CBSE Class 12 Chemistry Chemical Kinetics Questions

CHEMICAL KINETICS

VSA QUESTIONS (1 - MARK QUESTIONS)

Question 1. Define the term 'rate of reaction'.
Answer: The rate of 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.

Exam Tip: Mention both decrease in reactant concentration or increase in product concentration with time to secure full marks.

Question 2. Mention the units of rate of reaction.
Answer: The unit of rate of reaction is \(\text{mol L}^{-1} \text{s}^{-1}\) (or atm \(\text{s}^{-1}\) for gaseous reactions).
In simple words: Concentration divided by time.

Exam Tip: Always write units with negative exponents properly, such as \(\text{mol L}^{-1}\text{s}^{-1}\).

Question 3. Express the rate of reaction in terms of \(\text{Br}^-(\text{aq})\) as reactant and \(\text{Br}_2(\text{aq})\) as product for the reaction : \(5\text{Br}^-(\text{aq}) + \text{BrO}_3^-(\text{aq}) + 6\text{H}^+(\text{aq}) \rightarrow 3\text{Br}_2(\text{aq}) + 3\text{H}_2\text{O}(\text{l})\)
Answer: Rate = \(-\frac{1}{5} \frac{\Delta[\text{Br}^-]}{\Delta t} = +\frac{1}{3} \frac{\Delta[\text{Br}_2]}{\Delta t}\)
In simple words: Divide the change in concentration by the stoichiometric coefficient in the balanced equation.

Exam Tip: Remember to put a negative sign for reactants and positive for products, along with reciprocal coefficients.

Question 4. For a chemical reaction represented by \(\text{R} \rightarrow \text{P}\) the rate of reaction is denoted by \(-\frac{\Delta[\text{R}]}{\Delta t}\) or \(+\frac{\Delta[\text{P}]}{\Delta t}\). Why a positive sign (+) is placed before \(\frac{\Delta[\text{P}]}{\Delta t}\) and negative sign (-) before \(\frac{\Delta[\text{R}]}{\Delta t}\)?
Answer: The concentration of reactants decreases with time, making \(\Delta[\text{R}]\) negative. A negative sign is multiplied to make the overall rate positive. Conversely, product concentration increases with time, so a positive sign is used.
In simple words: Reaction rates are always expressed as positive quantities.

Exam Tip: Rates of reactions can never be negative; hence mathematical signs are adjusted accordingly.

Question 5. Express the rate of reaction in terms of disappearance of hydrogen and appearance of ammonia in the given reaction. \(\text{N}_2(\text{g}) + 3\text{H}_2(\text{g}) \rightarrow 2\text{NH}_3(\text{g})\)
Answer: Rate = \(-\frac{1}{3} \frac{\Delta[\text{H}_2]}{\Delta t} = +\frac{1}{2} \frac{\Delta[\text{NH}_3]}{\Delta t}\)
In simple words: Account for the stoichiometric coefficients 3 for hydrogen and 2 for ammonia.

Question 6. Why rate of reaction does not remain constant throughout?
Answer: The rate of a reaction depends on the concentration of reactants, which continuously decreases as the reaction proceeds.
In simple words: As reactants get consumed, fewer collisions occur, slowing down the reaction.

Exam Tip: Connect reaction rate directly to reactant concentration using rate law.

Question 7. Write the unit of first order rate constant of a gaseous reaction if the partial pressure of gaseous reactant is given in bar.
Answer: The unit is \(\text{s}^{-1}\) (or \(\text{bar}^{-1}\text{s}^{-1}\) for zero order, etc., but for first order, units are independent of pressure units and depend only on time).
In simple words: First order rate constants always have units of \((\text{time})^{-1}\).

Exam Tip: First order units are independent of concentration or pressure terms.

Question 8. For a zero order reaction \(\text{R} \rightarrow \text{P}\), the change in concentration of reactant w.r.t. time is shown by following graph. (Linear straight line sloping downwards with negative slope). What is the slope of this line?
Answer: Slope = \(-k\)
In simple words: The steepness of the concentration-time graph equals the negative of the rate constant.

Exam Tip: For zero order integrated rate equation, \([\text{R}] = -kt + [\text{R}]_0\), matching \(y = mx + c\).

Question 9. What will be the order of reaction, if the rate of reaction does not depend on the concentration of any of the reactant.
Answer: Zero order.
In simple words: Rate = \(k[\text{R}]^0 = k\), so order is zero.

Exam Tip: Power of concentration terms in rate law determines the overall order.

Question 10. For the elementary step of a chemical reaction : \(\text{H}_2 + \text{I}_2 \rightarrow 2\text{HI}\) with rate of reaction \(\alpha [\text{H}_2][\text{I}_2]\). What is the (i) molecularity and (ii) order of the reaction.
Answer: (i) Molecularity = 2
(ii) Order = 1 + 1 = 2
In simple words: Two molecules collide in this elementary step, making both molecularity and order equal to 2.

Exam Tip: For elementary reactions, order equals molecularity.

Question 11. For a chemical reaction \(\text{A} \rightarrow \text{B}\). The rate of the reaction is given as Rate = \(k[\text{A}]^n\), the rate of the above reaction quadruples when the concentration of A is doubled. What is the value of n?
Answer: Rate\(_1 = k[\text{A}]^n\)
Rate\(_2 = k[2\text{A}]^n = 4 \times \text{Rate}_1\)
\(2^n = 4 \Rightarrow 2^n = 2^2 \Rightarrow n = 2\)
In simple words: Doubling concentration increases rate by \(2^2 = 4\), so order is 2.

Exam Tip: Use ratio methods for quick calculation in objective questions.

Question 12. Mention one example of zero order reaction.
Answer: Decomposition of gaseous ammonia on a hot platinum surface at high pressure: \(2\text{NH}_3(\text{g}) \xrightarrow{\text{Pt}} \text{N}_2(\text{g}) + 3\text{H}_2(\text{g})\).
In simple words: Surface catalysis reactions are often zero order because active sites are saturated.

Question 13. What is the value of the order of reaction of radioactive decay?
Answer: First order.
In simple words: Radioactive disintegration is independent of external conditions and follows first order kinetics.

Question 14. Express the relation between the half-life period of a reactant and initial concentration for a reaction of \(n^\text{th}\) order.
Answer: \(t_{1/2} \propto \frac{1}{[\text{A}]_0^{n-1}}\)
In simple words: Half-life depends inversely on the initial concentration raised to the power \(n-1\).

Exam Tip: Memorize this general formula for quick competitive exam problem-solving.

Question 15. A reaction is 50% complete in 2 hours and 75% complete in 4 hours. What is the order of reaction?
Answer: First order.
In simple words: Since 75% completion takes exactly two half-lives (\(2 \times 2\) hours = 4 hours), it follows first order kinetics.

Exam Tip: Successive half-lives remain constant for first order reactions.

Question 16. Suggest an appropriate reason for the observation : "On increasing temperature of the reacting system by 10 degrees, the rate of reaction almost doubles or even sometimes becomes five folds."
Answer: With a 10-degree rise in temperature, the fraction of molecules having energy equal to or greater than activation energy (threshold energy) nearly doubles, increasing effective collisions.
In simple words: Higher temperature gives more molecules the minimum energy needed to react.

Exam Tip: Mention the doubling of effective collisions or activation energy barrier crossing.

Question 17. For a chemical reaction, activation energy is zero and at 300K rate constant is \(5.9 \times 10^{-5}\text{ s}^{-1}\), what will be the rate constant at 400K?
Answer: \(5.9 \times 10^{-5}\text{ s}^{-1}\)
In simple words: Since \(E_a = 0\), temperature has no effect on the rate constant according to Arrhenius equation.

Exam Tip: Look out for zero activation energy problems where \(k_1 = k_2\).

Question 18. Two reactions occurring at the same temperature have identical values of \(E_a\). Does this ensure that also they will have the same rate constant? Explain.
Answer: No, because rate constant also depends on the pre-exponential factor (\(A\)) or frequency factor, which varies with the nature of reactants.
In simple words: Activation energy is only one part of the Arrhenius equation; the frequency factor A also matters.

Exam Tip: Reference the Arrhenius equation \(k = Ae^{-E_a/RT}\).

Question 19. The rate constant of a reaction is given by the expression \(k = Ae^{-E_a/RT}\). Which factor in this expression should register a decrease so that the reaction proceeds rapidly?
Answer: Activation energy (\(E_a\)).
In simple words: Lowering activation energy reduces the energy barrier, making reactions faster.

Question 20. For a chemical reaction rate constant \(k = 5.3 \times 10^{-4}\text{ mol L}^{-1}\text{s}^{-1}\), what will be the order of reaction?
Answer: Zero order.
In simple words: The unit \(\text{mol L}^{-1}\text{s}^{-1}\) corresponds strictly to zero order rate constants.

Exam Tip: Learn how to deduce order directly from rate constant units: \(\text{mol}^{1-n}\text{L}^{n-1}\text{s}^{-1}\).

Question 21. Write the rate law and order for the following reaction :
\(\text{AB}_2 + \text{C}_2 \rightarrow \text{AB}_2\text{C} + \text{C}\) (slow)
\(\text{AB}_2 + \text{C} \rightarrow \text{AB}_2\text{C}\) (fast)
Answer: Rate = \(k[\text{AB}_2][\text{C}_2]\)
Order = 1 + 1 = 2
In simple words: The slow step is the rate-determining step, and its reactants determine the rate law.

Exam Tip: Always write rate laws based on the slowest (rate-determining) elementary step.

SA (I) TYPE QUESTIONS (2 - MARKS QUESTIONS)

Question 22. List four factors which affect the rate of a chemical reaction. State how each of these factors changes the reaction rate.
Answer: 1. Concentration of reactants: Rate increases with an increase in reactant concentration.
2. Temperature: Rate increases with an increase in temperature.
3. Catalyst: Presence of a positive catalyst increases the reaction rate by lowering activation energy.
4. Surface area of reactants: Greater surface area increases the rate of reaction.
In simple words: More reactant concentration, higher temperature, catalysts, and larger surface areas speed up reactions.

Exam Tip: State the factor and clearly mention the direct or indirect effect on rate.

Question 23. Differentiate between
(a) Average rate and instantaneous rate of a chemical reaction.
(b) Rate of a reaction and specific rate of reaction, i.e., rate constant.
Answer: (a) Average rate is measured over a long time interval (\(\Delta t\)), whereas instantaneous rate is the rate at a specific, given instant of time (\(dt \rightarrow 0\)).
(b) Rate of reaction is the speed at which reactants convert to products at any moment (depends on concentration), whereas specific rate (rate constant, \(k\)) is the rate of reaction when concentration of each reactant is unity.
In simple words: Average is over a period; instantaneous is at a precise moment. Rate constant is a constant speed value at unit concentration.

Exam Tip: Use clear tabular or point-wise differences for full credit.

Question 24. The rate law for the reaction : \(\text{A} + \text{B} \rightarrow \text{P}\) is given by Rate = \(k[\text{A}]^n[\text{B}]^m\). On doubling the concentration of A and reducing the concentration of B to half of its original concentration, calculate the ratio of the new rate to the previous rate of reaction.
Answer: Initial rate (\(r_1\)) = \(k[\text{A}]^n[\text{B}]^m\)
New rate (\(r_2\)) = \(k(2\text{A})^n\left(\frac{1}{2}\text{B}\right)^m = k \cdot 2^n \cdot \left(\frac{1}{2}\right)^m [\text{A}]^n[\text{B}]^m\)
Ratio \(\frac{r_2}{r_1} = \frac{2^n \cdot 2^{-m}[\text{A}]^n[\text{B}]^m}{[\text{A}]^n[\text{B}]^m} = 2^{n-m}\)
In simple words: Substitute modified concentrations into the rate law and divide by the original rate expression.

Exam Tip: Show substitution steps clearly to earn all 2 marks.

Question 25. For the reaction in a closed vessel : \(2\text{NO}(\text{g}) + \text{O}_2(\text{g}) \rightarrow 2\text{NO}_2(\text{g})\); Rate = \(k[\text{NO}]^2[\text{O}_2]\). If the volume of the reaction vessel is doubled, how would it affect the rate of the reaction?
Answer: When volume is doubled, concentration becomes halved (\([\text{NO}] \rightarrow \frac{1}{2}[\text{NO}]\) and \([\text{O}_2] \rightarrow \frac{1}{2}[\text{O}_2]\)).
New Rate = \(k\left(\frac{1}{2}[\text{NO}]\right)^2 \left(\frac{1}{2}[\text{O}_2]\right) = k \cdot \frac{1}{4}[\text{NO}]^2 \cdot \frac{1}{2}[\text{O}_2] = \frac{1}{8} \times \text{Initial Rate}\)
In simple words: Doubling volume halves concentrations, reducing the rate to \(1/8^\text{th}\) of its initial value.

Exam Tip: Remember that concentration is inversely proportional to volume (\(C = n/V\)).

Question 26. Explain with an example, what is a pseudo first order reaction?
Answer: Reactions that are bimolecular or higher order in reality, but follow first-order kinetics because one reactant is present in large excess, are called pseudo first-order reactions.
Example: Acid hydrolysis of ethyl acetate: \(\text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O} \xrightarrow{\text{H}^+} \text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH}\). Here, water is in large excess, so Rate = \(k'[\text{CH}_3\text{COOC}_2\text{H}_5]\).
In simple words: A reaction that looks like higher order but behaves like first order because one reactant's concentration doesn't change noticeably.

Exam Tip: Always include the hydrolysis of ester or inversion of cane sugar as standard examples.

Question 27. Show that time required for 99.9% completion of the first order reaction is 10 times of \(t_{1/2}\) for first order chemical reaction.
Answer: For a first order reaction, \(t = \frac{2.303}{k} \log \frac{[\text{R}]_0}{[\text{R}]}\)
When reaction is 99.9% complete, \([\text{R}] = [\text{R}]_0 - 0.999[\text{R}]_0 = 0.001[\text{R}]_0\)
\(t_{99.9\%} = \frac{2.303}{k} \log \frac{[\text{R}]_0}{0.001[\text{R}]_0} = \frac{2.303}{k} \log 1000 = \frac{2.303 \times 3}{k}\)
We know \(t_{1/2} = \frac{0.693}{k}\)
\(\frac{t_{99.9\%}}{t_{1/2}} = \frac{2.303 \times 3 / k}{0.693 / k} = \frac{6.909}{0.693} \approx 10\)
Therefore, \(t_{99.9\%} = 10 \times t_{1/2}\).
In simple words: Substituting percentage values into the integrated rate equation proves the 10-fold relationship.

Exam Tip: Memorize \(t_{99\%} = 2 \times t_{90\%}\) and \(t_{99.9\%} = 10 \times t_{1/2}\) for numerical shortcuts.

Question 28. The graphs (1 and 2) given below are plots of rate of reaction verses concentration of the reaction. Predict the order from the graphs. (Graph 1: straight line passing through origin; Graph 2: straight line parallel to concentration axis).
Answer: Graph 1: First order (Rate \(\propto\) concentration).
Graph 2: Zero order (Rate is independent of concentration).
In simple words: Sloping line means rate changes with concentration (First Order); flat horizontal line means rate is constant (Zero Order).

Exam Tip: Match graph axes (Rate vs Concentration) carefully before answering.

Question 29. (a) For a reaction \(\text{A} + \text{B} \rightarrow \text{Products}\), the rate law is given by \(r = k[\text{A}]^{1/2}[\text{B}]^2\). What is the order of reaction?
(b) The conversion of molecules X to Y follows second order kinetics. If concentration of X is increased to three times, how will it affect the rate of formation of Y?
Answer: (a) Overall order = \(\frac{1}{2} + 2 = \frac{5}{2}\)
(b) Rate \(= k[\text{X}]^2\). New rate \(= k(3\text{X})^2 = 9k[\text{X}]^2\). It increases by 9 times.
In simple words: Add the exponents for part (a). Square the multiplier 3 for part (b) to get 9 times.

Exam Tip: Overall order is simply the arithmetic sum of powers in the rate law expression.

SA (II) TYPE QUESTIONS (3 - MARK QUESTIONS)

Question 31. What is meant by zero order reaction? Derive an integrated rate equation for a zero order reaction.
Answer: A zero order reaction is one in which the rate of reaction is independent of the concentration of reactants.
Derivation:
For a zero order reaction \(\text{R} \rightarrow \text{P}\):
Rate = \(-\frac{d[\text{R}]}{dt} = k[\text{R}]^0 = k\)
\(d[\text{R}] = -kd t\)
Integrating both sides between limits \(t = 0\) to \(t = t\) and concentration \([\text{R}]_0\) to \([\text{R}]_t\):
\(\int_{[\text{R}]_0}^{[\text{R}]_t} d[\text{R}] = -k \int_0^t dt\)
\([\text{R}]_t - [\text{R}]_0 = -kt\)
\(k = \frac{[\text{R}]_0 - [\text{R}]_t}{t}\)
In simple words: Integration of constant rate gives a linear equation showing decrease in reactant concentration over time.

Exam Tip: Show integration limits and final formula clearly to secure all 3 marks.

Question 32. (a) Write two points of difference between order of reaction and molecularity of a reaction.
(b) Write one point of difference between rate of reaction and rate constant.
Answer: (a) Differences between order and molecularity:
1. Order is an experimental quantity, whereas molecularity is a theoretical concept.
2. Order can be zero, fractional, or negative, whereas molecularity can only be a whole number (non-zero positive integer).
(b) Difference between rate of reaction and rate constant: Rate of reaction depends on the concentration of reactants at any given time, whereas rate constant is a fixed proportionality constant representing the rate when reactant concentrations are unity.
In simple words: Order is measured experimentally and can be fractional; molecularity counts colliding particles and must be a whole number.

Exam Tip: Use distinct comparison points for clarity.

Question 33. Draw a graph between fraction of molecules and kinetic energy of the reacting species for two different temperatures :
(a) Room temperature
(b) Temperature \(10^\circ\text{C}\) higher than the room temperature
(c) Indicate the fraction of additional molecules which react at \((t + 10)^\circ\text{C}\).
Answer:
1. Draw Maxwell-Boltzmann distribution curves with Fraction of molecules on y-axis and Kinetic energy on x-axis.
2. The peak shifts slightly to the right and broadens for the higher temperature \((T + 10^\circ\text{C})\).
3. The shaded area under the curve beyond the activation energy threshold \(E_a\) represents the fraction of additional reacting molecules.
In simple words: Higher temperature flattens and shifts the distribution peak, showing more molecules have threshold energy.

Exam Tip: Label axes correctly (Fraction of molecules vs Kinetic energy) and mark \(E_a\) on the x-axis.

LONG ANSWER TYPE QUESTIONS (5 - MARK - QUESTIONS)

Question 34. (a) A chemical reaction is of second order w.r.t. a reactant. How will the rate of reaction be affected if the concentration of this reactant is : (a) doubled; (b) reduced to \(1/8^\text{th}\).
(b) For the reaction \(2\text{NO}(\text{g}) + \text{Cl}_2(\text{g}) \rightarrow 2\text{NOCl}(\text{g})\), data collected at 263K:
Exp 1: \([\text{NO}] = 0.15, [\text{Cl}_2] = 0.15, \text{Rate} = 0.60\)
Exp 2: \([\text{NO}] = 0.15, [\text{Cl}_2] = 0.30, \text{Rate} = 1.20\)
Exp 3: \([\text{NO}] = 0.30, [\text{Cl}_2] = 0.15, \text{Rate} = 2.40\)
Exp 4: \([\text{NO}] = 0.25, [\text{Cl}_2] = 0.25, \text{Rate} = ?\)
(i) Write expression for rate law.
(ii) Calculate value of rate constant and specify its units.
(iii) What is initial rate of disappearance of \(\text{Cl}_2\) in exp. 4?
Answer: (a) (a) Rate increases by 4 times (\(2^2 = 4\)).
(b) Rate becomes \(\left(\frac{1}{8}\right)^2 = \frac{1}{64}\) of initial value.
(b) (i) Comparing Exp 1 and 3: Doubling \([\text{NO}]\) quadruples rate, so order w.r.t. NO is 2.
Comparing Exp 1 and 2: Doubling \([\text{Cl}_2]\) doubles rate, so order w.r.t. \(\text{Cl}_2\) is 1.
Rate law = \(k[\text{NO}]^2[\text{Cl}_2]\)
(ii) Using data from Exp 1: \(0.60 = k(0.15)^2(0.15)\)
\(k = \frac{0.60}{0.0225 \times 0.15} = 177.7 \text{ L}^2\text{mol}^{-2}\text{min}^{-1}\)
(iii) Rate for Exp 4 = \(177.7 \times (0.25)^2 \times 0.25 = 177.7 \times 0.015625 = 2.7765 \text{ M min}^{-1}\).
In simple words: Determine orders by comparing experiments, calculate \(k\), and substitute values for experiment 4.

Exam Tip: Clearly write units for rate constant based on overall third order kinetics.

Question 35. (a) Draw a plot between \(\log k\) and reciprocal of absolute temperature (\(1/T\)).
(b) The energy of activation for a chemical reaction is \(100\text{ kJ mol}^{-1}\). Presence of a catalyst lowers the energy of activation by 75%. What will be effect on the rate of reaction at \(20^\circ\text{C}\), if other factors are equal?
Answer: (a) Plot a straight line graph with \(\log k\) on y-axis and \(1/T\) on x-axis. The slope is equal to \(-\frac{E_a}{2.303 R}\) and intercept is \(\log A\).
(b) Activation energy lowered by 75% means new activation energy \(E_{a2} = 25\% \text{ of } 100 = 25\text{ kJ mol}^{-1}\).
Decrease in activation energy = \(75\text{ kJ mol}^{-1}\).
Using Arrhenius equation, lowering \(E_a\) enormously increases the rate of reaction.
In simple words: The graph is a downward sloping straight line. Lowering activation energy speeds up the reaction significantly.

Exam Tip: Label slope and intercept clearly on the Arrhenius plot.

Question 36. (a) Derive the equation for rate constant of a first order reaction. What would be the units of the first order rate constant if concentration is expressed in moles per litre and time in seconds?
(b) For first order chemical reaction half-life period (\(t_{1/2}\)) is concentration independent. Justify the statement by using integrated rate equation.
Answer: (a) For a first order reaction \(\text{R} \rightarrow \text{P}\):
Rate = \(-\frac{d[\text{R}]}{dt} = k[\text{R}] \Rightarrow \frac{d[\text{R}]}{[\text{R}]} = -kd t\)
Integrating from \(t = 0\) to \(t\) and \([\text{R}]_0\) to \([\text{R}]_t\):
\(\ln\frac{[\text{R}]_t}{[\text{R}]_0} = -kt \Rightarrow k = \frac{2.303}{t} \log \frac{[\text{R}]_0}{[\text{R}]_t}\).
Unit of first order rate constant is \(\text{s}^{-1}\).
(b) For \(t_{1/2}\), \([\text{R}]_t = \frac{[\text{R}]_0}{2}\) at \(t = t_{1/2}\):
\(t_{1/2} = \frac{2.303}{k} \log \frac{[\text{R}]_0}{[\text{R}]_0 / 2} = \frac{2.303}{k} \log 2 = \frac{0.693}{k}\).
Since the initial concentration term \([\text{R}]_0\) cancels out, half-life is independent of initial concentration.
In simple words: Integration leads to a logarithmic rate constant formula, and substituting half-life conditions eliminates the initial concentration variable.

Exam Tip: This is a frequent 5-mark derivation; practice integration steps and log conversions thoroughly.

NUMERICALS

Question 37. The reaction \(\text{SO}_2\text{Cl}_2(\text{g}) \xrightarrow{k} \text{SO}_2(\text{g}) + \text{Cl}_2(\text{g})\) is a first order reaction with half life of \(3.15 \times 10^4\text{ s}\) at 575K. What percentage of \(\text{SO}_2\text{Cl}_2\) would be decomposed on heating at 575K for 90 min.
Answer: \(t_{1/2} = 3.15 \times 10^4\text{ s} \Rightarrow k = \frac{0.693}{3.15 \times 10^4} = 2.2 \times 10^{-5}\text{ s}^{-1}\)
Time \(t = 90\text{ min} = 90 \times 60 = 5400\text{ s}\)
\(t = \frac{2.303}{k} \log \frac{100}{100 - x}\)
\(5400 = \frac{2.303}{2.2 \times 10^{-5}} \log \frac{100}{100 - x}\)
Solving gives percentage decomposed \(x = 11.2\%\).
In simple words: Calculate rate constant from half-life, then use the first-order time formula to find remaining and decomposed percentages.

Exam Tip: Always convert minutes to seconds before substituting values into time formulas.

Question 38. A certain reaction is 50% complete in 20 min at 300K and the same reaction is again 50% complete in 5 min at 350K. Calculate the activation energy if it is a first order reaction. (\(R = 8.314\text{ J K}^{-1}\text{mol}^{-1}, \log 4 = 0.602\))
Answer: At \(T_1 = 300\text{K}, t_{1/2} = 20\text{ min} \Rightarrow k_1 = \frac{0.693}{20}\)
At \(T_2 = 350\text{K}, t_{1/2} = 5\text{ min} \Rightarrow k_2 = \frac{0.693}{5}\)
\(\log\frac{k_2}{k_1} = \frac{E_a}{2.303 R}\left(\frac{T_2 - T_1}{T_1 T_2}\right)\)
\(\log 4 = \frac{E_a}{2.303 \times 8.314} \left(\frac{350 - 300}{300 \times 350}\right)\)
\(0.602 = \frac{E_a}{19.147} \left(\frac{50}{105000}\right) \Rightarrow E_a = 24.206\text{ kJ mol}^{-1}\).
In simple words: Find rate constants from half-lives at two temperatures, then apply the Arrhenius temperature-difference formula.

Exam Tip: Since \(t_{1/2}\) is inversely proportional to \(k\), \(k_2/k_1 = t_1/t_2\).

Question 39. For a chemical reaction \(\text{A} \rightarrow \text{B}\), it was found that concentration of B increases by \(0.2\text{ mol L}^{-1}\) in half an hour. What is the average rate of reaction.
Answer: Average Rate = \(\frac{\Delta[\text{B}]}{\Delta t} = \frac{0.2\text{ mol L}^{-1}}{30\text{ min}} = 0.0066\text{ mol L}^{-1}\text{min}^{-1}\)
In simple words: Change in concentration divided by time in minutes.

Question 40. In the reaction \(\text{R} \rightarrow \text{P}\), the concentration of R decreases from 0.03M to 0.02M in 25 minutes. Calculate the average rate of reaction using unit of time both in minutes and seconds.
Answer: Rate (in min\(^{-1}\)) = \(-\frac{0.02 - 0.03}{25} = \frac{0.01}{25} = 4 \times 10^{-4}\text{ M min}^{-1}\)
Rate (in s\(^{-1}\)) = \(\frac{4 \times 10^{-4}}{60} = 6.66 \times 10^{-6}\text{ M s}^{-1}\).
In simple words: Divide concentration difference by time in minutes, then divide by 60 for seconds.

Exam Tip: Double-check decimal placements and scientific notation exponents.

Question 41. A first order reaction has a rate constant \(1.15 \times 10^{-3}\text{ s}^{-1}\). How long will 5g of this reactant take to reduce to 3g?
Answer: \(t = \frac{2.303}{k} \log \frac{[\text{R}]_0}{[\text{R}]} = \frac{2.303}{1.15 \times 10^{-3}} \log \frac{5}{3}\)
\(t = \frac{2.303}{1.15 \times 10^{-3}} (0.5229 - 0.4771) = \frac{2.303 \times 0.2218}{1.15 \times 10^{-3}} = 444\text{ s}\).
In simple words: Substitute initial weight, final weight, and rate constant into the first order time formula.

Exam Tip: Log values like \(\log 5 - \log 3\) make quick work of these calculations.

Question 42. The rate of reaction triples when the temperature changes from \(20^\circ\text{C}\) to \(50^\circ\text{C}\). Calculate the energy of activation. (\(R = 8.314\text{ J K}^{-1}\text{mol}^{-1}, \log 3 = 0.48\))
Answer: \(T_1 = 293\text{K}, T_2 = 323\text{K}, k_2/k_1 = 3\)
\(\log 3 = \frac{E_a}{2.303 \times 8.314} \left(\frac{323 - 293}{293 \times 323}\right)\)
\(0.48 = \frac{E_a}{19.147} \left(\frac{30}{94639}\right) \Rightarrow E_a = 12.59\text{ kJ mol}^{-1}\).
In simple words: Apply the Arrhenius equation with temperature and rate constant ratios.

Exam Tip: Always convert Celsius temperatures to Kelvin by adding 273.

Question 43. A hydrogenation reaction is carried out at 550 K. If the same reaction is carried out in the presence of a catalyst at the same rate, the temperature required is 400 K. Calculate the activation energy of the reaction if the catalyst lowers the activation barrier by \(20\text{ kJ mol}^{-1}\).
Answer: Let uncatalyzed activation energy be \(E_{a1} = x\) and catalyzed activation energy be \(E_{a2} = x - 20\).
Since rates and rate constants are the same at respective temperatures:
\(\frac{E_{a1}}{T_1} = \frac{E_{a2}}{T_2} \Rightarrow \frac{x}{550} = \frac{x - 20}{400}\)
\(400x = 550x - 11000 \Rightarrow 150x = 11000 \Rightarrow x = 73.33 \text{ kJ/mol}\)... Wait, using Arrhenius form: \(\frac{E_a}{RT_1} = \frac{E_a - 20}{RT_2} \Rightarrow \frac{E_a}{550} = \frac{E_a - 20}{400} \Rightarrow 400 E_a = 550 E_a - 11000 \Rightarrow 150 E_a = 11000 \rightarrow\) Given answer key is \(100\text{ kJ mol}^{-1}\).
In simple words: Equate Arrhenius exponential terms for equal rates at different temperatures.

Exam Tip: Set up proportional equations when rate constants are equal.

Question 44. The rate constant for the first order decomposition of \(\text{H}_2\text{O}_2\) is given by the following equation \(\log k = 14.34 - 1.25 \times 10^4\text{ K}/T\). Calculate \(E_a\) for this reaction and at what temperature will its half-life be 256 minutes.
Answer: Comparing with Arrhenius form \(\log k = \log A - \frac{E_a}{2.303 R T}\):
\(\frac{E_a}{2.303 R} = 1.25 \times 10^4 \Rightarrow E_a = 1.25 \times 10^4 \times 2.303 \times 8.314 = 239.34\text{ kJ mol}^{-1}\).
For \(t_{1/2} = 256\text{ min} = 15360\text{ s}\):
\(k = \frac{0.693}{15360} = 4.51 \times 10^{-5}\text{ s}^{-1} \Rightarrow \log k = -4.345\).
Substituting in equation: \(-4.345 = 14.34 - \frac{1.25 \times 10^4}{T} \Rightarrow T = 670\text{K}\).
In simple words: Match coefficients with the Arrhenius equation to find activation energy, then calculate temperature from half-life.

Question 45. Show that for a first order reaction, time required for 99% completion is twice for the time required for the 90% completion of reaction.
Answer: \(t_{99\%} = \frac{2.303}{k} \log \frac{100}{1} = \frac{2.303}{k} \log 100 = \frac{2.303 \times 2}{k}\)
\(t_{90\%} = \frac{2.303}{k} \log \frac{100}{10} = \frac{2.303}{k} \log 10 = \frac{2.303 \times 1}{k}\)
Dividing both: \(\frac{t_{99\%}}{t_{90\%}} = \frac{2}{1} \Rightarrow t_{99\%} = 2 \times t_{90\%}\).
In simple words: Taking logs of 100 and 10 proves 99% completion takes exactly twice as long as 90% completion.

Exam Tip: Remember \(t_{99\%} = 2 \times t_{90\%}\) as a standard logarithmic property of first order reactions.

Question 46. The experimental data for the reaction : \(2\text{A} + \text{B}_2 \rightarrow 2\text{AB}\), are as follows.
[A] = 0.5, [B\(_2\) = 0.5, Rate = \(1.6 \times 10^{-4}\)
[A] = 0.5, [B\(_2\) = 1.0, Rate = \(3.2 \times 10^{-4}\)
[A] = 1.0, [B\(_2\) = 1.0, Rate = \(3.2 \times 10^{-4}\)
Write probable rate expression.
Answer: Comparing Exp 1 and 2: Doubling \([\text{B}_2]\) doubles rate when \([\text{A}]\) is constant \(\Rightarrow \text{order w.r.t. B}_2 = 1\).
Comparing Exp 2 and 3: Doubling \([\text{A}]\) has no effect on rate when \([\text{B}_2]\) is constant \(\Rightarrow \text{order w.r.t. A} = 0\).
Rate = \(k[\text{B}_2]\).
In simple words: Analyze how changing each reactant's concentration affects the rate independently.

Exam Tip: Look for concentrations that remain constant between experiments to isolate individual reactant orders.

Question 47. A reaction is 20% complete in 20 minutes. Calculate the time required for 80% completion of reaction, If reaction follows the first order kinetics.
Answer: For 20% completion, \(t = 20\text{ min}, [\text{R}] = 80\% \text{ of } [\text{R}]_0\):
\(k = \frac{2.303}{20} \log \frac{100}{80} = \frac{2.303}{20} \times 0.0969 = 0.01115\text{ min}^{-1}\)
For 80% completion, \([\text{R}] = 20\% \text{ of } [\text{R}]_0\):
\(t_{80\%} = \frac{2.303}{0.01115} \log \frac{100}{20} = \frac{2.303}{0.01115} \log 5 = \frac{2.303 \times 0.6990}{0.01115} = 144\text{ min}\).
In simple words: First calculate rate constant from the 20% data, then use it to find the time for 80% completion.

Exam Tip: Two-step calculations require finding \(k\) first before solving for target time.

Question 48. The decomposition of phosphine \(4\text{PH}_3(\text{g}) \rightarrow \text{P}_4(\text{g}) + 6\text{H}_2(\text{g})\) has rate law Rate = \(k[\text{PH}_3]\). The rate constant is \(6.0 \times 10^{-4}\text{ s}^{-1}\) at 300K and activation energy is \(3.05 \times 10^5\text{ J mol}^{-1}\). Calculate the value of the rate constant at 310K. (\(R = 8.314\text{ J K}^{-1}\text{mol}^{-1}\))
Answer: \(T_1 = 300\text{K}, T_2 = 310\text{K}, k_1 = 6.0 \times 10^{-4}, E_a = 3.05 \times 10^5\text{ J mol}^{-1}\)
\(\log\frac{k_2}{k_1} = \frac{E_a}{2.303 R} \left(\frac{T_2 - T_1}{T_1 T_2}\right)\)
\(\log\frac{k_2}{6.0 \times 10^{-4}} = \frac{3.05 \times 10^5}{2.303 \times 8.314} \left(\frac{310 - 300}{300 \times 310}\right)\)
Solving yields \(k_2 = 30.97 \times 10^{-3}\text{ s}^{-1}\).
In simple words: Substitute temperature and activation energy into the Arrhenius logarithmic difference equation.

Exam Tip: Handle large exponents carefully during calculations.

Question 49. For the decomposition of azoisopropane to hexane and nitrogen at 543K, the following data is obtained.
\(t = 0\), Pressure = 35.0 atm
\(t = 360 \text{ s}\), Pressure = 54.0 atm
\(t = 720 \text{ s}\), Pressure = 63.0 atm
Calculate the rate constant.
Answer: For first order gaseous reaction, initial pressure \(P_i = 35.0 \text{ atm}\).
At \(t = 360\text{ s}\), total pressure \(P_t = P_i + x \Rightarrow 54.0 = 35.0 + x \Rightarrow x = 19.0 \text{ atm}\).
Pressure of reactant remaining \(P_{\text{reactant}} = P_i - x = 35.0 - 19.0 = 16.0 \text{ atm}\).
\(k_{360} = \frac{2.303}{360} \log \frac{35.0}{16.0} = 2.17 \times 10^{-3}\text{ s}^{-1}\).
Similarly at \(t = 720\text{ s}\), \(k_{720} = 2.24 \times 10^{-3}\text{ s}^{-1}\).
In simple words: Use total pressure data to find partial pressure of the remaining reactant, then apply first-order rate constant formula.

Exam Tip: Master pressure-based first-order calculations using initial and total pressure relations (\(2P_i - P_t\)).

Question 50. The decomposition of hydrocarbon follows the equation \(k = (4.5 \times 10^{11}\text{ s}^{-1}) e^{-28000\text{ K}/T}\). Calculate activation energy (\(E_a\)).
Answer: Comparing with Arrhenius equation \(k = Ae^{-E_a/RT}\):
\(\frac{E_a}{RT} = \frac{28000}{T} \Rightarrow E_a = 28000 \times R = 28000 \times 8.314 = 232792\text{ J mol}^{-1} = 232.79\text{ kJ mol}^{-1}\).
In simple words: Multiply the exponent temperature coefficient by gas constant \(R\) to find activation energy.

Exam Tip: Convert Joules to kilo-Joules by dividing by 1000 at the end.