JEE Chemistry Chemical Kinetics MCQs Set 13

Answer: (a) A high activation energy usually implies a fast reaction

Explanation:
1. The Arrhenius equation is \(k=Ae^{-E_a/RT}\).
2. A high activation energy makes \(e^{-E_a/RT}\) small, so the reaction is usually slow, not fast.
3. The other three statements describe correctly the effect of temperature, the sensitivity to \(E_a\) and the meaning of A.

Teacher's Note:
1. The larger the barrier, the fewer molecules cross it.
2. Rate constants of reactions with large \(E_a\) change most with temperature.

Answer: (a) 0.5 mol^\(-1\) dm³

Explanation:
1. For first order, \(k_1=\dfrac{0.693}{t_{1/2}}=\dfrac{0.693}{40}=0.0173\text{ s}^{-1}\).
2. For zero order, \(t_{1/2}=\dfrac{[A]_0}{2k_0}\), so \(k_0=\dfrac{1.386}{2\times20}=0.0347\text{ mol dm}^{-3}\text{ s}^{-1}\).
3. \(\dfrac{k_1}{k_0}=\dfrac{0.0173}{0.0347}=0.5\) mol^\(-1\) dm³.

Teacher's Note:
1. The units of the ratio follow from the units of each constant.
2. Zero order half-life depends on the initial concentration, first order does not.

Answer: (a) 1

Explanation:
1. The pressure of X falls from 800 to 400 mm Hg in the first 100 min and from 400 to 200 mm Hg in the next 100 min.
2. The time taken to halve the partial pressure is constant (100 min), which is the characteristic of a first order reaction.

Teacher's Note:
1. Constant half-life means first order.
2. For zero order the half-life would decrease and for second order it would increase as the concentration falls.

Answer: (b) \(E_b < E_f\)

Explanation:
1. \(\Delta H = E_f - E_b\).
2. For an endothermic reaction \(\Delta H\) is positive, so \(E_f > E_b\), that is, \(E_b < E_f\).

Teacher's Note:
1. In an endothermic reaction the products lie higher in energy than the reactants.
2. The forward barrier is therefore higher than the backward barrier.

Answer: (d) \(-20\)

Explanation:
1. \(\Delta H = E_f - E_b = 180 - 200 = -20\) kJ mol^\(-1\).
2. In the presence of the catalyst both barriers fall by 100 kJ mol^\(-1\): \(E_f = 80\) and \(E_b = 100\), so the difference is still \(-20\) kJ mol^\(-1\).
3. A catalyst does not change \(\Delta H\).

Teacher's Note:
1. A catalyst changes the path but not the initial and final energies.
2. Therefore \(\Delta H\) is unchanged.

Answer: (a) L mol^\(-1\) s^\(-1\)

Explanation:
1. The half-life of B is independent of its concentration, so the reaction is first order in B.
2. Doubling [A] doubles the rate, so the reaction is first order in A.
3. The overall order is 2, so the unit of k is \(\text{mol}^{1-n}\text{L}^{n-1}\text{s}^{-1}\) with n = 2, that is L mol^\(-1\) s^\(-1\).

Teacher's Note:
1. Half-life independent of concentration means first order.
2. Add the individual orders to get the overall order before choosing the unit.

Answer: (c) 2

Explanation:
1. The rate is given by the slow step: rate = \(k_2[\ce{NOBr2}][\ce{NO}]\).
2. The intermediate is eliminated using the fast equilibrium, \([\ce{NOBr2}] = K[\ce{NO}][\ce{Br2}]\).
3. Rate = \(k_2K[\ce{NO}]^2[\ce{Br2}]\), so the order with respect to NO is 2.

Teacher's Note:
1. The rate law cannot contain an intermediate.
2. Substitute the equilibrium expression for the intermediate.

Answer: (b) 46.06 minutes

Explanation:
1. \(k=\dfrac{0.693}{t_{1/2}}=\dfrac{0.693}{6.93}=0.1\text{ min}^{-1}\).
2. \(t=\dfrac{2.303}{k}\log\dfrac{100}{1}=\dfrac{2.303}{0.1}\times2=46.06\) minutes.

Teacher's Note:
1. For 99% completion \([A]_0/[A]=100\), so the logarithm equals 2.
2. This takes about 6.64 half-lives.

Answer: (b) 55.14 kJ/mol

Explanation:
1. For first order, \(k\propto 1/t_{1/2}\), so \(k_2/k_1 = 20/5 = 4\).
2. \(\log\dfrac{k_2}{k_1}=\dfrac{E_a}{2.303R}\left(\dfrac1{T_1}-\dfrac1{T_2}\right)\) with \(T_1=300\) K and \(T_2=320\) K gives \(0.602=\dfrac{E_a}{2.303\times8.314}\times\dfrac{20}{300\times320}\).
3. \(E_a\approx 55\) kJ/mol; the option nearest to this is 55.14 kJ/mol.

Teacher's Note:
1. A rise of 20 K increased the rate four times.
2. Use absolute temperatures in the Arrhenius equation.

Answer: (d) 3

Explanation:
1. For order n, \(t_{1/2}\propto P^{1-n}\), so \(\dfrac{950}{235}=\left(\dfrac{500}{250}\right)^{n-1}\).
2. Taking logarithms, \(\log 95-\log 23.5 = 1.97-1.37 = 0.60=(n-1)\log2 = (n-1)(0.30)\).
3. So \(n-1=2\) and \(n=3\).

Teacher's Note:
1. The half-life increased about four times when the pressure was halved.
2. That matches \(2^{n-1}=4\), i.e. a third order reaction.

Answer: (c) \(\dfrac{nA_0}{n+1}\)

Explanation:
1. At any time, \([A]=A_0-\dfrac{[B]}{n}\), because each mole of A gives n moles of B.
2. At the point of intersection \([A]=[B]\), so \([B]=A_0-\dfrac{[B]}{n}\).
3. \([B]\left(1+\dfrac1n\right)=A_0\), hence \([B]=\dfrac{nA_0}{n+1}\).

Teacher's Note:
1. The rising curve is B and the falling curve is A.
2. Use the stoichiometry to link the two concentrations.

Answer: (b) 8

Explanation:
1. Let the pressure of \(\ce{A2}\) that reacts be x.
2. Total pressure = \((100-x)+x+x/2=100+x/2\), so \(x/2=20\) and \(x=40\) mm.
3. Rate of disappearance of \(\ce{A2}\) = \(40/5=8\) mm min^\(-1\).

Teacher's Note:
1. One mole of A2 gives 1.5 moles of gas, so the net increase is half of the pressure consumed.
2. So the pressure consumed is twice the pressure rise.

Answer: (a) Zero

Explanation:
1. Let the initial pressure be \(P_0\) and let 2x of A react. Total pressure \(=P_0-2x+3x+x=P_0+2x\).
2. Doubling means \(2x=P_0\), that is all of A has been consumed in 3 hours.
3. Only a zero order reaction can go to completion in a finite time; a first or second order reaction only approaches completion, so the order may be zero.

Teacher's Note:
1. The pressure rise equals the amount of A reacted when all of it is used up.
2. The rate of a zero order reaction does not depend on concentration, so it reaches zero concentration in finite time.

Answer: (d) 0.4 mol/min

Explanation:
1. The rate expression has no concentration term, so the reaction is zero order and the rate is constant.
2. [A] falls by 2 mol every 5 min, so \(k = 2/5 = 0.4\) mol/min.
3. The rate at 20 minutes is the same, 0.4 mol/min.

Teacher's Note:
1. A rate equal to k indicates zero order.
2. The rate does not change as the reaction proceeds.

Answer: (b) 15 min

Explanation:
1. Let \([B]_0=b\). Then \([A]_0=4b\).
2. After time t, \([A]=4b\times2^{-t/5}\) and \([B]=b\times2^{-t/15}\).
3. Equating, \(4=2^{t/5-t/15}=2^{2t/15}\), so \(\dfrac{2t}{15}=2\) and \(t=15\) min.

Teacher's Note:
1. A decays faster than B because of its shorter half-life.
2. The initial excess of A is removed after it has halved twice more than B has.