JEE Chemistry Chemical Kinetics MCQs Set 19

Answer: (d) The half-life of a reaction is half the time required for the reaction to go to completion

Explanation:
1. A first order reaction never reaches 100% completion in finite time, so there is no definite time of completion.
2. The first three statements are standard facts.
3. Statement d is false.

Teacher's Note:
1. Successive half-lives are equal for a first order reaction.
2. A first order reaction only approaches completion.

Answer: (c) both a bimolecular and a second order reaction

Explanation:
1. One molecule of ester and one of NaOH take part in the rate-determining step, so it is bimolecular.
2. The rate depends on the concentration of each reactant to the first power, so it is second order.

Teacher's Note:
1. In alkaline hydrolysis the base is not in excess in the rate law.
2. So both molecularity and order are two.

Answer: (b) \( C_t\,e^{k_1 t} = C_0 \)

Explanation:
1. For first order, \( \ln\dfrac{C_0}{C_t} = k_1 t \), so \( C_t = C_0 e^{-k_1 t} \).
2. Rearranged, \( C_t\,e^{k_1 t} = C_0 \).
3. The other options have the wrong sign in the exponent.

Teacher's Note:
1. Concentration decays exponentially in a first order reaction.
2. The exponent is negative.

Answer: (c) 23.1 min

Explanation:
1. \( k = \text{rate}/[A] = 1.5 \times 10^{-2}/0.5 = 0.03 \) min⁻¹.
2. \( t_{1/2} = 0.693/0.03 = 23.1 \) min.

Teacher's Note:
1. For first order, rate = k[A].
2. Find k first and then the half-life.

Answer: (c) 30 min

Explanation:
1. The concentration halves in 15 min, so \( t_{1/2} = 15 \) min.
2. 0.1 M to 0.025 M is a fall by a factor 4, which is two half-lives.
3. Time = 2 × 15 = 30 min.

Teacher's Note:
1. Get the half-life from the first data pair.
2. Count how many halvings the second change needs.

Answer: (d) 347 s

Explanation:
1. \( k = \text{rate}/[A] = 2.0 \times 10^{-5}/0.01 = 2.0 \times 10^{-3} \) s⁻¹.
2. \( t_{1/2} = 0.693/(2.0 \times 10^{-3}) = 346.5 \) s, which is about 347 s.

Teacher's Note:
1. Rate divided by concentration gives k for first order.
2. Then use \( t_{1/2} = 0.693/k \).

Answer: (b) 2.0

Explanation:
1. Halving [A] changes the rate from 2.4 to 0.6, a factor of 4.
2. \( (1/2)^n = 1/4 \), so n = 2.

Teacher's Note:
1. Rate falls by the same power of the concentration change.
2. A fall to one fourth for halving means second order.

Answer: (c) 2

Explanation:
1. \( t_{1/2} \propto a^{1-n} \).
2. Doubling a halves \( t_{1/2} \), so \( 2^{1-n} = 1/2 \).
3. Hence 1 − n = −1 and n = 2.

Teacher's Note:
1. Half-life that falls when concentration rises means order above 1.
2. Use the exponent 1 − n.

Answer: (c) \( 3.47 \times 10^{-4} \) M min⁻¹

Explanation:
1. 0.1 M to 0.025 M is two half-lives, so \( t_{1/2} = 20 \) min and \( k = 0.693/20 = 0.03465 \) min⁻¹.
2. Rate = k[A] = 0.03465 × 0.01 = \( 3.47 \times 10^{-4} \) M min⁻¹.

Teacher's Note:
1. A fall by a factor of 4 is two half-lives.
2. For first order, rate = k[A].

Answer: (d) \( k_1 = k_2 e^{Ea_1/RT} \)

Explanation:
1. With equal A, \( k_1 = A e^{-Ea_1/RT} \) and \( k_2 = A e^{-Ea_2/RT} \).
2. \( \dfrac{k_1}{k_2} = e^{(Ea_2 - Ea_1)/RT} = e^{Ea_1/RT} \) since \( Ea_2 - Ea_1 = Ea_1 \).
3. So \( k_1 = k_2 e^{Ea_1/RT} \).

Teacher's Note:
1. Divide the two Arrhenius expressions with the same A.
2. Then substitute Ea2 = 2Ea1.

Answer: (a) 53.6 kJ mol⁻¹

Explanation:
1. \( \log 2 = \dfrac{E_a}{2.303 \times 8.314}\left(\dfrac{310 - 300}{300 \times 310}\right) \).
2. \( E_a = 2.303 \times 8.314 \times 0.301 \times 9300 = 53{,}600 \) J mol⁻¹.
3. \( E_a = 53.6 \) kJ mol⁻¹.

Teacher's Note:
1. Use the Arrhenius equation in log form with the two temperatures.
2. Keep R in J K⁻¹ mol⁻¹.

Answer: (c) Assertion is true but the Reason is false.

Explanation:
1. The Assertion is true: rate is the change in concentration of a reactant or product per unit time.
2. The rate generally changes as the concentrations fall, and it stays constant only for a zero order reaction, so the Reason is false.

Teacher's Note:
1. Rate usually falls as the reactants are used up.
2. Only zero order reactions have a constant rate.

Answer: (c) Assertion is true but the Reason is false.

Explanation:
1. Molecularity is a whole number counting the molecules in an elementary step. It is rarely more than 3, so the Assertion is true.
2. Order can be zero, fractional or negative, so the Reason is false.

Teacher's Note:
1. Molecularity is theoretical and always a whole number.
2. Order is experimental and can be fractional.

Answer: (a) Both Assertion and Reason are true and the Reason is a correct explanation of the Assertion.

Explanation:
1. The elementary reaction \( \ce{H2 + I2 -> 2HI} \) involves one molecule of each reactant, so two molecules collide and it is bimolecular.
2. The Reason states exactly this and so explains the Assertion.

Teacher's Note:
1. Molecularity is the number of molecules colliding in the elementary step.
2. Count the reactant molecules in the step.

Answer: (c) Assertion is true but the Reason is false.

Explanation:
1. The rate constant depends only on temperature (and catalyst), so it stays the same when the concentration is doubled. The Assertion is true.
2. The rate, not the rate constant, varies with concentration, so the Reason is false.

Teacher's Note:
1. Rate depends on concentration and k does not.
2. Doubling the concentration doubles the rate of a first order reaction.

Answer: (c) Assertion is true but the Reason is false.

Explanation:
1. For first order, rate = k[A], so doubling [A] doubles the rate. The Assertion is true.
2. The rate constant stays the same, so the Reason is false.

Teacher's Note:
1. Concentration changes the rate and not the rate constant.
2. Only temperature or a catalyst changes k.

Answer: (c) Assertion is true but the Reason is false.

Explanation:
1. \( t_{1/2} = 0.693/k \) does not depend on concentration, so the Assertion is true.
2. \( t_{3/4} \) is two half-lives, so \( t_{3/4} = 2\,t_{1/2} \). The Reason states the reverse and is false.

Teacher's Note:
1. 75% completion takes two half-lives.
2. So t3/4 is twice t1/2, not half of it.

Answer: (c) Assertion is true but the Reason is false.

Explanation:
1. \( \Delta H = Ea(\text{forward}) - Ea(\text{backward}) > 0 \) for an endothermic reaction, so the Assertion is true.
2. Threshold energy is the energy of the activated complex, which is the same for the forward and backward reactions. The Reason is false.

Teacher's Note:
1. Both directions pass through the same activated complex.
2. Their activation energies differ because the starting energies differ.

Answer: (c) Assertion is true but the Reason is false.

Explanation:
1. A low barrier means more molecules can react, so the reaction is fast and the Assertion is true.
2. Reactions with a low Ea are less sensitive to temperature, so their temperature coefficient is small and the Reason is false.

Teacher's Note:
1. Higher Ea means a stronger effect of temperature.
2. So a low Ea gives a small temperature coefficient.

Answer: (c) Assertion is true but the Reason is false.

Explanation:
1. For many reactions the temperature coefficient is 2 to 3, so the Assertion is true.
2. Collision frequency rises by only about 1 to 2% for a 10°C rise, as it is proportional to \( \sqrt{T} \). The rate increases mainly because more molecules cross the activation barrier, so the Reason is false.

Teacher's Note:
1. Rate depends on the fraction of molecules with energy above Ea.
2. Collision frequency changes only slightly with temperature.

Answer: (c) Assertion is true but the Reason is false.

Explanation:
1. Both the forward and backward rate constants rise with temperature, so the rate of any elementary reaction increases. The Assertion is true.
2. Heating shifts equilibrium forward only for endothermic reactions, so the Reason is false.

Teacher's Note:
1. Higher temperature increases rate constants.
2. Direction of equilibrium shift depends on the sign of the heat change.

Answer: (c) Assertion is true but the Reason is false.

Explanation:
1. A catalyst speeds up both the forward and backward reactions, so the Assertion is true.
2. It lowers the activation energy of both directions by the same amount, so the Reason is false.

Teacher's Note:
1. A catalyst lowers the barrier in both directions.
2. Equilibrium is therefore reached faster but K does not change.