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Class 12 Chemistry Part 1

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Previous year questions: Chemical Kinetics

Chemical Kinetics32 min readLesson 7 of 7

Q1 (CBSE Board 2023, 2 marks)

A first order reaction has a half-life of 60 minutes. Find the time needed for 90 percent of the reactant to be used up. Given log 2 = 0.3010, log 3 = 0.4771 and log 10 = 1.

Answer: The rate constant is 0.693 divided by the half-life, which is 0.693 divided by 60, or 0.01155 per minute. For first order kinetics the time is 2.303 divided by k times log of the initial amount divided by the amount left. When 90 percent is consumed, 10 percent remains, so the ratio is 10 and its logarithm is 1. The time is 2.303 divided by 0.01155 times 1, which is about 199 minutes, or 3 hours 19 minutes.

Q2 (CBSE Board 2024, 2 marks)

For a first order reaction with rate constant 1.25 times 10^-3 s^-1, find the time needed for the reactant mass to fall from 5 g to 2.5 g.

Answer: The mass falls from 5 g to 2.5 g, which is exactly one half, so the time needed is one half-life. For a first order reaction the half-life is 0.693 divided by k, which is 0.693 divided by 1.25 times 10^-3, equal to 554 s. The time is therefore about 554 s or 9.2 minutes. The same result follows from the integrated equation with the ratio of masses equal to two.

Q3 (CBSE Board 2024, 3 marks)

Initial rate data for the reaction 2NO + Br2 giving 2NOBr are: experiment 1 with 0.05 M NO and 0.05 M Br2 gives 1.0 times 10^-3; experiment 2 with 0.05 M NO and 0.15 M Br2 gives 3.0 times 10^-3; experiment 3 with 0.15 M NO and 0.05 M Br2 gives 9.0 times 10^-3, all rates in mol L^-1 s^-1. (a) Find the order in NO and in Br2. (b) Calculate the rate constant. (c) Find the rate when NO is 0.4 M and Br2 is 0.2 M.

Answer: (a) Compare experiments 1 and 2: Br2 becomes three times larger while NO stays fixed, and the rate also becomes three times larger, so the order in Br2 is one. Compare experiments 1 and 3: NO becomes three times larger while Br2 is fixed, and the rate becomes nine times larger, which is three squared, so the order in NO is two. The rate law is rate = k[NO]^2[Br2] and the overall order is three. (b) Substituting experiment 1 gives k = 1.0 times 10^-3 divided by (0.05 squared times 0.05), which is 1.0 times 10^-3 divided by 1.25 times 10^-4, equal to 8.0 L^2 mol^-2 s^-1. (c) At 0.4 M NO and 0.2 M Br2 the rate is 8.0 times 0.16 times 0.2, which is 0.256 mol L^-1 s^-1.

Q4 (CBSE Board 2025, 2 marks)

In the decomposition of a reactant A, the concentration was measured as 0.40 M at time zero, 0.20 M after 1 hour, 0.10 M after 2 hours and 0.05 M after 3 hours. Predict the order of the reaction and write the rate law.

Answer: The concentration halves in every one hour interval: 0.40 to 0.20, then to 0.10, then to 0.05. A constant half-life that does not depend on the starting concentration is the signature of a first order reaction. The order is therefore one and the rate law is rate = k[A]. The rate constant is 0.693 divided by the half-life in hours, which is 0.693 h^-1, or about 1.93 times 10^-4 s^-1. The data also fit the integrated first order equation with a constant k for every interval.

Q5 (CBSE Board 2026, 3 marks)

The decomposition of hydrogen peroxide is catalysed by iodide ion: 2H2O2 giving 2H2O + O2. The proposed mechanism is step I, H2O2 + I- giving H2O + IO- (slow), and step II, H2O2 + IO- giving H2O + I- + O2 (fast). (a) Write the rate law. (b) State the order with respect to H2O2, with respect to I- and the overall order. (c) What is the molecularity of step II?

Answer: (a) The slow step controls the rate, so the rate law is written from step I: rate = k[H2O2][I-]. The intermediate IO- appears only after the slow step and does not enter the rate law. (b) The order with respect to H2O2 is one and the order with respect to I- is one, so the overall order is two. (c) Step II involves one H2O2 molecule and one IO- ion, so two species collide in that elementary step and its molecularity is two.

Q6 (CBSE Board 2026, 3 marks)

The half-life of a radioactive element is 1.5 times 10^10 years. Calculate the time in which the activity of the element is reduced to 75 percent of its original value. Given log 2 = 0.30, log 3 = 0.48 and log 4 = 0.60.

Answer: Radioactive decay is first order, so the rate constant is 0.693 divided by the half-life. The activity falls to 75 percent, so 25 percent has decayed and 75 percent remains. The time is 2.303 divided by k times log of 100 over 75, which is 2.303 divided by k times log of 4 by 3. Using the given logarithms, log 4 minus log 3 is 0.60 minus 0.48, which is 0.12. Therefore the time is the half-life times 2.303 times 0.12 divided by 0.693, which simplifies to 1.5 times 10^10 times 0.399, or about 6.0 times 10^9 years.

Q7 (CBSE SQP 2022-23, 1 mark)

Which radioactive isotope has the longer half-life, 15O or 19O? The rate constants are 5.63 times 10^-3 s^-1 and 2.38 times 10^-2 s^-1 respectively. Options: 15O, 19O, both equal, data insufficient.

Answer: For first order decay the half-life equals 0.693 divided by k, which means the smaller rate constant belongs to the longer-lived isotope. The rate constant of 15O is about four times smaller than that of 19O, so 15O has the longer half-life. The correct option is 15O. The two half-lives work out to about 123 s and 29 s, so the difference is large and easy to see.

Q8 (CBSE SQP 2022-23, 2 marks)

Explain how and why the rate of a given reaction is affected when (a) a catalyst is added and (b) the temperature at which the reaction takes place is decreased.

Answer: (a) A catalyst provides an alternative pathway of lower activation energy. More collisions can then cross the barrier at the same temperature, so the rate increases. The catalyst is not consumed and it speeds both the forward and the reverse reactions without changing the equilibrium constant. (b) Decreasing the temperature lowers the average kinetic energy and shifts the Maxwell-Boltzmann distribution to lower energies. The fraction of collisions with energy above the activation energy falls sharply, so the rate constant decreases according to the Arrhenius equation and the reaction slows down.

Q9 (CBSE SQP 2023-24, 1 mark)

The initial concentration of A is 1.5 M and after 120 s it is 0.75 M. The reaction follows zero order kinetics. What is the rate constant? Options: 6.25 times 10^-3 mol L^-1 s^-1, 6.25 times 10^-3 s^-1, 5.78 times 10^-3 mol L^-1 s^-1, 5.78 times 10^-3 s^-1.

Answer: For a zero order reaction the rate constant is the fall in concentration divided by the time, k = (1.5 - 0.75) divided by 120, which is 0.75 divided by 120, or 6.25 times 10^-3. The unit of k for a zero order reaction is mol L^-1 s^-1, so the correct option is 6.25 times 10^-3 mol L^-1 s^-1. The half-life of this reaction would be the initial concentration divided by twice the rate constant, which is 120 s, matching the data.

Q10 (CBSE SQP 2023-24, 3 marks)

At 200 K the rate constant of a reaction is 0.02 s^-1, and at 500 K it becomes 0.20 s^-1. Work out the activation energy. Take 2.303R as 19.15 J K^-1 mol^-1.

Answer: The two temperature form of the Arrhenius equation gives log of k2 by k1 equal to Ea divided by 2.303R times the difference of the temperatures divided by their product. The ratio of rate constants is 0.20 divided by 0.02, which is 10, so its logarithm is 1. The difference is 300 K and the product is 100000 K squared, so the ratio is 0.0030 per K. Rearranging for Ea gives 19.15 times 1 divided by 0.0030, which is about 6.38 times 10^3 J mol^-1, or 6.38 kJ mol^-1. The small activation energy explains why this rate constant changes only by a factor of ten across a 300 K range.