Chemical Kinetics - MCQs with Explanations

1. For $A+B \rightarrow C+D$, $\Delta H = -20 \text{ kJ mol}^{-1}$. The activation energy of the forward reaction is $85 \text{ kJ mol}^{-1}$. The activation energy for backward reaction is:

• (a) 65 kJ mol^-1
• (b) 105 kJ mol^-1
• (c) 85 kJ mol^-1
• (d) 40 kJ mol^-1

Correct Answer: (b)

Explanation: For any chemical reaction, the activation energy of the forward reaction is equal to the activation energy for the backward reaction plus the enthalpy change. Using the values provided, $85 = E_{a(back)} - 20$, which gives $E_{a(back)} = 105 \text{ kJ mol}^{-1}$.

2. The half-life period of a radioactive element is 140 days. After 560 days, one gram of the element will reduce to:

• (a) 1/2 g
• (b) 1/4 g
• (c) 1/8 g
• (d) 1/16 g

Correct Answer: (d)

Explanation: First, calculate the number of half-lives ($n$) that have passed: $560 / 140 = 4$. The remaining concentration is given by the formula $[A]/[A]_0 = (1/2)^n$, so $(1/2)^4 = 1/16$. Given the initial mass is 1 g, the element reduces to $1/16 \text{ g}$.

3. For an endothermic reaction where $2N_2O_5 \rightarrow 4NO_2 + O_2$ and $2NO_2 + \frac{1}{2}O_2 \rightarrow N_2O_5$ have activation energies $E_1$ and $E_2$ respectively, then:

• (a) $E_1 > E_2$
• (b) $E_1 < E_2$
• (c) $E_1 = 2E_2$
• (d) $\sqrt{E_1 E_2^2} = 1$

Correct Answer: (a)

Explanation: In an endothermic reaction, the enthalpy of the products is higher than the enthalpy of the reactants. Consequently, the activation energy for the forward process ($E_1$) must be greater than that of the backward process ($E_2$).

4. In the reaction $A+2B \rightarrow C+2D$, the initial rate $-d[A]/dt$ was found to be $2.6 \times 10^{-2} \text{ M s}^{-1}$. The value of $-d[B]/dt$ at $t=0$ is:

• (a) $2.6 \times 10^{-2}$
• (b) $5.2 \times 10^{-2}$
• (c) $1.0 \times 10^{-1}$
• (d) $6.5 \times 10^{-3}$

Correct Answer: (b)

Explanation: According to the stoichiometry, the rate can be expressed as $-d[A]/dt = -\frac{1}{2} d[B]/dt$. Therefore, $-d[B]/dt = 2 \times (-d[A]/dt)$, which equals $2 \times 2.6 \times 10^{-2} = 5.2 \times 10^{-2} \text{ M s}^{-1}$.

5. The mechanism for $P \rightarrow R$ is $P \xrightarrow{k_1} Q$ (fast) and $Q+P \xrightarrow{k_2} R$ (slow). The rate law is:

• (a) $k_1[P][Q]$
• (b) $k_1k_2[P]$
• (c) $k_1k_2[P]^2$
• (d) $k_1k_2[Q]$

Correct Answer: (c)

Explanation: The rate-determining step is the slowest one, giving $Rate = k_2[P][Q]$. From the fast equilibrium step, $k_1 = [Q]/[P]$, or $[Q] = k_1[P]$. Substituting this into the rate equation yields $Rate = k_1k_2[P]^2$.

6. The rate of $Cl_3CCHO + NO \rightarrow CHCl_3 + NO + CO$ is $Rate = k[Cl_3CCHO][NO]$. The units of $k$ are:

• (a) L² mol^-2 s^-1
• (b) mol L^-1 s^-1
• (c) L mol^-1 s^-1
• (d) s^-1

Correct Answer: (c)

Explanation: This is a second-order reaction because the sum of the powers of the concentration terms in the rate law is 2. The unit for a second-order rate constant is $Conc.^{-1} \text{ time}^{-1}$, which translates to $L \text{ mol}^{-1} \text{ s}^{-1}$.

7. For $A+B \rightarrow \text{products}$, the rate is $k[A]^{1.5}[B]^{2.5}$. The order of the reaction is:

• (a) -1
• (b) +1
• (c) 3.75
• (d) 4

Correct Answer: (d)

Explanation: The overall order of a reaction is defined as the sum of the powers to which the molar concentration terms are raised in the rate law. Here, $1.5 + 2.5 = 4$.

8. In the reaction $CH_3COCH_3 \rightarrow C_2H_4 + H_2 + CO$, initial pressure was 0.40 atm and after 10 min, it was 0.50 atm. The rate constant for the first order reaction is:

• (a) 0.0133 min^-1
• (b) 0.4 s^-1
• (c) 10 s^-1
• (d) 0.6 min^-1

Correct Answer: (a)

Explanation: The increase in pressure allows us to find the extent of reaction ($x = 0.05$). Plugging this into the first-order rate equation $k = (2.303/10) \log(0.40/0.35)$ yields $k \approx 0.0133 \text{ min}^{-1}$.

9. A plot of $\log k$ vs $1/T$ gives a straight line with a slope of -5632. The energy of activation for this reaction is:

• (a) 127.67 kJ mol^-1
• (b) 107.84 kJ mol^-1
• (c) 86 kJ mol^-1
• (d) 246.8 kJ mol^-1

Correct Answer: (b)

Explanation: According to the Arrhenius equation, the slope of a $\log k$ versus $1/T$ plot is equal to $-E_a / 2.303R$. Calculating $E_a = 5632 \times 2.303 \times 8.314$ results in approximately $107.84 \text{ kJ mol}^{-1}$.

10. For $X \rightarrow Y$, the graph of product concentration $(x)$ vs $(t)$ is a straight line through the origin. The graph of $-d[X]/dt$ vs time would be:

• (a) straight line with negative slope
• (b) straight line with positive slope
• (c) a straight line parallel to x-axis
• (d) a hyperbola

Correct Answer: (c)

Explanation: A straight-line plot of concentration versus time passing through the origin is characteristic of a zero-order reaction. In such reactions, the rate ($-d[X]/dt$) is constant and independent of time, resulting in a line parallel to the time axis.

11. For the first-order decomposition of $N_2O_5$ with rate constants $k$ and $k'$ for different formulations, it is observed that:

• (a) $k = k'$
• (b) $k = 2k'$
• (c) $k = k'/2$
• (d) $k = k'^2$

Correct Answer: (b)

Explanation: The rate of disappearance of a reactant is independent of how the equation is written. By comparing the rate expressions for the two stoichiometry options provided, we find that $k = 2k'$.

12. In the energy diagram provided, which represents the activation energy for the backward reaction?

• (a) A
• (b) B
• (c) C
• (d) D

Correct Answer: (a)

Explanation: For the backward reaction, activation energy is defined as the energy difference between the products and the activated complex. In the diagram, this corresponds to segment A.

13. The rate constant is $2.3 \times 10^{-2} \text{ mol}^{-2} L^2 min^{-1}$. The order of reaction is:

• (a) zero
• (b) 1
• (c) 2
• (d) 3

Correct Answer: (d)

Explanation: The units of the rate constant are given by $(L/\text{mol})^{n-1} \text{ time}^{-1}$. Since $n-1 = 2$, it follows that $n=3$, indicating a third-order reaction.

14. If a catalytic reaction follows paths P, Q, and R, the catalytic efficiency order is:

• (a) $P > Q > R$
• (b) $Q > P > R$
• (c) $P > R > Q$
• (d) $R > Q > P$

Correct Answer: (d)

Explanation: A catalyst is more efficient if it lowers the activation energy to a greater extent. Comparing the energy peaks for paths P, Q, and R, path R has the lowest barrier, making it the most efficient, while P has the highest.

15. Which equation is correct for the reaction $N_{2(g)} + 3H_{2(g)} \rightarrow 2NH_{3(g)}$?

• (a) $3d[H_2]/dt = 2d[N_2]/dt$
• (b) $3d[NH_3]/dt = -2d[H_2]/dt$
• (c) $2d[NH_3]/dt = -3d[H_2]/dt$
• (d) $2d[N_2]/dt = 1/3 d[H_2]/dt$

Correct Answer: (b)

Explanation: The rate of the reaction is given by $-\frac{1}{3} d[H_2]/dt = \frac{1}{2} d[NH_3]/dt$. Cross-multiplying the denominators leads to the relation $3 d[NH_3]/dt = -2 d[H_2]/dt$.

16. For a second-order reaction, 20% is completed in 500 seconds. Time for 60% completion is:

• (a) 500 sec
• (b) 1000 sec
• (c) 3000 sec
• (d) 1500 sec

Correct Answer: (c)

Explanation: Using the integrated rate law for a second-order reaction $k = (1/t) \cdot x/a(a-x)$, we first solve for $k$ using the 20% data. Then, solving for $t$ at 60% completion gives 3000 seconds.

17. The reaction $A+B \rightarrow C+D + 40 \text{ kJ}$ has an activation energy of 18 kJ. The activation energy for $C+D \rightarrow A+B$ is:

• (a) 58 kJ
• (b) 40 kJ
• (c) -18 kJ
• (d) 22 kJ

Correct Answer: (a)

Explanation: The reaction is exothermic with $\Delta H = -40 \text{ kJ}$. The activation energy for the reverse process is the sum of the forward activation energy and the absolute value of the enthalpy change: $40 + 18 = 58 \text{ kJ}$.

18. The unit of rate constant for a zero order reaction is:

• (a) s^-1
• (b) mol L s^-1
• (c) mol L^-1 s^-1
• (d) no unit

Correct Answer: (c)

Explanation: For a zero-order reaction, the rate is equal to the rate constant ($Rate = k$), so the unit is $\text{mol L}^{-1} \text{ s}^{-1}$.

19. In the reaction $2A+B \rightarrow A_2B$, reactant B will disappear at:

• (a) half the rate as A will decrease
• (b) the same rate as A will decrease
• (c) twice the rate as A will decrease
• (d) half the rate as A₂B will form

Correct Answer: (a)

Explanation: The rate of disappearance is related by stoichiometry: $-\frac{1}{2} d[A]/dt = -d[B]/dt$. This means B disappears at half the rate of A.

20. For $2N_2O_5 \rightarrow 4NO_2 + O_2$ with $Rate = k[N_2O_5]$, which statement is true?

• (a) Order 1, molecularity 1
• (b) Order 1, molecularity 2
• (c) Order 2, molecularity 2
• (d) Order 2, molecularity 1

Correct Answer: (b)

Explanation: The rate law indicates the order is 1. The chemical equation shows that two molecules of reactant are involved, so the molecularity is 2.

21. Which of the following reactions are of the first order? (I. tert-butyl peroxide decomp., II. Rearrangement of N-chloroacetanilide, III. Persulphate-Iodide reaction, IV. Ester saponification).

• (a) I and IV
• (b) I and II
• (c) I and III
• (d) All

Correct Answer: (b)

Explanation: Reactions I and II follow first-order kinetics, while III and IV follow different orders.

22. In a pseudo first-order ester hydrolysis reaction, the net rate equation is given. The equilibrium constant ($K_{eq}$) is:

• (a) $1.3 \times 10^{-7}$
• (b) $1.3 \times 10^{-6}$
• (c) $1.33 \times 10^{-9}$
• (d) 1.33

Correct Answer: (d)

Explanation: The equilibrium constant is the ratio of the forward and backward rate constants ($K = k_f / k_b$). Dividing $4 \times 10^{-4}$ by $3 \times 10^{-4}$ gives 1.33.

23. Which reactant sample reacts at the highest rate?

• (a) 1 mol A and 1 mol B in 1 L
• (b) 2 mol A and 2 mol B in 2 L
• (c) 0.2 mol A and 0.2 mol B in 0.1 L
• (d) equal rates

Correct Answer: (c)

Explanation: Rate depends on concentration ($n/V$). Sample (c) has the highest concentration ($2 \text{ M}$), leading to the fastest rate.

24. A substance undergoes parallel first order reactions to B and C. If $k_1 = 1.26 \times 10^{-4} s^{-1}$ and $k_2 = 3.8 \times 10^{-5} s^{-1}$, the percentage distributions are:

• (a) 80%, 20%
• (b) 76.83%, 23.17%
• (c) 90%, 10%
• (d) 63.94%, 36.06%

Correct Answer: (b)

Explanation: The percentage for B is calculated as $[k_1 / (k_1 + k_2)] \times 100$, which gives 76.83%. The remainder is 23.17% for C.

25. Plotting $\log k$ vs $1/T$ gives a straight line with intercept $OX=5$ and $\tan \theta = (1/2.303)$. $E_a$ is:

• (a) 5 cal
• (b) 5/2.303 cal
• (c) -2 cal
• (d) 2.303 x 2 cal

Correct Answer: (c)

Explanation: The slope of the line is $-E_a / 2.303R$. Solving for $E_a$ based on the given slope leads to -2 cal.

26. If $E_a = 65 \text{ kJ}$, how much faster is the reaction at $25^{\circ}C$ than at $0^{\circ}C$?

• (a) 2 times
• (b) 16 times
• (c) 11 times
• (d) 6 times

Correct Answer: (c)

Explanation: Using the Arrhenius ratio formula, the rate constant ratio $k_2 / k_1$ is calculated to be 11.04, or approximately 11 times faster.

27. First-order $k$ at 600 K is $1.60 \times 10^{-5} s^{-1}$. If $E_a = 209 \text{ kJ/mol}$, $k$ at 700 K is:

• (a) 6.36 x 10^-3 s^-1
• (b) 6.24 x 10^-3 s^-1
• (c) 6.57 x 10^-3 s^-1
• (d) 6.86 x 10^-3 s^-1

Correct Answer: (a)

Explanation: Using the logarithmic form of the Arrhenius equation to solve for $k_2$ at 700 K yields $6.36 \times 10^{-3} \text{ s}^{-1}$.

28. Which statement about Arrhenius equation is incorrect?

• (a) If $E_a = 0$, $k=A$
• (b) High $E_a$ makes rate independent of temperature
• (c) Same $E_a$ does not mean same rate
• (d) Higher $E_a$ means smaller $k$

Correct Answer: (b)

Explanation: A high activation energy actually makes the rate very sensitive to and highly dependent on temperature changes.

29. $^{14}C$ half-life is 5730 years. An artifact wood has 80% $^{14}C$. Its age is:

• (a) 2865 years
• (b) 1845 years
• (c) 1765 years
• (d) 2345 years

Correct Answer: (b)

Explanation: For first-order decay, age $t = (2.303/k) \log(100/80)$. Calculating $k = 0.693/5730$, the age is approximately 1845 years.

30. Doubling initial concentration for a first-order reaction causes $t_{1/2}$ to:

• (a) increase 2x
• (b) decrease 4x
• (c) remain same
• (d) decrease half

Correct Answer: (c)

Explanation: The half-life of a first-order reaction ($t_{1/2} = 0.693/k$) is independent of the initial concentration of reactants.

31. 1st order: $0.1 \text{ M} \rightarrow 0.025 \text{ M}$ in 40 min. Rate at $0.01 \text{ M}$ reactant concentration:

• (a) 3.47 x 10^-4 M min^-1
• (b) 3.47 x 10^-5 M min^-1
• (c) 1.735 x 10^-6 M min^-1
• (d) 1.735 x 10^-4 M min^-1

Correct Answer: (a)

Explanation: First, calculate $k = 0.03466 \text{ min}^{-1}$ using the 40-minute drop. The rate at $0.01 \text{ M}$ is $k \times 0.01 = 3.47 \times 10^{-4} \text{ M min}^{-1}$.

32. $E_a$ can be calculated by plotting:

• (a) k vs T
• (b) k vs 1/log T
• (c) log k vs 1/T
• (d) log k vs 1/log T

Correct Answer: (c)

Explanation: Plotting $\log k$ versus $1/T$ yields a straight line where the slope represents $-E_a / 2.303R$.

33. Unit of $k$ for second order reaction:

• (a) mol^-1 L^-1 s^-1
• (b) mol^-1 L s^-1
• (c) mol L^-1 s^-1
• (d) mol L^-1 s

Correct Answer: (b)

Explanation: For second order ($n=2$), the unit is $\text{mol}^{-1} \text{ L s}^{-1}$.

34. In first order $R \rightarrow Q$ with initial conc. $0.5 \text{ M}$, $t_{1/2}$ is:

• (a) log 2 / k
• (b) log 2 / k(0.5)^1/2
• (c) ln 2 / k
• (d) 0.693 / 0.5 k

Correct Answer: (c)

Explanation: For a first-order reaction, $t_{1/2} = \ln 2 / k$, regardless of the initial concentration.

35. Which statement for reaction order is incorrect?

• (a) Determined experimentally
• (b) Not influenced by stoichiometry
• (c) Sum of power of conc. terms
• (d) Always a whole number

Correct Answer: (d)

Explanation: The order of a reaction can be zero, fractional, or negative; it is not restricted to being a whole number.

36. A 10 K rise in temperature doubles the rate constant because:

• (a) collision frequency increases 2-3x
• (b) fraction of molecules with threshold energy increases 2-3x
• (c) $E_a$ is lowered 2-3x
• (d) none

Correct Answer: (b)

Explanation: A small temperature rise significantly increases the number of molecules possessing energy equal to or greater than the threshold energy.

37. 1st order: $k = 1.5 \times 10^{-2} s^{-1}$, $t = 10$ min. Final conc. if initial is 100 moles?

• (a) 10^-7
• (b) 10^-5
• (c) 10^-6
• (d) 10^-2

Correct Answer: (d)

Explanation: Solving the first-order integrated rate equation for final concentration $[A]$ gives $10^{-2} \text{ mol L}^{-1}$.

38. A plot of $\ln K_{eq}$ vs $1/T$ with a negative slope indicates the reaction is:

• (a) exothermic
• (b) endothermic
• (c) negligible enthalpy change
• (d) highly spontaneous

Correct Answer: (a)

Explanation: The negative slope indicates that the enthalpy change ($\Delta H$) for the reaction is negative, confirming it is exothermic.

39. If $Rate = k[A]^c [B]^d$, the total order is:

• (a) $x+y$
• (b) $m+n$
• (c) $c+d$
• (d) $x/y$

Correct Answer: (c)

Explanation: The overall order is the sum of the exponents of the concentration terms in the experimentally determined rate law.

40. 1st order: time to reduce to 1/4 is 20 min. Time to reduce to 1/16 is:

• (a) 20 min
• (b) 10 min
• (c) 80 min
• (d) 40 min

Correct Answer: (d)

Explanation: Reducing concentration to $1/4$ requires 2 half-lives (20 min), so 1 half-life is 10 min. Reducing to $1/16$ ($1/2^4$) requires 4 half-lives, which equals 40 minutes.

41. A catalyst:

• (a) increases average kinetic energy
• (b) decreases activation energy
• (c) increases collision frequency
• (d) alters reaction mechanism

Correct Answer: (d)

Explanation: A catalyst functions by altering the reaction mechanism to provide an alternative pathway with a lower activation energy barrier.

42. For $2N_2O_5 \rightarrow 4NO_2 + O_2$, the rate of reaction is:

• (a) $\frac{1}{2} d[N_2O_5]/dt$
• (b) $2 d[N_2O_5]/dt$
• (c) $\frac{1}{4} d[NO_2]/dt$
• (d) $4 d[NO_2]/dt$

Correct Answer: (c)

Explanation: The rate is expressed by dividing the change in concentration of each species by its stoichiometric coefficient: $\frac{1}{4} d[NO_2]/dt$.

43. If forward $E_a = 1640 \text{ kJ/mol}$ and $\Delta H = -120 \text{ kJ/mol}$, the activation energy for reverse reaction is:

• (a) -120 kJ mol^-1
• (b) +152 kJ mol^-1
• (c) +120 kJ mol^-1
• (d) +1760 kJ mol^-1

Correct Answer: (d)

Explanation: For an exothermic reaction, reverse $E_a$ is $E_a(f) - \Delta H = 1640 - (-120) = 1760 \text{ kJ mol}^{-1}$.

44. Rate doubles for every $10^\circ C$ rise. For a $60^\circ C$ increase, the rate increases by:

• (a) 20 times
• (b) 32 times
• (c) 64 times
• (d) 128 times

Correct Answer: (c)

Explanation: The rate increases by a factor of $2^{(T_2-T_1)/10} = 2^{60/10} = 2^6 = 64$ times.

45. Match the plots with their slopes (P. C vs t (0 order), Q. log C vs t (1st order), R. Rate vs C (0 order), S. ln Rate vs ln C (1st order)).

• (a) P-4, Q-1, R-2, S-3
• (b) P-3, Q-4, R-2, S-1
• (c) P-2, Q-1, R-3, S-4
• (d) P-3, Q-2, R-4, S-1

Correct Answer: (b)

Explanation: For zero order, $C = C_0 - kt$ (slope $-k$); for first order, $\log C = \log C_0 - (k/2.303)t$ (slope $-k/2.303$). Rate is constant for zero order (slope 0), and for first order, the log plot has unity slope.