AmbifulWorksheets
Practice that never repeats
CBSE · Class 12 · Chemistry

Chemical Kinetics

Standard difficulty · 20 questions · 40 marks

Name: Date: Score: ____ / 40

Show your working for Sections B and C. Answers are on the last page — try every question first!

Section A — Choose the correct answer8 × 1 = 8 marks
1.
The rate of a zero-order reaction:
  1. is proportional to the square of concentration
  2. is always zero
  3. is proportional to concentration
  4. does not depend on the concentration of the reactant
2.
For rate = k[A][B]2, the overall order of the reaction is:
  1. 3
  2. 1
  3. 2
  4. 4
3.
The unit of the rate constant of a third-order reaction, with time in seconds, is:
  1. s−1
  2. mol L−1 s−1
  3. L mol−1 s−1
  4. L2 mol−2 s−1
4.
A first-order reaction has k = 0.013 s−1. Its half-life is:
  1. 53.31 s
  2. 0.009009 s
  3. 106.62 s
  4. 76.92 s
5.
A zero-order reaction has [A]0 = 0.7 mol/L and k = 0.007 mol L−1 s−1. Its half-life is:
  1. 50 s
  2. 0.02 s
  3. 99 s
  4. 100 s
6.
A first-order reaction has a half-life of 48 min. The time for it to be 93.75% complete is:
  1. 384 min
  2. 240 min
  3. 48 min
  4. 192 min
7.
For a reaction that is of order 0 in A, the concentration of A is made 4 times. The rate becomes:
  1. unchanged
  2. 16 times
  3. 4 times
  4. 8 times
8.
Assertion (A): A first-order reaction with k = 0.019 min−1 has the same half-life when it starts at 5 mol/L as at 8 mol/L.
Reason (R): The half-life of a first-order reaction, 0.693/k, does not depend on the initial concentration.
  1. Both A and R are true, and R is the correct explanation of A.
  2. Both A and R are true, but R is not the correct explanation of A.
  3. A is true, but R is false.
  4. A is false, but R is true.
Section B — Short answer8 × 2 = 16 marks
9.
Distinguish between order and molecularity of a reaction.
10.
For N2 + 3H2 → 2NH3, H2 is used up at 0.024 mol L−1 s−1. Find the rate of formation of NH3 and the rate of the reaction.
11.
In a first-order reaction the concentration of the reactant falls to 1/10 of its initial value in 30 min. Find the rate constant. (log 2 = 0.3010)
12.
The half-life of a first-order reaction is 26 min. How long does it take for 1/64 of the reactant to remain? What fraction is left after 130 min?
13.
For A + B → products, initial rates are: [A] = 0.1, [B] = 0.1: 2 × 10−3; [A] = 0.3, [B] = 0.1: 0.018; [A] = 0.1, [B] = 0.3: 6 × 10−3 (concentrations in mol/L, rates in mol L−1 s−1). Find the order with respect to A and B.
14.
A zero-order reaction has k = 0.09 mol L−1 min−1 and starts at 1.7 mol/L. Find the concentration after 5 min, and the half-life.
15.
The rate constant of a reaction becomes 4 times larger when the temperature rises from 310 K to 330 K. Find the activation energy. (R = 8.314 J mol−1 K−1, log 2 = 0.3010)
16.
A first-order reaction has k = 0.074 s−1. How long does it take for 50% of the reactant to react?
Section C — Word problems4 × 4 = 16 marks
17.
Derive the integrated rate equation for a first-order reaction and show that its half-life does not depend on the initial concentration.
18.
For 2A + B → C, the initial rates were: experiment 1: [A] = 0.1, [B] = 0.1 mol/L, rate = 0.01 mol L−1 s−1; experiment 2: [A] = 0.2, [B] = 0.1 mol/L, rate = 0.01 mol L−1 s−1; experiment 3: [A] = 0.1, [B] = 0.2 mol/L, rate = 0.02 mol L−1 s−1. Find the order with respect to each reactant, write the rate law, and find k.
19.
A first-order reaction has a half-life of 46 min. Find (a) the rate constant, (b) the time for 93.75% completion, (c) the time for the concentration to fall to 1/10 of its initial value. (log 10 = 1)
20.
The rate constant of a reaction is k at 310 K and 10k at 330 K. Find its activation energy, and explain why a small rise in temperature can increase the rate so much. (R = 8.314 J mol−1 K−1, log 2 = 0.3010)
AmbifulWorksheets
Practice that never repeats
CBSE · Class 12 · Chemistry

Answer key

Chemical Kinetics · AW-12C-KIN-2026-V1 · For parents — fold back or tear off before handing the sheet to your child.

QAnswerWorking
1(d) does not depend on the concentration of the reactant—
2(a) 3add the powers
3(d) L2 mol−2 s−1(mol L−1)^(1 − n) s−1 with n = 3
4(a) 53.31 st½ = 0.693/k
5(a) 50 st½ = [A]0/(2k)
6(d) 192 min6.25% left = 4 half-lives
7(a) unchangedrate ∝ [A]0: 40 = 1
8(a) Both A and R are true, and R is the correct explanation of A.A is true: t½ = 0.693/k = 36.47 min for any starting concentration. R is true and explains A.
9Order is the sum of the powers of concentrations in the experimental rate law; it can be zero or a fraction. Molecularity is the number of species colliding in an elementary step; it is a whole number (1, 2 or 3).—
100.016 mol L−1 s−1; rate of reaction 0.008 mol L−1 s−1rate = −(1/3) d[H2]/dt = (1/2) d[NH3]/dt
110.0768 min−1k = (2.303/t) log(10) = (2.303/30) × 1
12156 min; 1/32each half-life halves it; 1/64 = (1/2)6
13order 2 in A, 1 in B; overall 3[A] × 3 changes the rate × 9; [B] × 3 changes it × 3
141.25 mol/L; 9.44 min[A] = [A]0 − kt; t½ = [A]0/2k
1559 kJ/mollog(k2/k1) = (Ea/2.303R)(1/T1 − 1/T2)
169.36 s50% reacted = 1 half-life; t½ = 0.693/k
17−d[R]/dt = k[R] ⇒ d[R]/[R] = −k dt ⇒ ln[R] = −kt + ln[R]0 ⇒ k = (2.303/t) log([R]0/[R]). At t½, [R] = [R]0/2, so k = (2.303/t½) log 2 ⇒ t½ = 0.693/k, which contains no concentration term.—
18order 0 in A, 1 in B; rate = k[B]; k = 0.1 s−1compare experiments 1 and 2 for A, 1 and 3 for B
19(a) 0.0151 min−1; (b) 184 min; (c) 153 mink = 0.693/t½; t = (2.303/k) log([A]0/[A])
2097.9 kJ/mol; the fraction of molecules with energy above Ea rises sharply with temperatureEa = 2.303R log(k2/k1) × T1T2/(T2 − T1)