CBSE · NCERT · Class 12 Chemistry · Chapter 3

NCERT Solutions: Class 12 Chemistry Chapter 3 - Chemical Kinetics

27 textbook Q&A27 verifiedFree Content

Chapter-wise NCERT intext questions and exercise answers for Chemical Kinetics, grounded in the official textbook.

Questions are taken verbatim from the NCERT textbook; answers were grounded against the chapter's content during generation. Items needing review are marked.
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Exercises 27
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1Exercises27 questions
Q.3.1From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants. (i) 3NO(g) → N2O (g) Rate = k[NO]^2 (ii) H2O2 (aq) + 3I– (aq) + 2H+ → 2H2O (l) + I3– Rate = k[H2O2][I-] (iii) CH3CHO (g) → CH4 (g) + CO(g) Rate = k [CH3CHO]^3/2 (iv) C2H5Cl (g) → C2H4 (g) + HCl (g) Rate = k [C2H5Cl]v
Solution

The order is the sum of powers of concentration terms in the rate law. For an nth order reaction, units of k are (mol L^-1)^(1-n) s^-1. Applying this gives the stated orders and dimensions.

Answer:

(i) second order, k: L mol^-1 s^-1; (ii) second order, k: L mol^-1 s^-1; (iii) order 3/2, k: L^1/2 mol^-1/2 s^-1; (iv) first order, k: s^-1.

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Q.3.2For the reaction: 2A + B → A2B the rate = k[A][B]^2 with k = 2.0 × 10^-6 mol^-2 L^2 s^-1. Calculate the initial rate of the reaction when [A] = 0.1 mol L^-1, [B] = 0.2 mol L^-1. Calculate the rate of reaction after [A] is reduced to 0.06 mol L^-1.v
Solution

Initial rate = (2.0 x 10^-6)(0.1)(0.2)^2 = 8.0 x 10^-9 mol L^-1 s^-1. From 2A + B → A2B, when [A] falls by 0.04 M, [B] falls by 0.02 M, so [B] = 0.18 M. New rate = (2.0 x 10^-6)(0.06)(0.18)^2 = 3.89 x 10^-9 mol L^-1 s^-1.

Answer:

Initial rate = 8.0 x 10^-9 mol L^-1 s^-1; later rate = 3.89 x 10^-9 mol L^-1 s^-1.

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Q.3.4The decomposition of dimethyl ether leads to the formation of CH4, H2 and CO and the reaction rate is given by Rate = k [CH3OCH3]^3/2 The rate of reaction is followed by increase in pressure in a closed vessel, so the rate can also be expressed in terms of the partial pressure of dimethyl ether, i.e., Rate = k pCH3OCH3^3/2. If the pressure is measured in bar and time in minutes, then what are the units of rate and rate constants?v
Solution

When pressure is used instead of concentration, rate has units of pressure per time, i.e. bar min^-1. Since rate = k p^3/2, k has units (bar min^-1)/(bar^3/2) = bar^-1/2 min^-1.

Answer:

Rate: bar min^-1; k: bar^-1/2 min^-1.

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Q.3.5Mention the factors that affect the rate of a chemical reaction.v
Solution

Reaction rate depends on the number and effectiveness of collisions. Increasing concentration or gas pressure usually increases collision frequency. Higher temperature increases the fraction of molecules with sufficient energy. Catalysts lower activation energy. Nature and physical state of reactants, including surface area for solids, also matter.

Answer:

Concentration, temperature, pressure for gases, catalyst, nature of reactants and surface area affect reaction rate.

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Q.3.6A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is (i) doubled (ii) reduced to half ?v
Solution

For second order dependence on a reactant, rate is proportional to [A]^2. Doubling [A] multiplies rate by 2^2 = 4. Reducing [A] to half multiplies rate by (1/2)^2 = 1/4.

Answer:

(i) Rate becomes four times. (ii) Rate becomes one-fourth.

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Q.3.7What is the effect of temperature on the rate constant of a reaction? How can this effect of temperature on rate constant be represented quantitatively?v
Solution

For many reactions, the rate constant approximately doubles for a 10 K rise in temperature near room temperature. Quantitatively, k = A e^(-Ea/RT), where A is the pre-exponential factor, Ea is activation energy, R is the gas constant and T is absolute temperature. A plot of ln k against 1/T is linear with slope -Ea/R.

Answer:

The rate constant generally increases with temperature and is represented by the Arrhenius equation.

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Q.3.8In a pseudo first order reaction in water, the following results were obtained: t/s 0 30 60 90 [A]/ mol L^-1 0.55 0.31 0.17 0.085 Calculate the average rate of reaction between the time interval 30 to 60 seconds.v
Solution

Average rate of disappearance of A between 30 s and 60 s = -Δ[A]/Δt = -(0.17 - 0.31)/(60 - 30) = 0.14/30 = 4.67 x 10^-3 mol L^-1 s^-1.

Answer:

4.67 x 10^-3 mol L^-1 s^-1.

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Q.3.9A reaction is first order in A and second order in B. (i) Write the differential rate equation. (ii) How is the rate affected on increasing the concentration of B three times? (iii) How is the rate affected when the concentrations of both A and B are doubled?v
Solution

First order in A and second order in B gives rate = k[A][B]^2. Tripling B multiplies the rate by 3^2 = 9. Doubling both A and B multiplies the rate by 2 x 2^2 = 8.

Answer:

(i) Rate = k[A][B]^2. (ii) Rate becomes 9 times. (iii) Rate becomes 8 times.

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Q.3.10In a reaction between A and B, the initial rate of reaction (r0) was measured for different initial concentrations of A and B as given below: A/ mol L^-1 0.20 0.20 0.40 B/ mol L^-1 0.30 0.10 0.05 r0/mol L^-1s^-1 5.07 × 10^-5 5.07 × 10^-5 1.43 × 10^-4 What is the order of the reaction with respect to A and B?v
Solution

In experiments 1 and 2, [A] is constant while [B] changes from 0.30 to 0.10, but the rate is unchanged. Therefore order in B is zero. Comparing experiments 2 and 3, B does not matter, [A] doubles from 0.20 to 0.40 and rate changes by 1.43 x 10^-4/(5.07 x 10^-5) = 2.82. Thus 2^x = 2.82, giving x ≈ 1.5.

Answer:

Order with respect to A = 1.5; order with respect to B = 0.

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Q.3.11The following results have been obtained during the kinetic studies of the reaction: 2A + B → C + D Experiment [A]/mol L^-1 [B]/mol L^-1 Initial rate of formation of D/mol L^-1 min^-1 I 0.1 0.1 6.0 × 10^-3 II 0.3 0.2 7.2 × 10^-2 III 0.3 0.4 2.88 × 10^-1 IV 0.4 0.1 2.40 × 10^-2 Determine the rate law and the rate constant for the reaction.v
Solution

Comparing I and IV, [B] is constant, [A] increases 4 times and rate increases 4 times, so order in A is 1. Comparing II and III, [A] is constant, [B] doubles and rate becomes 4 times, so order in B is 2. Thus rate = k[A][B]^2. Using experiment I, k = 6.0 x 10^-3/(0.1 x 0.1^2) = 6.0 L^2 mol^-2 min^-1.

Answer:

Rate = k[A][B]^2; k = 6.0 L^2 mol^-2 min^-1.

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Q.3.12The reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the blanks in the following table: Experiment [A]/ mol L^-1 [B]/ mol L^-1 Initial rate/ mol L^-1 min^-1 I 0.1 0.1 2.0 × 10^-2 II – 0.2 4.0 × 10^-2 III 0.4 0.4 – IV – 0.2 2.0 × 10^-2v
Solution

Since the reaction is first order in A and zero order in B, rate = k[A]. From experiment I, k = (2.0 x 10^-2)/0.1 = 0.20 min^-1. For experiment II, [A] = rate/k = (4.0 x 10^-2)/0.20 = 0.2 M. For III, rate = 0.20 x 0.4 = 8.0 x 10^-2. For IV, [A] = (2.0 x 10^-2)/0.20 = 0.1 M.

Answer:

Experiment II [A] = 0.2 mol L^-1; Experiment III rate = 8.0 x 10^-2 mol L^-1 min^-1; Experiment IV [A] = 0.1 mol L^-1.

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Q.3.13Calculate the half-life of a first order reaction from their rate constants given below: (i) 200 s^-1 (ii) 2 min^-1 (iii) 4 years^-1v
Solution

For a first order reaction, t1/2 = 0.693/k. Therefore (i) t1/2 = 0.693/200 = 3.47 x 10^-3 s; (ii) t1/2 = 0.693/2 = 0.347 min; (iii) t1/2 = 0.693/4 = 0.173 year.

Answer:

(i) 3.47 x 10^-3 s; (ii) 0.347 min; (iii) 0.173 year.

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Q.3.14The half-life for radioactive decay of 14C is 5730 years. An archaeological artifact containing wood had only 80% of the 14C found in a living tree. Estimate the age of the sample.v
Solution

Radioactive decay is first order. k = 0.693/5730 year^-1. With N/N0 = 0.80, t = (2.303/k)log(N0/N) = (5730/0.693)ln(1/0.80) = 1.85 x 10^3 years.

Answer:

About 1.85 x 10^3 years.

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Q.3.16The rate constant for a first order reaction is 60 s^-1. How much time will it take to reduce the initial concentration of the reactant to its 1/16th value?v
Solution

For first order reaction, t = (2.303/k)log([A]0/[A]). Here [A] = [A]0/16, so t = (2.303/60)log16 = 4.62 x 10^-2 s.

Answer:

4.62 x 10^-2 s.

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Q.3.17During nuclear explosion, one of the products is 90Sr with half-life of 28.1 years. If 1mg of 90Sr was absorbed in the bones of a newly born baby instead of calcium, how much of it will remain after 10 years and 60 years if it is not lost metabolically.v
Solution

For first order radioactive decay, N = N0(1/2)^(t/t1/2). With N0 = 1 mg and t1/2 = 28.1 years, after 10 years N = (1/2)^(10/28.1) = 0.781 mg. After 60 years N = (1/2)^(60/28.1) = 0.228 mg.

Answer:

After 10 years: 0.781 mg; after 60 years: 0.228 mg.

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Q.3.18For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.v
Solution

For a first order reaction, t = (2.303/k)log(a/(a - x)). For 90% completion, t90 = (2.303/k)log(100/10) = 2.303/k. For 99% completion, t99 = (2.303/k)log(100/1) = 2(2.303/k). Hence t99 = 2t90.

Answer:

t99% = 2 t90%.

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Q.3.19A first order reaction takes 40 min for 30% decomposition. Calculate t1/2.v
Solution

For 30% decomposition, 70% remains. k = (2.303/40)log(100/70) = 8.92 x 10^-3 min^-1. Therefore t1/2 = 0.693/k = 77.7 min.

Answer:

77.7 min.

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Q.3.20For the decomposition of azoisopropane to hexane and nitrogen at 543 K, the following data are obtained. t (sec) 0 360 720 P(mm of Hg) 35.0 54.0 63.0 Calculate the rate constant.v
Solution

For A(g) → products with one extra mole of gas formed, partial pressure of reactant at time t is PA = 2P0 - Pt. With P0 = 35.0 mm, at 360 s PA = 70 - 54 = 16 mm. k = (2.303/360)log(35/16) = 2.17 x 10^-3 s^-1. At 720 s, PA = 70 - 63 = 7 mm, giving k = 2.24 x 10^-3 s^-1. Thus k ≈ 2.2 x 10^-3 s^-1.

Answer:

k ≈ 2.2 x 10^-3 s^-1.

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Q.3.21The following data were obtained during the first order thermal decomposition of SO2Cl2 at a constant volume. SO2Cl2(g) → SO2(g) + Cl2(g) Experiment Time/s^-1 Total pressure/atm 1 0 0.5 2 100 0.6 Calculate the rate of the reaction when total pressure is 0.65 atm.v
Solution

For SO2Cl2(g) → SO2(g) + Cl2(g), pressure of SO2Cl2 at time t is 2P0 - Pt. At 100 s, PSO2Cl2 = 1.0 - 0.6 = 0.4 atm. k = (2.303/100)log(0.5/0.4) = 2.23 x 10^-3 s^-1. When total pressure is 0.65 atm, PSO2Cl2 = 1.0 - 0.65 = 0.35 atm. Rate = kP = (2.23 x 10^-3)(0.35) = 7.8 x 10^-4 atm s^-1.

Answer:

7.8 x 10^-4 atm s^-1.

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Q.3.23The rate constant for the decomposition of hydrocarbons is 2.418 × 10^-5 s^-1 at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponential factor.v
Solution

Arrhenius equation is k = A e^(-Ea/RT), so A = k e^(Ea/RT). Substituting k = 2.418 x 10^-5 s^-1, Ea = 179900 J mol^-1, R = 8.314 J mol^-1 K^-1 and T = 546 K gives A = 3.93 x 10^12 s^-1.

Answer:

A = 3.93 x 10^12 s^-1.

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Q.3.24Consider a certain reaction A → Products with k = 2.0 × 10^-2 s^-1. Calculate the concentration of A remaining after 100 s if the initial concentration of A is 1.0 mol L^-1.v
Solution

The units of k indicate a first order reaction. [A] = [A]0e^-kt = 1.0 e^-(2.0 x 10^-2)(100) = e^-2 = 0.135 mol L^-1.

Answer:

0.135 mol L^-1.

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Q.3.25Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with t1/2 = 3.00 hours. What fraction of sample of sucrose remains after 8 hours ?v
Solution

For first order decay, fraction remaining = (1/2)^(t/t1/2). Thus fraction after 8 h = (1/2)^(8/3) = 0.157.

Answer:

0.157 of the sample remains.

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Q.3.26The decomposition of hydrocarbon follows the equation k = (4.5 × 10^11 s^-1) e^-28000K/T Calculate Ea.v
Solution

Compare k = Ae^-Ea/RT with k = (4.5 x 10^11 s^-1)e^-28000K/T. Therefore Ea/R = 28000 K, so Ea = 28000 x 8.314 = 2.33 x 10^5 J mol^-1 = 233 kJ mol^-1.

Answer:

Ea = 233 kJ mol^-1.

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Q.3.27The rate constant for the first order decomposition of H2O2 is given by the following equation: log k = 14.34 – 1.25 × 10^4 K/T Calculate Ea for this reaction and at what temperature will its half-period be 256 minutes?v
Solution

For Arrhenius form in common logarithms, log k = log A - Ea/(2.303RT). Thus Ea/(2.303R) = 1.25 x 10^4 K, so Ea = 2.303 x 8.314 x 1.25 x 10^4 = 2.39 x 10^5 J mol^-1. If t1/2 = 256 min, k = 0.693/(256 x 60) = 4.51 x 10^-5 s^-1. Substituting in log k = 14.34 - 1.25 x 10^4/T gives T ≈ 669 K.

Answer:

Ea = 239 kJ mol^-1; T ≈ 669 K.

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Q.3.28The decomposition of A into product has value of k as 4.5 × 10^3 s^-1 at 10°C and energy of activation 60 kJ mol^-1. At what temperature would k be 1.5 × 10^4 s^-1?v
Solution

Use log(k2/k1) = Ea/(2.303R)(1/T1 - 1/T2). Here k1 = 4.5 x 10^3 s^-1, k2 = 1.5 x 10^4 s^-1, T1 = 283 K and Ea = 60000 J mol^-1. Solving gives T2 = 297 K, or about 24°C.

Answer:

297 K, approximately 24°C.

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Q.3.29The time required for 10% completion of a first order reaction at 298K is equal to that required for its 25% completion at 308K. If the value of A is 4 × 10^10 s^-1. Calculate k at 318K and Ea.v
Solution

For equal times, k298/k308 = log(100/90)/log(100/75). This gives the ratio of rate constants. Using log(k308/k298) = Ea/(2.303R)(1/298 - 1/308), Ea ≈ 7.67 x 10^4 J mol^-1. Then Arrhenius equation k = Ae^-Ea/RT with A = 4 x 10^10 s^-1 and T = 318 K gives k ≈ 1.02 x 10^-2 s^-1.

Answer:

Ea ≈ 76.7 kJ mol^-1; k at 318 K ≈ 1.02 x 10^-2 s^-1.

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Q.3.30The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature.v
Solution

Use log(k2/k1) = Ea/(2.303R)(1/T1 - 1/T2). Here k2/k1 = 4, T1 = 293 K and T2 = 313 K. Thus Ea = 2.303R log4 x (T1T2)/(T2 - T1) = 5.29 x 10^4 J mol^-1 = 52.9 kJ mol^-1.

Answer:

52.9 kJ mol^-1.

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