JEE Chemistry Chemical Kinetics MCQs Set 36

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Chapter-wise Objective Questions: Chemical Kinetics

Navigate directly to the 50 objective questions for Chemical Kinetics using the digital viewer below. Each practice set includes verified answer keys, allowing students to instantly cross-check their work and identify areas requiring further revision.

Multiple Choice Questions

Question 1: In a gaseous reaction important in the upper atmosphere, \( \ce{H2O} \) and O react bimolecularly to form two OH radicals. \( \Delta H \) for this reaction is 72 kJ at 500 K and \( E_a \) is 77 kJ. The \( E_a \) for the bimolecular recombination of two OH radicals to form \( \ce{H2O} \) and O is:
(a) 3 kJ
(b) 4 kJ
(c) 5 kJ
(d) 7 kJ
Show Answer & Explanation

Answer: (c) 5 kJ

Explanation:
1. For the forward (endothermic) reaction, \( \Delta H=E_{a,f}-E_{a,b} \).
2. \( E_{a,b}=77-72=5 \) kJ.

Teacher's Note:
1. Reverse activation energy = forward Ea minus dH.

Question 2: The half-lives and initial concentrations of a first order and a zero order reaction are the same. The ratio of the initial rate of the first order reaction to that of the zero order reaction is:
(a) \( 2\times0.693 \)
(b) \( \frac{1}{0.693} \)
(c) \( \frac{2}{0.693} \)
(d) 6.93
Show Answer & Explanation

Answer: (a) \( 2\times0.693 \)

Explanation:
1. First order: \( \text{rate}_1=k_1[A]_0=\frac{0.693}{t_{1/2}}[A]_0 \).
2. Zero order: \( t_{1/2}=\frac{[A]_0}{2k_0} \), so \( \text{rate}_0=k_0=\frac{[A]_0}{2t_{1/2}} \).
3. Ratio = \( \frac{0.693}{1/2}=2\times0.693 \).

Teacher's Note:
1. Write each rate in terms of the common half-life.

Question 3: For the first order reactions (A) \( \xrightarrow{K_1} \) product and (B) \( \xrightarrow{K_2} \) product, the ratio \( K_1/K_2 \) if 90% of (A) has reacted in time t while 99% of (B) has reacted in time 2t is:
(a) 2
(b) 1/2
(c) \( 2\left(\frac{1-2\log3}{2-\log11-2\log3}\right) \)
(d) 1
Show Answer & Explanation

Answer: (d) 1

Explanation:
1. \( K_1t=\ln\frac{100}{10}=\ln10 \).
2. \( K_2(2t)=\ln\frac{100}{1}=\ln100=2\ln10 \), so \( K_2t=\ln10 \).
3. Hence \( K_1/K_2=1 \).

Teacher's Note:
1. Use \( k t \) = ln(initial/remaining) for each reaction.

Question 4: For a reaction \( A\to \) Product, the rate law is \( -\frac{d[A]}{dt}=K[A] \) (first order). The concentration of A left after time \( t=1/K \) is:
(a) \( [A]_0e \)
(b) \( \frac{[A]_0}{e} \)
(c) \( \frac{[A]_0}{e^2} \)
(d) \( \frac{1}{[A]_0} \)
Show Answer & Explanation

Answer: (b) \( \frac{[A]_0}{e} \)

Explanation:
1. Integrating, \( [A]=[A]_0e^{-Kt} \).
2. At \( t=1/K \): \( [A]=[A]_0e^{-1}=\frac{[A]_0}{e} \).

Teacher's Note:
1. The mean life \( 1/K \) is the time to fall to \( 1/e \).

Question 5: The reaction \( \ce{A(g) + 2B(g) -> C(g) + D(g)} \) is an elementary process. In an experiment, the initial partial pressures of A and B are \( P_A=0.6 \) and \( P_B=0.8 \) atm. The rate of reaction relative to the initial rate when \( P_C \) becomes 0.2 atm is:
(a) \( \frac14 \)
(b) \( \frac16 \)
(c) \( \frac1{10} \)
(d) 2
Show Answer & Explanation

Answer: (b) \( \frac16 \)

Explanation:
1. Rate = \( kP_AP_B^2 \) (elementary). Initial: \( 0.6\times0.8^2=0.384 \).
2. When \( P_C=0.2 \): \( A \) has lost 0.2 and \( B \) has lost 0.4, so \( P_A=0.4 \) and \( P_B=0.4 \).
3. New rate = \( 0.4\times0.4^2=0.064 \); ratio = \( \frac{0.064}{0.384}=\frac16 \).

Teacher's Note:
1. \( B \) is consumed twice as fast as \( A \).

Question 6: For the reaction \( A\to B \), the rate constant is \( 1.2\times10^{-2} \) M s\(^{-1}\). The concentration of B after 10 min, if we start with 10 M of A, is:
(a) 2.8 M
(b) 8.2 M
(c) 2.7 M
(d) 7.2 M
Show Answer & Explanation

Answer: (d) 7.2 M

Explanation:
1. The units of k (M per \( s \)) show a zero order reaction.
2. \( [B]=kt=1.2\times10^{-2}\times600=7.2 M \) (less than \( 10 M \), so \( A \) is not exhausted).

Teacher's Note:
1. Units of \( k \) reveal the order: \( M \) per \( s \) means zero order.

Question 7: Two substances A (\( t_{1/2}=5 \) min) and B (\( t_{1/2}=15 \) min) follow first order kinetics and are taken so that initially \( [A]=4[B] \). The time after which the concentrations of both will be equal is:
(a) 15 min
(b) 20 min
(c) 10 min
(d) 25 min
Show Answer & Explanation

Answer: (a) 15 min

Explanation:
1. \( [A]=4[B]_0\,2^{-t/5} \) and \( [B]=[B]_0\,2^{-t/15} \).
2. Equal when \( 4=2^{t/5-t/15}=2^{2t/15} \), so \( \frac{2t}{15}=2 \) and \( t=15 \min \).

Teacher's Note:
1. Write each concentration using powers of 1/2.

Question 8: For the reaction \( A+B\to C+D \) the rate law is rate = \( k[A]^{1/2}[B]^{1/2} \). Starting with 1 M each of A and B, the time taken for the concentration of A to become 0.25 M is: (\( k=2.303\times10^{-3} \) s\(^{-1}\))
(a) 300 s
(b) 900 s
(c) 600 s
(d) 1200 s
Show Answer & Explanation

Answer: (c) 600 s

Explanation:
1. A and \( B \) are used up equally, so \( [A] = [B] \) throughout and rate = \( k[A] \): effectively first order.
2. \( t=\frac{2.303}{k}\log\frac{1}{0.25}=\frac{2.303}{2.303\times10^{-3}}\times0.602\approx600 s \).

Teacher's Note:
1. With equal concentrations the half-order terms combine to first order.

Question 9: A first order reaction is 50% complete in 20 minutes at 27 \(^\circ\)C and in 5 min at 47 \(^\circ\)C. The energy of activation is:
(a) 43.85 kJ per mol
(b) 11.97 kJ per mol
(c) 55.14 kJ per mol
(d) 6.65 kJ per mol
Show Answer & Explanation

Answer: (c) 55.14 kJ per mol

Explanation:
1. \( k\propto\frac1{t_{1/2}} \), so \( \frac{k_2}{k_1}=\frac{20}{5}=4 \).
2. \( E_a=\frac{2.303R\,T_1T_2}{T_2-T_1}\log4=\frac{2.303\times8.314\times300\times320}{20}\times0.602\approx55\ \text{kJ per mol} \).

Teacher's Note:
1. Use the Arrhenius equation between two temperatures.

Question 10: Under the same reaction conditions, an initial concentration of 1.386 mol dm\(^{-3}\) of a substance becomes half in 40 seconds and 20 seconds through first order and zero order kinetics respectively. The ratio \( \frac{k_1}{k_0} \) of the rate constants for first order (\( k_1 \)) and zero order (\( k_0 \)) reactions is:
(a) 0.5 mol\(^{-1}\) dm\(^3\)
(b) 1.0 mol dm\(^{-3}\)
(c) 2.0 mol\(^{-1}\) dm\(^3\)
(d) 0.1 mol dm\(^{-3}\)
Show Answer & Explanation

Answer: (a) 0.5 mol\(^{-1}\) dm\(^3\)

Explanation:
1. First order: \( k_1=\frac{0.693}{40}=0.01733 s ^{-1} \).
2. Zero order: \( k_0=\frac{[A]_0}{2t_{1/2}}=\frac{1.386}{40}=0.03465 \) mol \( dm ^{-3} s ^{-1} \).
3. \( \frac{k_1}{k_0}=0.5 \) mol\( ^{-1} dm ^3 \).

Teacher's Note:
1. Zero order half-life is \( [A]0 / 2k \).

Question 11: The concentration of R in the reaction \( R\to P \) was measured as a function of time: [R] = 1.0, 0.75, 0.40, 0.10 M at t = 0.0, 0.05, 0.12, 0.18 min. The order of the reaction is:
(a) 1
(b) 2
(c) 4
(d) Zero
Show Answer & Explanation

Answer: (d) Zero

Explanation:
1. Fall in concentration per unit time: \( \frac{0.25}{0.05}=5 \), \( \frac{0.60}{0.12}=5 \), \( \frac{0.90}{0.18}=5 \).
2. The rate is constant, so the reaction is zero order.

Teacher's Note:
1. Constant rate of decrease means zero order.

Question 12: A chemical reaction is first order in A and second order in B. Experiment 1: rate 0.10 M s\(^{-1}\), [A] = 1.0 M, [B] = 0.20 M. Experiment 2: [A] = 2.0 M, [B] = 0.60 M. The rate in experiment 2 is:
(a) 0.20 M s\(^{-1}\)
(b) 1.8 M s\(^{-1}\)
(c) 1.2 M s\(^{-1}\)
(d) 0.36 M s\(^{-1}\)
Show Answer & Explanation

Answer: (b) 1.8 M s\(^{-1}\)

Explanation:
1. Rate = \( k[A][B]^2 \).
2. \( \frac{r_2}{r_1}=\frac{2.0}{1.0}\times\left(\frac{0.60}{0.20}\right)^2=2\times9=18 \).
3. \( r_2=18\times0.10=1.8 M s ^{-1} \).

Teacher's Note:
1. Scale the rate by each factor raised to its order.

Question 13: In a reaction carried out at 400 K, 0.0001% of the total number of collisions are effective. The energy of activation of the reaction is: (take R = 2 cal per mol per K)
(a) Zero
(b) 11.05 kcal per mol
(c) 7.37 kcal per mol
(d) 9.212 kcal per mol
Show Answer & Explanation

Answer: (b) 11.05 kcal per mol

Explanation:
1. Fraction of effective collisions = \( e^{-E_a/RT}=10^{-6} \).
2. \( E_a=2.303\times RT\times6=2.303\times2\times400\times6\approx11050 \) cal.

Teacher's Note:
1. 0.0001% equals \( 10^{-6} \) as a fraction.

Question 14: A solution of \( \ce{N2O5} \) in \( \ce{CCl4} \) yields, by decomposition at 45 \(^\circ\)C, 5 mL of \( \ce{O2} \) 20 minutes after the start and 10 mL of \( \ce{O2} \) after a very long time. The decomposition obeys first order kinetics. The volume of \( \ce{O2} \) evolved 40 minutes after the start is:
(a) 7.25 mL
(b) 1.5 mL
(c) 6 mL
(d) 7.5 mL
Show Answer & Explanation

Answer: (d) 7.5 mL

Explanation:
1. Half of the total oxygen (5 of 10 mL) is formed in \( 20 \min \), so \( t_{1/2}=20 \min \).
2. After \( 40 \min \) (two half-lives), 75% has decomposed: \( 0.75\times10=7.5 \) mL.

Teacher's Note:
1. Volume of \( O2 \) is proportional to the amount decomposed.

Question 15: Two first-order reactions have half-lives in the ratio 3 : 2. The ratio of the time intervals \( t_1:t_2 \), where \( t_1 \) is the time for 25% completion of the first reaction and \( t_2 \) is the time for 75% completion of the second, is:
(a) 0.311 : 1
(b) 0.420 : 1
(c) 0.273 : 1
(d) 0.119 : 1
Show Answer & Explanation

Answer: (a) 0.311 : 1

Explanation:
1. \( k_1:k_2=2:3 \) (inverse of half-lives).
2. \( t_1=\frac{\ln(1/0.75)}{k_1}=\frac{0.2877}{k_1} \) and \( t_2=\frac{\ln4}{k_2}=\frac{1.386}{k_2} \).
3. \( \frac{t_1}{t_2}=0.2075\times\frac{k_2}{k_1}=0.2075\times1.5=0.311 \).

Teacher's Note:
1. Use \( k = 0.693 \) / half-life for each.

Question 16: A certain reaction obeys the rate equation (in integrated form) \( \left[C^{(1-n)}-C_0^{(1-n)}\right]=(n-1)kt \), where n is the order, \( C_0 \) is the initial concentration and C the concentration after time t. The unit of k for n = 3 is:
(a) sec\(^{-1}\)
(b) litre mol\(^{-1}\) sec\(^{-1}\)
(c) litre\(^2\) mol\(^{-2}\) sec\(^{-1}\)
(d) mol litre\(^{-1}\) sec\(^{-1}\)
Show Answer & Explanation

Answer: (c) litre\(^2\) mol\(^{-2}\) sec\(^{-1}\)

Explanation:
1. For \( n = 3 \) the left side has units of \( C^{-2} \), that is litre\(^2\) mol\(^{-2}\).
2. \( k=\frac{C^{-2}}{t} \) has units litre\(^2\) mol\( ^{-2} \sec ^{-1} \).

Teacher's Note:
1. Unit of \( k = (conc)^{1-n} \) per time.

Question 17: The half-life of a first order reaction is one hour. The time interval during which the concentration is reduced to \( \frac78 \) of its initial value is:
(a) 6.32 minutes
(b) 17.4 minutes
(c) 11.6 minutes
(d) 23.2 minutes
Show Answer & Explanation

Answer: (c) 11.6 minutes

Explanation:
1. \( k=\frac{0.693}{60}=0.01155 \min ^{-1} \).
2. \( t=\frac1k\ln\frac87=\frac{0.1335}{0.01155}\approx11.6 \min \).

Teacher's Note:
1. Remaining 7/8 means the ratio initial/remaining is 8/7.

Question 18: The time elapsed between 33% and 67% completion of a first order reaction is 30 minutes. The time needed for 25% completion is:
(a) 12.3 min
(b) 15.5 min
(c) 18.5 min
(d) 16.5 min
Show Answer & Explanation

Answer: (a) 12.3 min

Explanation:
1. \( kt=\ln\frac{67\text{ left}}{33\text{ left}}\): between the two stages, \( k\times30=\ln\frac{0.67}{0.33}=0.708 \), so \( k=0.0236 \min ^{-1} \).
2. For 25% completion, \( t=\frac{\ln(1/0.75)}{k}=\frac{0.2877}{0.0236}\approx12.2 \min \), which matches the nearest option \( 12.3 \min \).

Teacher's Note:
1. At 33% completion 67% remains; at 67% completion 33% remains.

Question 19: At a certain temperature, the half-life periods for the catalytic decomposition of \( \ce{NH3} \) were found to be: pressure 50, 100, 200 mm Hg with half-life 3.52, 1.76, 0.88 hours. The pressure at which the half-life period is 1.5 hours is:
(a) 206 mm Hg
(b) 89 mm Hg
(c) 160 mm Hg
(d) 117 mm Hg
Show Answer & Explanation

Answer: (d) 117 mm Hg

Explanation:
1. Half-life is inversely proportional to pressure: \( t_{1/2}\times P=176 \) in every case, the behaviour of a zero order reaction.
2. For \( t_{1/2}=1.5 h \), \( P=\frac{176}{1.5}\approx117 \) mm Hg.

Teacher's Note:
1. If \( t1/2 x P \) is constant the reaction is zero order.

Question 20: The kinetics of \( \ce{A(g) -> 2B(g) + C(g)} \) is followed by measuring the total pressure. The rate constant is \( 1\times10^{-3} \) s\(^{-1}\) (first order) and the initial pressure of A is 0.5 atm. The rate of reaction \( -\frac{d[A]}{dt} \) when the total pressure is 0.7 atm is:
(a) \( 2.0\times10^{-4} \) atm s\(^{-1}\)
(b) \( 4.0\times10^{-4} \) atm s\(^{-1}\)
(c) \( 5.0\times10^{-4} \) atm s\(^{-1}\)
(d) \( 7.0\times10^{-4} \) atm s\(^{-1}\)
Show Answer & Explanation

Answer: (b) \( 4.0\times10^{-4} \) atm s\(^{-1}\)

Explanation:
1. If \( x \) atm of \( A \) decomposes, total pressure = \( 0.5+2x=0.7 \), so \( x = 0.1 \) and \( P_A=0.4 \) atm.
2. Rate = \( kP_A=10^{-3}\times0.4=4.0\times10^{-4} \) atm \( s ^{-1} \).

Teacher's Note:
1. Total pressure rises by \( 2x \) because one \( A \) gives three gas molecules.

Question 21: For \( \ce{N2 + 3H2 -> 2NH3} \) with \( \Delta H=-22 \) kcal, the activation energy \( E_a \) is 70 kcal. Hence \( E_a \) of \( \ce{2NH3 -> N2 + 3H2} \) is:
(a) 70 kcal
(b) 92 kcal
(c) -70 kcal
(d) -92 kcal
Show Answer & Explanation

Answer: (b) 92 kcal

Explanation:
1. The forward reaction releases 22 kcal, so the reverse is endothermic by 22 kcal.
2. \( E_{a,rev}=E_{a,fwd}+22=92 \) kcal.

Teacher's Note:
1. Reverse Ea = forward Ea + heat released by the forward reaction.

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Chemical Kinetics Objective Questions & Solutions for JEE Chemistry

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