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Class 12 Chemistry Chapter 3: Chemical Kinetics worksheet with answer key

A 20-mark worksheet on Chemical Kinetics for about 40 minutes of class or homework: 5 multiple-choice, 3 very short answer and 3 short answer questions, written from the NCERT chapter and checked by a second reviewer. Print the PDF for the class; the answer key carries the marking scheme. Teachers may photocopy it freely.

Question 1

The gas-phase reaction 2NO(g) + 2H₂(g) → N₂(g) + 2H₂O(g) follows Rate = k[NO]²[H₂]. With concentration in mol L⁻¹ and time in seconds, the unit of k is

(a) s⁻¹(b) mol⁻¹ L s⁻¹(c) mol L⁻¹ s⁻¹(d) mol⁻² L² s⁻¹

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Answer: (d) mol⁻² L² s⁻¹

The order is 2 + 1 = 3, so k = Rate/[conc]³ has unit (mol L⁻¹ s⁻¹)/(mol L⁻¹)³ = mol⁻² L² s⁻¹.

Question 2

A zero order reaction has k = 0.005 mol L⁻¹ min⁻¹. If the initial concentration is 0.40 mol L⁻¹, the half-life is

(a) 80 min(b) 40 min(c) 20 min(d) 138.6 min

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Answer: (b) 40 min

For zero order t½ = [R]₀/2k = 0.40/(2 × 0.005) = 40 min. (138.6 min is 0.693/k, the first order formula wrongly applied.)

Question 3

For CHCl₃ + Cl₂ → CCl₄ + HCl, the experimental rate law is Rate = k[CHCl₃][Cl₂]^1/2. The overall order of the reaction is

(a) 1(b) 2(c) 3/2(d) 1/2

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Answer: (c) 3/2

Order = sum of the powers in the rate law = 1 + ½ = 3/2, although the coefficients in the equation add to 2.

Question 4

In the iodide-catalysed decomposition of H₂O₂, the fast step is the elementary reaction H₂O₂ + IO⁻ → H₂O + I⁻ + O₂. The molecularity of this step is

(a) 2(b) 1(c) 3(d) 4

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Answer: (a) 2

Two species, H₂O₂ and IO⁻, must collide together; the step is bimolecular.

Question 5

Which statement about order and molecularity is correct?

(a) Molecularity is found by measuring the rate at different concentrations.(b) Molecularity can be zero for a reaction on a metal surface.(c) Order has meaning only for elementary reactions.(d) Order can be zero or a fraction, but molecularity cannot.

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Answer: (d) Order can be zero or a fraction, but molecularity cannot.

Order is an experimental quantity and may be zero or fractional; molecularity counts colliding species in an elementary step, so it is a whole number and has no meaning for a complex reaction.

Question 6 · 2 marks

A first order reaction has k = 6.93 × 10⁻³ min⁻¹. How long will it take for 87.5% of the reactant to be used up?

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t½ = 0.693/k = 0.693/(6.93 × 10⁻³) = 100 min. 87.5% used means 12.5% = ⅛ remains: three half-lives. t = 3 × 100 = 300 min.

t½ = 100 min (1)
⅛ remaining is three half-lives, 300 min (1)

Question 7 · 2 marks

State the two conditions a collision must meet to give products, and name the factor introduced in collision theory to account for one of them.

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An effective collision needs (i) energy equal to or more than the threshold energy, and (ii) proper orientation of the colliding molecules. The probability or steric factor P accounts for orientation: Rate = P Z_AB e^(−Ea/RT).

both conditions: enough energy (threshold) and proper orientation (1)
P, the probability (steric) factor, for orientation (1)

Question 8 · 2 marks

For 4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g), NO is formed at 3.6 × 10⁻³ mol L⁻¹ s⁻¹ at an instant. Calculate (i) the rate of disappearance of O₂ and (ii) the rate of formation of H₂O at that instant.

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Rate = −¼Δ[NH₃]/Δt = −⅕Δ[O₂]/Δt = ¼Δ[NO]/Δt = ⅙Δ[H₂O]/Δt. Rate = ¼ × 3.6 × 10⁻³ = 9.0 × 10⁻⁴ mol L⁻¹ s⁻¹. (i) −Δ[O₂]/Δt = 5 × 9.0 × 10⁻⁴ = 4.5 × 10⁻³ mol L⁻¹ s⁻¹. (ii) Δ[H₂O]/Δt = 6 × 9.0 × 10⁻⁴ = 5.4 × 10⁻³ mol L⁻¹ s⁻¹.

O₂ disappears at 4.5 × 10⁻³ mol L⁻¹ s⁻¹ (5/4 of the NO rate) (1)
H₂O forms at 5.4 × 10⁻³ mol L⁻¹ s⁻¹ (6/4 of the NO rate) (1)

Question 9 · 3 marks

The reaction H₂(g) + 2ICl(g) → I₂(g) + 2HCl(g) is proposed to go in two steps: (1) H₂ + ICl → HI + HCl (slow); (2) HI + ICl → I₂ + HCl (fast). (i) Write the rate law this mechanism predicts and its order. (ii) Name the intermediate and give the molecularity of each step. (iii) Why is the order not 3, though three molecules appear in the overall equation?

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(i) The slow step is rate determining: Rate = k[H₂][ICl], second order. (ii) HI is the intermediate, formed in step 1 and used in step 2; both steps are bimolecular. (iii) The overall equation is not an elementary step; the order follows the slow step, which involves one H₂ and one ICl, not the stoichiometric coefficients.

Rate = k[H₂][ICl], second order (1)
HI the intermediate; both steps bimolecular (1)
order is set by the slow step, not by the balanced equation (1)

Question 10 · 3 marks

A reaction has Ea = 46.0 kJ mol⁻¹ and k = 1.5 × 10⁻⁴ s⁻¹ at 300 K. Taking A as constant, calculate k at 320 K. (2.303 R = 19.15 J K⁻¹ mol⁻¹, antilog 0.500 = 3.16)

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log(k₂/k₁) = (Ea/2.303R)(T₂ − T₁)/(T₁T₂) = (46000/19.15) × 20/(300 × 320) = 2402 × 2.083 × 10⁻⁴ = 0.500. k₂/k₁ = 3.16, so k₂ = 3.16 × 1.5 × 10⁻⁴ = 4.74 × 10⁻⁴ s⁻¹.

the two-temperature Arrhenius relation written (1)
log(k₂/k₁) = 0.500 (1)
k₂ = 4.74 × 10⁻⁴ s⁻¹ (1)

Question 11 · 3 marks

A compound X decomposes by first order kinetics. Its concentration falls from 0.80 mol L⁻¹ to 0.20 mol L⁻¹ in 46.2 min. Calculate (i) the rate constant, (ii) the half-life, and (iii) the time for the concentration to fall to 0.05 mol L⁻¹. (log 2 = 0.3010)

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(i) k = (2.303/t) log([R]₀/[R]) = (2.303/46.2) log 4 = (2.303 × 0.6020)/46.2 = 3.0 × 10⁻² min⁻¹. (ii) t½ = 0.693/k = 0.693/0.030 = 23.1 min. (iii) 0.80 → 0.05 is a sixteenfold fall: t = (2.303/k) log 16 = 2 × 46.2 = 92.4 min (four half-lives).

k = 3.0 × 10⁻² min⁻¹ with the formula (1)
t½ = 23.1 min (1)
92.4 min for 0.05 mol L⁻¹ (1)

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