CBSE Class 12 Chemistry Chemical Kinetics NEET JEE Questions Set 01

Single Correct Answer

Question 1. For the chemical reaction, N₂(g) + 3H₂(g) ⇄ 2NH₃(g), the correct equation is (NEET 2019)
(a) \( 3\frac{d[\mathrm{H_2}]}{dt} = 2\frac{d[\mathrm{NH_3}]}{dt} \)
(b) \( -\frac{1}{3}\frac{d[\mathrm{H_2}]}{dt} = -\frac{1}{2}\frac{d[\mathrm{NH_3}]}{dt} \)
(c) \( -\frac{d[\mathrm{N_2}]}{dt} = 2\frac{d[\mathrm{NH_3}]}{dt} \)
(d) \( -\frac{d[\mathrm{N_2}]}{dt} = \frac{1}{2}\frac{d[\mathrm{NH_3}]}{dt} \)
Answer:
Ans. (d)
Rate of reaction \( = -\frac{d[\mathrm{N_2}]}{dt} = -\frac{1}{3}\frac{d[\mathrm{H_2}]}{dt} = \frac{1}{2}\frac{d[\mathrm{NH_3}]}{dt} \)

Teacher's Note:
1. Divide each rate by its stoichiometric coefficient; put a minus sign for reactants and a plus sign for products.

Question 2. The decomposition of N₂O₅ in CCl₄ at 318 K is studied by monitoring the concentration of N₂O₅ in the solution. Initially, the concentration of N₂O₅ is 2.4 mol L⁻¹ and after 200 minutes, it is reduced to 2.00 mol L⁻¹. What is the rate of production of NO₂ during this period in mol L⁻¹ min⁻¹? (Kerala PMT 2015)
(a) 4 × 10⁻³
(b) 2 × 10⁻³
(c) 1 × 10⁻³
(d) 2 × 10⁻⁴
(e) 5 × 10⁻³
Answer:
Ans. (a)
2N₂O₅ → 4NO₂ + O₂
Rate of reaction \( = -\frac{1}{2}\frac{d[\mathrm{N_2O_5}]}{dt} = +\frac{1}{4}\frac{d[\mathrm{NO_2}]}{dt} \) ...(i)
Rate of disappearance of N₂O₅ \( \left(-\frac{d[\mathrm{N_2O_5}]}{dt}\right) = \frac{(2.4-2.0)}{200\ \text{min}} \) mol L⁻¹ = \( \frac{0.4}{200} \) mol L⁻¹ min⁻¹ = 2 × 10⁻³ mol L⁻¹ min⁻¹
From eqn. (i), rate of production of NO₂ \( \left(+\frac{d[\mathrm{NO_2}]}{dt}\right) = -4\times\frac{1}{2}\frac{d[\mathrm{N_2O_5}]}{dt} \) = 2 × (2 × 10⁻³) mol L⁻¹ min⁻¹
= 4 × 10⁻³ mol L⁻¹ min⁻¹

Teacher's Note:
1. Every 2 mol of N₂O₅ gives 4 mol of NO₂, so NO₂ forms twice as fast as N₂O₅ disappears.

Question 3. In the synthesis of ammonia from nitrogen and hydrogen gases, if 6 × 10⁻² mole of hydrogen disappears in 10 minutes, the number of moles of ammonia formed in 0.3 minutes is (J & K CET 2009)
(a) 1.8 × 10⁻²
(b) 1.2 × 10⁻³
(c) 4 × 10⁻²
(d) 3.6 × 10⁻²
Answer:
Ans. (b)
N₂ + 3H₂ → 2NH₃
Rate \( = -\frac{d[\mathrm{N_2}]}{dt} = -\frac{1}{3}\frac{d[\mathrm{H_2}]}{dt} = +\frac{1}{2}\frac{d[\mathrm{NH_3}]}{dt} \)
Rate of disappearance of H₂, i.e., \( -\frac{d[\mathrm{H_2}]}{dt} = \frac{6\times10^{-2}\ \text{mol}}{10\ \text{min}} \) = 6 × 10⁻³ mol min⁻¹
Rate of formation of NH₃, i.e., \( \frac{d[\mathrm{NH_3}]}{dt} = \frac{2}{3}\left(-\frac{d[\mathrm{H_2}]}{dt}\right) = \frac{2}{3}\times 6\times10^{-3} \) mol min⁻¹ = 4 × 10⁻³ mol min⁻¹
Now, if dt = 0.3 min, NH₃ formed = (4 × 10⁻³ mol min⁻¹)(0.3 min) = 1.2 × 10⁻³ mol

Teacher's Note:
1. First convert the rate of H₂ to the rate of NH₃ using the ratio 2 : 3, then multiply by the new time interval.

Question 4. The rate of a gaseous reaction is generally expressed in terms of \( \frac{dP}{dt} \). If it were expressed in terms of change in number of moles per unit time \( \left(\frac{dn}{dt}\right) \) or in terms of change in molar concentration per unit time \( \left(\frac{dC}{dt}\right) \), which of the following relationship will hold good?
(a) \( \frac{dC}{dt} = \frac{dn}{dt} = \frac{dP}{dt} \)
(b) \( \frac{dC}{dt} = \frac{1}{V}\left(\frac{dn}{dt}\right) = \frac{1}{RT}\left(\frac{dP}{dt}\right) \)
(c) \( \frac{dC}{dt} = \frac{dn}{dt} = \frac{V}{RT}\left(\frac{dP}{dt}\right) \)
(d) None of these
Answer:
Ans. (b)
\( C = \frac{n}{V} \quad \therefore \frac{dC}{dt} = \frac{1}{V}\frac{dn}{dt} \)
As PV = nRT, i.e., \( P = \frac{n}{V}RT = CRT \) or \( C = \frac{P}{RT} \quad \therefore \frac{dC}{dt} = \frac{d}{dt}\left(\frac{P}{RT}\right) = \frac{1}{RT}\frac{dP}{dt} \)

Teacher's Note:
1. Concentration of a gas is n/V = P/RT, so at constant V and T the three rates differ only by the constant factors 1/V and 1/RT.

Question 5. In the reversible reaction, \( \mathrm{2NO_2 \underset{k_2}{\overset{k_1}{\rightleftharpoons}} N_2O_4} \), the rate of disappearance of NO₂ is equal to (AMU Engg. 2015)
(a) \( \frac{2k_1}{k_2}[\mathrm{NO_2}]^2 \)
(b) 2k₁ [NO₂]² − 2k₂ [N₂O₄]
(c) 2k₁ [NO₂]² − k₂ [N₂O₄]
(d) (2k₁ − k₂) [NO₂]
Answer:
Ans. (b)
Rate of reaction \( = -\frac{1}{2}\frac{d[\mathrm{NO_2}]}{dt} = k_1[\mathrm{NO_2}]^2 - k_2[\mathrm{N_2O_4}] \)
\( \therefore \) Rate of disappearance of NO₂ \( = -\frac{d[\mathrm{NO_2}]}{dt} = 2k_1[\mathrm{NO_2}]^2 - 2k_2[\mathrm{N_2O_4}] \)

Teacher's Note:
1. For a reversible step the net rate is forward rate minus backward rate; the factor 2 comes from the coefficient of NO₂.

Question 6. Consider the decomposition of N₂O₅ as \( \mathrm{N_2O_5 \longrightarrow 2NO_2 + \frac{1}{2}O_2} \). The rate of reaction is given by
\( -\frac{d[\mathrm{N_2O_5}]}{dt} = \frac{1}{2}\frac{d[\mathrm{NO_2}]}{dt} = 2\frac{d[\mathrm{O_2}]}{dt} = k_1[\mathrm{N_2O_5}] \)
Therefore, \( -\frac{d[\mathrm{N_2O_5}]}{dt} = k_1[\mathrm{N_2O_5}] \)
\( +\frac{d[\mathrm{NO_2}]}{dt} = 2k_1[\mathrm{N_2O_5}] = k_1'[\mathrm{N_2O_5}] \)
\( +\frac{d[\mathrm{O_2}]}{dt} = \frac{1}{2}k_1[\mathrm{N_2O_5}] = k_1''[\mathrm{N_2O_5}] \)
Choose the correct option
(a) k₁ = k₁′ = k₁″
(b) k₁ = 2k₁′ = k₁″
(c) 2k₁ = k₁′ = 4k₁″
(d) 4k₁ = 2k₁′ = k₁″
Answer:
Ans. (c)
\( -\frac{d[\mathrm{N_2O_5}]}{dt} = \frac{1}{2}\frac{d[\mathrm{NO_2}]}{dt} = 2\frac{d[\mathrm{O_2}]}{dt} \)
\( k_1[\mathrm{N_2O_5}] = \frac{1}{2}k_1'[\mathrm{N_2O_5}] = 2k_1''[\mathrm{N_2O_5}] \)
or \( k_1 = \frac{1}{2}k_1' = 2k_1'' \) or 2k₁ = k₁′ = 4k₁″

Teacher's Note:
1. The numerical value of a rate constant depends on which species the rate is written for; always state the species.

Question 7. The rate constant is numerically the same for three reactions of first, second and third order respectively. Which one of the following is true for the rate of these three reactions if concentration of the reactant is same and greater than 1 M?
(a) r₁ = r₂ = r₃
(b) r₁ > r₂ > r₃
(c) r₁ < r₂ < r₃
(d) There can be no definite order
Answer:
Ans. (c)
r₁ = k[A], r₂ = k[A]², r₃ = k[A]³
If [A] > 1 M, r₃ > r₂ > r₁ or r₁ < r₂ < r₃

Teacher's Note:
1. A number greater than 1 grows with higher powers; if [A] were less than 1 M the order would reverse.

Question 8. For the reaction, Ag⁺ + 2NH₃ ⇄ [Ag(NH₃)₂]⁺, the net rate of reaction is given by
\( \frac{dx}{dt} \) = 2 × 10⁷ [Ag⁺] [NH₃]² − 1 × 10⁻² [Ag(NH₃)₂]⁺
Then which of the following statement/s is/are correct?
(a) Rate constant for forward reaction = 2 × 10⁷
(b) Rate constant for backward reaction = 1 × 10⁻²
(c) Equilibrium constant of the reaction = 2 × 10⁹
(d) All the above
Answer:
Ans. (d)
Net reaction = Rate of forward reaction − Rate of backward reaction = k_f [Reactants] − k_b [Products]
Comparing with the given equation, k_f = 2 × 10⁷, k_b = 1 × 10⁻²
Eqm. const. = k_f/k_b = (2 × 10⁷)/(10⁻²) = 2 × 10⁹

Teacher's Note:
1. For an elementary reversible reaction, K = k_f/k_b.

Question 9. For the reaction 2H₂(g) + 2NO(g) → N₂(g) + 2H₂O(g), the observed rate expression is, rate = k_f [NO]² [H₂]. The rate expression for the reverse reaction is (JEE Main 2020)
(a) k_b [N₂] [H₂O]²
(b) k_b [N₂] [H₂O]
(c) k_b [N₂] [H₂O]²/[H₂]
(d) k_b [N₂] [H₂O]² [NO]
Answer:
Ans. (c)
2H₂(g) + 2NO(g) ⇄ N₂(g) + 2H₂O(g). Equilibrium constant, \( K_{eq} = \frac{[\mathrm{N_2}][\mathrm{H_2O}]^2}{[\mathrm{H_2}]^2[\mathrm{NO}]^2} = \frac{k_f}{k_b} \) ...(i) or k_f = k_b × K_eq
At equilibrium, rate of forward reaction = rate of backward reaction. But rate of forward reaction = k_f [NO]² [H₂] (Given)
∴ Rate of backward reaction = k_f [NO]² [H₂] = k_b · K_eq · [NO]² [H₂]
\( = k_b \times \frac{[\mathrm{N_2}][\mathrm{H_2O}]^2}{[\mathrm{H_2}]^2[\mathrm{NO}]^2}\times[\mathrm{NO}]^2[\mathrm{H_2}] = \) \( k_b \times \frac{[\mathrm{N_2}][\mathrm{H_2O}]^2}{[\mathrm{H_2}]} \)

Teacher's Note:
1. When the forward rate law is not from the stoichiometry, get the reverse rate law from K_eq = k_f/k_b rather than from the products' coefficients.

Question 10. For the reaction A + B → C + D, doubling the concentration of both the reactants increases the reaction rate by 8 times and doubling the concentration of only B simply doubles the reaction rate. The rate law is given as (AIPMT Prelim 2012)
(a) r = k [A]^1/2 [B]^1/2
(b) r = k [A] [B]²
(c) r = k [A]² [B]
(d) r = k [A] [B]
Answer:
Ans. (c)
(i) r = k a^α b^β   (ii) 8r = k (2a)^α (2b)^β   (iii) 2r = k a^α (2b)^β
Eqn. (ii)/Eqn. (iii) gives \( \frac{8}{2} = 2^{\alpha} \) or 2^α = 4 = 2² or α = 2.
Eqn. (iii)/Eqn. (i) gives 2 = 2^β or β = 1. Hence, rate = k [A]² [B].

Teacher's Note:
1. Doubling everything gives 2^{(total order)}; here 8 = 2³, so the overall order is 3, and B alone accounts for 1.

Question 11. If rate = k [H⁺]ⁿ and it becomes 100 times when the pH changes from 2 to 1, then the order of reaction is
(a) 0
(b) 1
(c) 2
(d) 3
Answer:
Ans. (c)
At pH = 2, i.e., [H⁺] = 10⁻² M, r = k (10⁻²)ⁿ
At pH = 1, i.e., [H⁺] = 10⁻¹ M, 100 r = k (10⁻¹)ⁿ
\( \therefore \frac{100r}{r} = \frac{10^{-n}}{10^{-2n}} \) or 100 = 10ⁿ. Hence, n = 2.

Teacher's Note:
1. A fall of one pH unit means [H⁺] becomes 10 times; a 100-fold rate rise then means order 2.

Question 12. A hypothetical reaction, A₂ + B₂ → 2AB, follows the mechanism as given below:
A₂ ⇄ A + A (Fast)
A + B₂ → AB + B (Slow)
A + B → AB (Fast)
The overall order of reaction is
(a) zero
(b) 1
(c) \( 1\frac{1}{2} \)
(d) 2
Answer:
Ans. (c)
As rate depends upon the slow step, Rate = k [A] [B₂]
For the equilibrium, A₂ ⇄ 2A, \( K_c = \frac{[\mathrm{A}]^2}{[\mathrm{A_2}]} \) or \( [\mathrm{A}] = \sqrt{K_c[\mathrm{A_2}]} \)
\( \therefore \) Rate = k K_c^1/2 [A₂]^1/2 [B₂] = k′ [A₂]^1/2 [B₂]
Hence, order \( = \frac{1}{2} + 1 = 1\frac{1}{2} \).

Teacher's Note:
1. An intermediate (here A) must not appear in the final rate law; replace it using the fast pre-equilibrium.

Question 13. The rate of the reaction A → Products, at the initial concentration of 3.24 × 10⁻² M is nine times its rate at another initial concentration of 1.2 × 10⁻³ M. The order of the reaction is (Kerala CET 2011)
(a) \( \frac{1}{2} \)
(b) \( \frac{3}{4} \)
(c) \( \frac{3}{2} \)
(d) \( \frac{2}{3} \)
(e) \( \frac{1}{3} \)
Answer:
Ans. (d)
9r = k (3.24 × 10⁻²)ⁿ ...(i)
r = k (1.2 × 10⁻³)ⁿ ...(ii)
Dividing (i) by (ii), \( 9 = \left(\frac{3.24\times10^{-2}}{1.2\times10^{-3}}\right)^n \) or 9 = (27)ⁿ
or (3)² = (3³)ⁿ = (3)³ⁿ. Hence, 3n = 2 or n = 2/3.

Teacher's Note:
1. Take the ratio of the two rate equations so that k cancels; then write both sides as powers of the same base.

Question 14. The time for half life period of a certain reaction A → Products is 1 hour. When the initial concentration of the reactant 'A', is 2.0 mol L⁻¹, how much time does it take for its concentration to come from 0.50 to 0.25 mol L⁻¹ if it is a zero order reaction?
(a) 0.25 h
(b) 1 h
(c) 4 h
(d) 0.5 h
Answer:
Ans. (a)
For a zero order reaction, \( k = \frac{[\mathrm{A}]_0}{2t_{1/2}} = \frac{2.0\ \text{mol L}^{-1}}{2\times1\ \text{hr}} \) = 1.0 mol L⁻¹ hr⁻¹
\( t = \frac{1}{k}\{[\mathrm{A}]_0-[\mathrm{A}]\} = \frac{1}{1.0}(0.50-0.25) \) = 0.25 hr

Teacher's Note:
1. For zero order, the time depends only on the amount that reacts (here 0.25 mol L⁻¹), not on the starting value.

Question 15. The figure below depicts the change in the concentration of the species A and B for the reaction A → B, as a function of time. The point of intersection of the two curves represents
(a) t_1/4
(b) t_1/3
(c) t_1/2
(d) The method cannot be used to predict fractional life time

Answer:
Ans. (c)
At the point of intersection, [A] = [B], i.e., half of the reactant has reacted. Hence, it represents t_1/2.

Teacher's Note:
1. For A → B (1 : 1), [A] = [B] only when exactly half of A has changed.

Question 16. Half-lives of a first order and a zero order reaction are same. Then the ratio of the initial rates of the first order reaction to that of zero order reaction is (Kerala PET 2010)
(a) \( \frac{1}{0.693} \)
(b) 2 × 0.693
(c) 0.693
(d) \( \frac{2}{0.693} \)
(e) 6.93
Answer:
Ans. (b)
t_1/2 for 1st order reaction \( = \frac{0.693}{k} \); t_1/2 for zero order reaction \( = \frac{[\mathrm{A}]_0}{2k'} \)
Initial rate for 1st order reaction (r₁) = k [A]₀; Initial rate for zero order reaction (r₀) = k′
\( \therefore \frac{r_1}{r_0} = \frac{k[\mathrm{A}]_0}{k'} \). But \( \frac{0.693}{k} = \frac{[\mathrm{A}]_0}{2k'} \) (Given)
\( \therefore \frac{k[\mathrm{A}]_0}{k'} = 2\times0.693 \), i.e., \( \frac{r_1}{r_0} = 2\times0.693 \)

Teacher's Note:
1. Equate the two half-life expressions first, then rearrange to get k[A]₀/k′.

Question 17. For a first order reaction, the time taken to reduce the initial concentration by a factor of \( \frac{1}{4} \) is 20 minutes. The time required to reduce initial concentration by a factor of 1/16 is (Kerala PMT 2011)
(a) 20 min
(b) 10 min
(c) 80 min
(d) 40 min
(e) 5 min
Answer:
Ans. (d)
Time taken to reduce by factor of \( \frac{1}{4} \) = 20 min. \( \therefore \) Time taken to reduce by factor of \( \frac{1}{2} \), i.e., t_1/2 = 10 min
\( a \xrightarrow{t_{1/2}} \frac{a}{2} \xrightarrow{t_{1/2}} \frac{a}{4} \xrightarrow{t_{1/2}} \frac{a}{8} \xrightarrow{t_{1/2}} \frac{a}{16} \)
Total time required = 4 × t_1/2 = 4 × 10 min = 40 min

Teacher's Note:
1. In first order kinetics, 1/4 left = 2 half-lives and 1/16 left = 4 half-lives.

Question 18. The rate constant of a second order reaction, 2A → Products, is 10⁻⁴ lit mol⁻¹ min⁻¹. The initial concentration of the reactant is 10⁻² mol lit⁻¹. What is the half-life (in min)?
(a) 10
(b) 1000
(c) 100
(d) 10⁶
Answer:
Ans. (d)
Half-life period of a second order reaction \( t_{1/2} = \frac{1}{\text{Rate constant}\times\text{Initial conc.}} = \frac{1}{ka} \)
\( = \frac{1}{10^{-4}\ \text{L mol}^{-1}\text{min}^{-1}\times10^{-2}\ \text{mol L}^{-1}} \) = 10⁶ min

Teacher's Note:
1. For second order, t_1/2 = 1/(k[A]₀): a small k and a dilute solution give a very long half-life.

Teacher's Note:
1. C is a solid, so it adds nothing to the gas pressure; use the final pressure to find P₀.

Question 20. Rate constant of a reaction is 0.0693 min⁻¹. Starting with 10 mol L⁻¹, rate of reaction after 10 minutes will be
(a) 0.0693 M min⁻¹
(b) 0.0693 × 2.5 M min⁻¹
(c) 0.0693 × 5 M min⁻¹
(d) 0.0693 × 10 M min⁻¹
Answer:
Ans. (c)
To calculate concentration after 10 min: \( k = \frac{2.303}{t}\log\frac{[\mathrm{A}]_0}{[\mathrm{A}]} \quad \therefore 0.0693 = \frac{2.303}{10}\log\frac{10}{[\mathrm{A}]} \)
or \( \log\frac{10}{[\mathrm{A}]} = 0.3010 \) or \( \frac{10}{[\mathrm{A}]} = 2 \) or [A] = 5 mol L⁻¹
Rate of reaction after 10 min = k [A] = 0.0693 × 5 mol L⁻¹ min⁻¹

Teacher's Note:
1. The unit min⁻¹ tells you the reaction is first order; 10 min here is exactly one half-life (0.693/0.0693).

Teacher's Note:
1. Compare two runs in which only one concentration changes; the stoichiometric coefficients (3, 2, 1) are a trap.

Question 22. At 500 K, the half-life period of a gaseous reaction at the initial pressure of 80 kPa is 350 sec. When the pressure is 40 kPa, the half-life period is 175 sec. The order of the reaction is (Kerala PMT 2007)
(a) zero
(b) one
(c) two
(d) three
(e) half
Answer:
Ans. (a)
When initial pressure is halved, the half-life period is also halved. This shows that t_1/2 ∝ initial pressure. This is so for reactions of zero order.

Teacher's Note:
1. Use t_1/2 ∝ [A]₀¹⁻ⁿ: t_1/2 proportional to [A]₀ means 1 − n = 1, so n = 0.

Question 23. A reaction P → Q is completed 25% in 25 min, 50% completed in 25 min if [P] is halved, 25% completed in 50 min if [P] is doubled. The order of reaction is
(a) 1
(b) 2
(c) 0
(d) 3
Answer:
Ans. (c)
By hit and trial method, if reaction is of zero order, \( k = \frac{1}{t}\{[\mathrm{A}]_0-[\mathrm{A}]\} \)
Case I. \( k = \frac{1}{25}\times\frac{25}{100} = 10^{-2} \)
Case II. \( k = \frac{1}{25}\times\frac{1}{2}\{[\mathrm{A}]_0-[\mathrm{A}]\} \) (as [P] is halved) \( = \frac{1}{25}\times\frac{1}{2}\times\frac{50}{100} = 10^{-2} \)
Case III. \( k = \frac{1}{50}\times2\{[\mathrm{A}]_0-[\mathrm{A}]\} \) (as [P] is doubled) \( = \frac{1}{50}\times2\times\frac{25}{100} = 10^{-2} \)
As k comes out to be constant, hence the given reaction is of zero order.

Teacher's Note:
1. In zero order, the amount reacted per minute is fixed, so the percentage reacted in a given time is inversely proportional to the initial concentration.

One or More Correct Answers

Question 24. In acidic medium, the rate of reaction between BrO₃⁻ and Br⁻ is given by the expression \( -\frac{d[\mathrm{BrO_3^-}]}{dt} = k[\mathrm{BrO_3^-}][\mathrm{Br^-}][\mathrm{H^+}]^2 \). It means
(a) Rate constant of the reaction depends upon the concentration of H⁺ ions.
(b) Rate of reaction is independent of the concentration of the acid added.
(c) The change in pH of the solution will affect the rate of reaction.
(d) Doubling the concentration of H⁺ ions will increase the reaction rate by 4 times.
Answer:
Ans. (c, d)
(a) is wrong because rate constant is constant at constant temperature, (b) is wrong because rate depends upon [H⁺]², (c) is correct because change in pH means change in [H⁺], (d) is correct because rate ∝ [H⁺]².

Teacher's Note:
1. Concentrations change the rate, never the rate constant; k changes only with temperature (or a catalyst).

Question 25. The rate constant of a reaction is given by k = 2.1 × 10¹⁰ exp (−2700/RT). It means that
(a) log k vs. 1/T will be a straight line with slope \( = -\frac{2700}{2.303R} \)
(b) log k vs. 1/T will be a straight line with intercept on log k axis = log (2.1 × 10¹⁰)
(c) The number of effective collisions are 2.1 × 10¹⁰ cm⁻³ sec⁻¹
(d) Half life of the reaction increases with increase of temperature.
Answer:
Ans. (a, b)
log k = log (2.1 × 10¹⁰) − \( \frac{2700}{2.303R}\cdot\frac{1}{T} \), so (a) and (b) are correct.
(c) is wrong because frequency factor gives total number of collisions and not effective collisions cm⁻³ sec⁻¹.
(d) is wrong because half-life of the reaction decreases with increase of temperature (as reaction becomes faster).

Teacher's Note:
1. Compare with log k = log A − E_a/(2.303RT): the intercept is log A, not A itself.

Question 26. In a hypothetical reaction X → Y, the activation energy for the forward and the backward reaction are 15 and 9 kJ mol⁻¹ respectively. The potential energy of X is 10 kJ mol⁻¹. Then
(a) Threshold energy of the reaction is 25 kJ
(b) The potential energy of Y is 16 kJ
(c) Heat of reaction is 6 kJ
(d) The reaction is endothermic.
Answer:
Ans. (a, b, c, d)
Energy of X = 10 kJ; threshold energy (E_t) = 10 + 15 = 25 kJ (a).
Energy of Y = E_t − 9 = 25 − 9 = 16 kJ (b).
Heat of reaction = 16 − 10 = + 6 kJ (c); as Y lies above X, the reaction is endothermic (d).
All (a) to (d) are correct.

Teacher's Note:
1. Draw a quick energy diagram: threshold energy = energy of reactant + E_a(forward).

Passage-Based Questions

Question 27. Comprehension 1: Arrhenius studied the effect of temperature on the rate of a reaction and postulated that rate constant varies with temperature exponentially as k = Ae^−E_a/RT. For most of the reactions, it was found that the temperature coefficient of the reaction lies between 2 to 3. The method is generally used for finding the activation energy of a reaction. Keeping temperature constant, the effect of catalyst on the activation energy has also been studied by studying how much the rate of reaction changes in the presence of catalyst. In most of the cases, it is observed that catalyst lowers the activation energy barrier and increases the rate of reaction.
The pre-exponential factor in the Arrhenius equation of a second order reaction has the units
(a) mol L⁻¹ s⁻¹
(b) L mol⁻¹ s⁻¹
(c) s⁻¹
(d) dimensionless
Answer:
Ans. (b)
The factor A has the same units as those of k. For a second order reaction, k has the units of L mol⁻¹ s⁻¹. (For a first order reaction, it has the units of s⁻¹)

Teacher's Note:
1. The exponential term is dimensionless, so A always carries the units of k.

Question 28. Comprehension 1: Arrhenius studied the effect of temperature on the rate of a reaction and postulated that rate constant varies with temperature exponentially as k = Ae^−E_a/RT. For most of the reactions, it was found that the temperature coefficient of the reaction lies between 2 to 3. The method is generally used for finding the activation energy of a reaction. Keeping temperature constant, the effect of catalyst on the activation energy has also been studied by studying how much the rate of reaction changes in the presence of catalyst. In most of the cases, it is observed that catalyst lowers the activation energy barrier and increases the rate of reaction.
Which of the following plot will be linear?
(a) ln k versus T with −ve slope
(b) k versus 1/T with −ve slope
(c) ln k versus 1/T with −ve slope
(d) ln k versus 1/T with +ve slope.
Answer:
Ans. (c)
\( \ln k = \ln A - \frac{E_a}{RT} \). Hence, plot of ln k versus 1/T is linear with slope = −E_a/R.

Teacher's Note:
1. Since E_a is positive, the slope −E_a/R is always negative.

Matrix Match Questions

Teacher's Note:
1. These orders are experimental facts worth memorising; they cannot be read off the balanced equations.

Integer and Numerical Answer Questions

Question 30. The order of the reaction 5Br⁻(aq) + BrO₃⁻(aq) + 6H⁺(aq) → 3Br₂(aq) + 3H₂O(l) is
Answer:
Ans. 4
Rate law equation for the given reaction is Rate = k [Br⁻] [BrO₃⁻] [H⁺]²
Hence, overall order of reaction is 4.

Teacher's Note:
1. The order comes from the experimental rate law (1 + 1 + 2), not from the coefficients 5, 1 and 6.

Question 31. The half-life period of a 3rd order reaction is found to be 1 hour when we start with a concentration of 0.3 mol L⁻¹ of the reactant. If we start with 0.1 mol L⁻¹, the half-life period in hours will be
Answer:
Ans. 9
\( t_{1/2} = k\frac{1}{[\mathrm{A}]_0^{\,n-1}} \). For 3rd order reaction, \( t_{1/2} = k\frac{1}{[\mathrm{A}]_0^{\,2}} \)
\( 1 = k\frac{1}{(0.3)^2} \) or k = 0.09
When [A]₀ = 0.1 mol L⁻¹, \( t_{1/2} = (0.09)\frac{1}{(0.1)^2} \) = 9 hours.

Teacher's Note:
1. Here k is just a proportionality constant; quicker: t_1/2 ∝ 1/[A]₀², and [A]₀ becomes 1/3, so t_1/2 becomes 9 times.

Question 32. In the reaction aA + bB → Products, if only concentration of A is doubled the rate is quadrupled but if further the vessel is reduced to half the volume, the rate becomes 8 times. The overall order of reaction is
Answer:
Ans. 3
r = k [A]^α [B]^β = k a^α b^β ...(i)
4r = k (2a)^α (b)^β ...(ii)
Now, if the volume of the vessel is reduced to half, concentration of both the reactants will be doubled. Hence, 8r = k (4a)^α (2b)^β ...(iii)
Dividing eqn. (ii) by eqn. (i), 4 = 2^α or α = 2
Dividing eqn. (iii) by eqn. (ii), 8 = 2^α × 2^β = 2² × 2^β or 2^β = 2 or β = 1.
Hence, overall order = 3.

Teacher's Note:
1. Halving the volume doubles every concentration, including the already-doubled A (which becomes 4a).

Question 33. Starting with an initial pressure of 5 atm of azoisopropane, 40% of it decomposes into nitrogen and hexane vapours in one hour. The pressure in atm exerted by the mixture in atm at this time will be
Answer:
Ans. 7
(CH₃)₂CHN=NCH(CH₃)₂(g) → N₂(g) + C₆H₁₄(g)
Initial pressure: 5 atm, 0, 0. After 1 hour: \( 5-\frac{40}{100}\times5 \) = 5 − 2 atm = 3 atm, 2 atm, 2 atm
Total pressure after one hour = 3 + 2 + 2 = 7 atm.

Teacher's Note:
1. Each mole of azoisopropane gives two moles of gas, so the pressure rises by the amount decomposed (2 atm).

Assertion–Reason Questions

Question 34. Statement-1: The dissociation of NH₃ on hot platinum surface may be reaction of zero order or first order.
Statement-2: Pressure of the gas affects the order of reaction.
(a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation of Statement-1.
(b) Statement-1 is True, Statement-2 is True; Statement-2 is NOT the correct explanation of Statement-1.
(c) Statement-1 is True, Statement-2 is False.
(d) Statement-1 is False, Statement-2 is True.
Answer:
Ans. (a)
At low pressure, it is first order reaction while at high pressure, it is a zero order reaction. Thus the pressure of the gas decides the order, and Statement-2 explains Statement-1.

Teacher's Note:
1. At high pressure the metal surface is fully covered, so the rate no longer depends on pressure (zero order).

Question 35. Statement-1: In the reaction, N₂ + 3H₂ → 2NH₃, the rate of reaction is different in terms of N₂, H₂ and NH₃.
Statement-2: Rate of disappearance of N₂ and H₂ and rate of formation of NH₃ are not equal to each other.
(a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation of Statement-1.
(b) Statement-1 is True, Statement-2 is True; Statement-2 is NOT the correct explanation of Statement-1.
(c) Statement-1 is True, Statement-2 is False.
(d) Statement-1 is False, Statement-2 is True.
Answer:
Ans. (d)
Correct Statement-1. The rate of reaction is same in terms of different reactants and products.
Rate \( = -\frac{d[\mathrm{N_2}]}{dt} = -\frac{1}{3}\frac{d[\mathrm{H_2}]}{dt} = \frac{1}{2}\frac{d[\mathrm{NH_3}]}{dt} \), though the individual rates of disappearance/formation are different.

Teacher's Note:
1. Distinguish the rate of reaction (one value) from the rates of change of individual species (which differ by their coefficients).

Question 36. Statement-1: Rate constant of a zero order reaction has same units as the rate of reaction.
Statement-2: Rate constant of a zero order reaction does not depend upon the units of concentration.
(a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation of Statement-1.
(b) Statement-1 is True, Statement-2 is True; Statement-2 is NOT the correct explanation of Statement-1.
(c) Statement-1 is True, Statement-2 is False.
(d) Statement-1 is False, Statement-2 is True.
Answer:
Ans. (c)
Correct Statement-2. k = conc./time = mol L⁻¹ s⁻¹, i.e., it depends upon units of concentration.

Teacher's Note:
1. Units of k are (concentration)¹⁻ⁿ time⁻¹; only for n = 1 does k not involve concentration units.

Question 37. Statement-1: In a zero order reaction, if concentration of the reactant is doubled, half-life period is also doubled.
Statement-2: The total time taken for a zero order reaction to complete is double of the half-life period.
(a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation of Statement-1.
(b) Statement-1 is True, Statement-2 is True; Statement-2 is NOT the correct explanation of Statement-1.
(c) Statement-1 is True, Statement-2 is False.
(d) Statement-1 is False, Statement-2 is True.
Answer:
Ans. (b)
Correct explanation. For zero order reaction, t_1/2 ∝ a (t_1/2 = a/2k).
Statement-2 is also true (t_100% = a/k = 2t_1/2), but it does not explain Statement-1.

Teacher's Note:
1. For zero order, t_1/2 = [A]₀/2k, so the half-life grows in proportion to the starting concentration.