always include the negative sign for decomposition reaction slopes.