JEE Chemistry Chemical Kinetics MCQs Set 18

Answer: (c) log k against 1/T

Explanation:
1. \( \log k = \log A - \dfrac{E_a}{2.303R}\cdot\dfrac{1}{T} \).
2. This is a straight line in log k against 1/T with slope \( -E_a/2.303R \).

Teacher's Note:
1. Compare with y = c + mx.
2. The variable multiplying the slope is 1/T.

Answer: (b) 349.3

Explanation:
1. \( \ln k = \ln A - E_a/RT \), so \( \ln A = \ln k + E_a/RT \).
2. \( \ln k = \ln(10^{-5}) = -11.51 \).
3. \( E_a/RT = 1.8 \times 10^6/(8.314 \times 600) = 360.8 \), so \( \ln A = 360.8 - 11.5 = 349.3 \).

Teacher's Note:
1. Rearrange the Arrhenius equation to get ln A.
2. Keep Ea in joules when R is in J.

Answer: (c) 1.5

Explanation:
1. The slow step gives rate = k[X][Y2].
2. From the fast equilibrium, \( [X]^2 = K[X_2] \), so \( [X] = K^{1/2}[X_2]^{1/2} \).
3. Rate = k'[X2]^{1/2}[Y2], so the overall order is 0.5 + 1 = 1.5.

Teacher's Note:
1. Write the rate from the slow step and eliminate the intermediate.
2. Use the fast equilibrium to express the intermediate through reactants.

Answer: (a) 138.6 s

Explanation:
1. \( t = \dfrac{2.303}{k}\log\dfrac{20}{5} = \dfrac{2.303}{10^{-2}} \times 0.602 \).
2. \( t = 138.6 \) s (this is two half-lives, 2 × 69.3 s).

Teacher's Note:
1. A fall from 20 g to 5 g is two half-lives.
2. t1/2 = 0.693/k = 69.3 s.

Answer: (c) 6x

Explanation:
1. One mole of \( \ce{Cr2O7^{2-}} \) is used for every 6 moles of \( \ce{Fe^{3+}} \) formed.
2. Rate of appearance of Fe³⁺ = 6x.

Teacher's Note:
1. Compare the coefficients of the two species.
2. The species with the larger coefficient changes faster.

Answer: (b) \( \dfrac{1}{t}\ln\dfrac{P_0}{3P_0 - 2P_t} \)

Explanation:
1. Let p be the pressure of A that reacts. Then B = p and C = p/2.
2. \( P_t = P_0 - p + p + p/2 = P_0 + p/2 \), so \( p = 2(P_t - P_0) \).
3. \( P_A = P_0 - p = 3P_0 - 2P_t \), so \( k = \dfrac{1}{t}\ln\dfrac{P_0}{3P_0 - 2P_t} \).

Teacher's Note:
1. Define the reacted pressure first and then write the total pressure.
2. The coefficient 2 on A matters in this problem.

Answer: (a) 0.02 M min⁻¹

Explanation:
1. Average rate of disappearance of A = 0.4/10 = 0.04 M min⁻¹.
2. Two moles of A give one mole of B, so the rate of appearance of B = 0.04/2 = 0.02 M min⁻¹.

Teacher's Note:
1. Use the coefficients to convert the rate of A into that of B.
2. B forms at half the rate A disappears.

Answer: (b) the half-life of a first order reaction does not depend on \( [A]_0 \); the half-life of a second order reaction does depend on \( [A]_0 \)

Explanation:
1. First order: \( t_{1/2} = 0.693/k \), independent of \( [A]_0 \).
2. Second order: \( t_{1/2} = 1/(k[A]_0) \), which depends on \( [A]_0 \).
3. Rates of both depend on concentration, and either can be catalysed, so the other options are wrong.

Teacher's Note:
1. Half-life independence of concentration is the signature of first order.
2. Both orders have concentration-dependent rates.

Answer: (b) is doubled

Explanation:
1. For zero order, \( t_{1/2} = [A]_0/2k \).
2. It is directly proportional to \( [A]_0 \), so doubling the concentration doubles the half-life.

Teacher's Note:
1. Zero order half-life is proportional to initial concentration.
2. First order half-life is independent of it.

Answer: (a) \( t_{1/2} = 2\,t_{1/4} \)

Explanation:
1. For zero order, time is proportional to the amount reacted: \( t = x/k \).
2. \( t_{1/2} = [A]_0/2k \) and \( t_{1/4} = [A]_0/4k \).
3. So \( t_{1/2} = 2\,t_{1/4} \).

Teacher's Note:
1. The amount reacted is linear in time for zero order.
2. Half the reaction takes twice as long as a quarter of it.

Answer: (a) +2

Explanation:
1. For zero order the rate equals k, so \( k_1 = k_2 \).
2. \( A_1 e^{-Ea_1/RT} = A_2 e^{-Ea_2/RT} \), so \( \ln\dfrac{A_1}{A_2} = \dfrac{Ea_1 - Ea_2}{RT} \).
3. With \( Ea_1 - Ea_2 = 2RT \), the value is +2.

Teacher's Note:
1. Equal rates for zero order mean equal rate constants.
2. Take the natural log of the Arrhenius expressions.

Answer: (a) 25.7 min

Explanation:
1. \( k = \dfrac{2.303}{60}\log\dfrac{100}{20} = \dfrac{2.303 \times 0.699}{60} = 0.02683 \) min⁻¹.
2. \( t_{1/2} = 0.693/0.02683 = 25.8 \) min, which is closest to 25.7 min.

Teacher's Note:
1. 80% completion leaves 20% of the reactant.
2. Find k first and then the half-life.

Answer: (a) 6 min

Explanation:
1. For first order, rate is proportional to concentration, so the concentration falls by a factor 10 in 20 min.
2. \( k = \dfrac{2.303}{20}\log 10 = 0.1151 \) min⁻¹.
3. \( t_{1/2} = 0.693/0.1151 = 6.0 \) min.

Teacher's Note:
1. Use the ratio of rates as the ratio of concentrations.
2. A ten-fold fall takes 20 min here.

Answer: (c) 10

Explanation:
1. \( \ln\dfrac{K_2}{K_1} = \dfrac{Ea_1 - Ea_2}{RT} = \dfrac{5750}{8.314 \times 300} = 2.305 \).
2. \( \log\dfrac{K_2}{K_1} = 2.305/2.303 \approx 1 \), so \( K_2/K_1 \approx 10 \).

Teacher's Note:
1. The reaction with the lower activation energy is faster.
2. Convert kJ to J before dividing by RT.

Answer: (a) 1

Explanation:
1. \( \ln\dfrac{K_2}{K_1} = \dfrac{E_a}{R}\left(\dfrac{1}{300} - \dfrac{1}{310}\right) = \dfrac{83000}{8.314} \times 1.075 \times 10^{-4} \).
2. This is 9983 × 1.075 × 10⁻⁴ = 1.07, which is nearly 1.

Teacher's Note:
1. Use 300 K and 310 K for 27°C and 37°C.
2. Keep Ea in J mol⁻¹.

Answer: (c) \( -\dfrac{d[A]}{dt} = -\dfrac{1}{2}\dfrac{d[B]}{dt} = k[A][B]^2 \)

Explanation:
1. One mole of A reacts with two moles of B, so B disappears twice as fast as A.
2. Rate = \( -\dfrac{d[A]}{dt} = -\dfrac{1}{2}\dfrac{d[B]}{dt} \), which is the form in option c.

Teacher's Note:
1. Divide the rate of change of each reactant by its coefficient.
2. Reactants carry a minus sign.

Answer: (b) −2

Explanation:
1. Rate \( \propto [B]^n \).
2. \( 2^n = 1/4 = 2^{-2} \), so n = −2.

Teacher's Note:
1. A decrease in rate on increasing concentration means a negative order.
2. Write 1/4 as a power of 2.

Answer: (b) dc/dt = k[A]

Explanation:
1. Rows 1 and 2: [B] doubles with no change in rate, so the order in B is zero.
2. Rows 1 and 3: [A] doubles and the rate doubles, so the order in A is one.
3. Rate = k[A].

Teacher's Note:
1. Compare rows with only one concentration changed.
2. Zero order in a reactant means it drops out of the rate law.

Answer: (a) 2

Explanation:
1. The slow step gives rate = k[NOBr2][NO].
2. From the fast equilibrium, \( [NOBr_2] = K[NO][Br_2] \).
3. Rate = kK[NO]²[Br2], so the order with respect to NO is 2.

Teacher's Note:
1. Eliminate the intermediate using the fast equilibrium.
2. NO enters both steps, so its powers add.

Answer: (c) low probability of simultaneous collision of all the reacting species

Explanation:
1. A reaction of order greater than 3 would need four or more molecules to collide at the same time in the right orientation.
2. The probability of such a collision is extremely small.

Teacher's Note:
1. Simultaneous collisions of many molecules are very unlikely.
2. That is why elementary steps are uni-, bi- or termolecular.

Answer: (a) the degree of dissociation is equal to \( 1 - e^{-kt} \) and the pre-exponential factor in the Arrhenius equation has the dimension of time⁻¹

Explanation:
1. For first order, \( [A] = [A]_0 e^{-kt} \), so the fraction reacted is \( 1 - e^{-kt} \).
2. The pre-exponential factor A has the same unit as k, which is time⁻¹.
3. A plot of 1/[A] against t is a straight line only for second order, and half-life of first order does not depend on concentration.

Teacher's Note:
1. Check each statement against the first order equations.
2. Linear 1/[A] against t belongs to second order.

Answer: (b) \( 366\left(\dfrac{\ln 2}{\ln 10}\right) \) min

Explanation:
1. \( k = \dfrac{\ln(100/10)}{366} = \dfrac{\ln 10}{366} \).
2. \( t_{1/2} = \dfrac{\ln 2}{k} = 366\left(\dfrac{\ln 2}{\ln 10}\right) \) min, which is about 110 min.

Teacher's Note:
1. 90% decomposition leaves 10% of the reagent.
2. Then use \( t_{1/2} = \ln 2/k \).