JEE Chemistry Chemical Kinetics MCQs Set 16

Multiple Choice Questions (MCQs) for JEE Chemistry: Chemical Kinetics

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Multiple Choice Questions

Question 1: For a first order reaction, the ratio of the time required for 99.9% completion to that for 50% completion is :
(a) 2
(b) 5
(c) 10
(d) none
Show Answer & Explanation

Answer: (c) 10

Explanation:
1. 99.9% completion takes \( \frac{2.303}{k}\log 1000 = \frac{6.909}{k} \).
2. 50% takes \( 0.693/k \).
3. Ratio = 6.909/0.693 ≈ 10.

Teacher's Note:
1. 99.9% is about 10 half-lives.

Question 2: From different sets of data of \( t_{1/2} \) at different initial concentrations a for a given reaction, the product \( t_{1/2} \times a \) is found to be constant. The order of the reaction is :
(a) 0
(b) 1
(c) 2
(d) 3
Show Answer & Explanation

Answer: (c) 2

Explanation:
1. \( t_{1/2} \propto 1/a \) corresponds to second order.

Teacher's Note:
1. Constant t₁/₂ × a means second order.

Question 3: The reaction \( \ce{2N2O5(g) -> 4NO2(g) + O2(g)} \) is first order in \( \ce{N2O5} \). Which graph gives a straight line?
(a) log \( p_{N_2O_5} \) against time with a negative slope
(b) \( p_{N_2O_5}^{-1} \) against time
(c) \( p_{N_2O_5} \) against time
(d) log \( p_{N_2O_5} \) against time with a positive slope
Show Answer & Explanation

Answer: (a) log \( p_{N_2O_5} \) against time with a negative slope

Explanation:
1. First order: \( \log p = \log p_0 - \dfrac{kt}{2.303} \).
2. A plot of log p against t is a straight line with negative slope.

Teacher's Note:
1. Take logs of the integrated first order law.

Question 4: The following data were obtained for the decomposition of ammonia at a certain temperature: pressure 50, 100, 200 mm with t₁/₂ = 3.64, 1.82, 0.91 respectively. The order of the reaction is :
(a) 0
(b) 1
(c) 2
(d) 3
Show Answer & Explanation

Answer: (c) 2

Explanation:
1. Doubling the pressure halves the half-life, so t₁/₂ ∝ 1/p.
2. This is a second order reaction.

Teacher's Note:
1. Check how t₁/₂ changes with initial pressure.

Question 5: A reaction has the rate constant x s⁻¹. By what factor is the rate increased if the initial concentration of A is tripled?
(a) 3
(b) 9
(c) x
(d) remains the same
Show Answer & Explanation

Answer: (a) 3

Explanation:
1. The unit s⁻¹ means first order, so rate = kC.
2. Tripling C triples the rate.

Teacher's Note:
1. The unit of k gives the order.

Question 6: Which statement is incorrect?
(a) Half-life of a first order reaction is independent of initial concentration
(b) Rate of reaction is constant for a first order reaction
(c) Unit of k for a second order reaction is mol⁻¹ L s⁻¹
(d) Half-life of a zero order reaction is proportional to the initial concentration
Show Answer & Explanation

Answer: (b) Rate of reaction is constant for a first order reaction

Explanation:
1. For a first order reaction the rate falls as the concentration falls (rate = kC).
2. The rate constant is fixed, not the rate.

Teacher's Note:
1. Do not confuse rate with rate constant.

Question 7: Hydrolysis of an ester in alkaline medium is :
(a) first order with molecularity one
(b) second order with molecularity two
(c) first order with molecularity two
(d) second order with molecularity one
Show Answer & Explanation

Answer: (b) second order with molecularity two

Explanation:
1. Rate = k[ester][OH⁻], so the order is 2.
2. Both species collide in the step, so the molecularity is 2.

Teacher's Note:
1. In alkaline hydrolysis neither reactant is in large excess.

Question 8: The expression which gives the time in which one-fourth of the reactant reacts in a first order reaction is :
(a) \( \dfrac{k}{2.303}\log\dfrac{4}{3} \)
(b) \( \dfrac{2.303}{k}\log 3 \)
(c) \( \dfrac{2.303}{k}\log\dfrac{3}{4} \)
(d) \( \dfrac{2.303}{k}\log\dfrac{4}{3} \)
Show Answer & Explanation

Answer: (d) \( \dfrac{2.303}{k}\log\dfrac{4}{3} \)

Explanation:
1. One fourth reacted means 3/4 is left, so \( a/(a-x) = 4/3 \).
2. \( t = \dfrac{2.303}{k}\log\dfrac{4}{3} \).

Teacher's Note:
1. Remaining fraction is 3/4.

Question 9: The rate constant of a zero order reaction is 0.2 mol dm⁻³ h⁻¹. If the concentration of the reactant after 30 minutes is 0.05 mol dm⁻³, its initial concentration was :
(a) 6.05 mol dm⁻³
(b) 0.15 mol dm⁻³
(c) 0.25 mol dm⁻³
(d) 4.00 mol dm⁻³
Show Answer & Explanation

Answer: (b) 0.15 mol dm⁻³

Explanation:
1. Amount used = 0.2 × 0.5 h = 0.1 mol dm⁻³.
2. Initial = 0.05 + 0.1 = 0.15 mol dm⁻³.

Teacher's Note:
1. Convert 30 min to hours.

Question 10: A reaction is of first order. After 100 minutes 75 g of reactant A are decomposed when 100 g are taken initially. The time required when 150 g of A are decomposed if the initial weight taken is 200 g is :
(a) 100 minutes
(b) 200 minutes
(c) 150 minutes
(d) 175 minutes
Show Answer & Explanation

Answer: (a) 100 minutes

Explanation:
1. 75% decomposition is 2 half-lives, so t₁/₂ = 50 min.
2. 150 g of 200 g is also 75%, so the time is again 100 min.

Teacher's Note:
1. Same fraction gives the same time.

Question 11: For which reaction are the unit of rate and the unit of rate constant identical?
(a) first order reaction
(b) zero order reaction
(c) second order reaction
(d) fractional order reaction
Show Answer & Explanation

Answer: (b) zero order reaction

Explanation:
1. Zero order: rate = k, so both have the same unit, mol L⁻¹ s⁻¹.

Teacher's Note:
1. The unit of k is (conc)^{1−n} time⁻¹.

Question 12: A first order reaction \( \ce{2A -> products} \) has a half-life of 20 min at [A]₀ = 0.2 mol L⁻¹. The rate constant is :
(a) 4 s⁻¹
(b) 20 s⁻¹
(c) 4 L mol⁻¹ min⁻¹
(d) 0.034 min⁻¹
Show Answer & Explanation

Answer: (d) 0.034 min⁻¹

Explanation:
1. \( k = 0.693/t_{1/2} = 0.693/20 = 0.0347 \) min⁻¹.

Teacher's Note:
1. First order k = 0.693 divided by the half-life.

Question 13: The correct statement about a first order reaction is :
(a) \( t_{completion} \) is finite
(b) \( t_{1/2} \propto 1/a \)
(c) unit of k is mol L⁻¹ s⁻¹
(d) \( t_{1/2} \times k \) is constant at constant temperature
Show Answer & Explanation

Answer: (d) \( t_{1/2} \times k \) is constant at constant temperature

Explanation:
1. \( t_{1/2} = 0.693/k \), so \( t_{1/2} \times k = 0.693 \), a constant.
2. Completion time is infinite, and the unit of k is s⁻¹.

Teacher's Note:
1. The product of k and t₁/₂ is 0.693.

Question 14: According to collision theory of reaction rates :
(a) every collision between reactants leads to a chemical reaction
(b) rate of reaction is proportional to the velocity of molecules
(c) all reactions which occur in the gaseous phase are zero order reactions
(d) rate of reaction is directly proportional to collision frequency
Show Answer & Explanation

Answer: (d) rate of reaction is directly proportional to collision frequency

Explanation:
1. Only effective collisions react, so (a) is false.
2. Rate is proportional to the collision frequency times the fraction of effective collisions.

Teacher's Note:
1. Effective collisions need energy and orientation.

Question 15: Activation energy of a reaction is :
(a) the energy released during the reaction
(b) the energy evolved when the activated complex is formed
(c) the minimum extra amount of energy needed to overcome the potential barrier of the reaction
(d) the energy needed to form one mole of the product
Show Answer & Explanation

Answer: (c) the minimum extra amount of energy needed to overcome the potential barrier of the reaction

Explanation:
1. Activation energy is the extra energy reactants must gain to cross the barrier.

Teacher's Note:
1. It is the height of the barrier above the reactants.

Question 16: Chemical reactions occur as a result of collisions between reacting molecules. Therefore the reaction rate is given by :
(a) total number of collisions occurring in a unit volume per second
(b) fraction of molecules which possess energy less than the threshold energy
(c) total number of effective collisions
(d) temperature
Show Answer & Explanation

Answer: (c) total number of effective collisions

Explanation:
1. Only collisions with enough energy and proper orientation lead to reaction.
2. Rate depends on the number of effective collisions.

Teacher's Note:
1. Effective collisions decide the rate.

Question 17: An endothermic reaction A → B has an activation energy of 15 kcal/mol and the heat of reaction is 5 kcal/mol. The activation energy of the reaction B → A is :
(a) 20 kcal/mol
(b) 15 kcal/mol
(c) 10 kcal/mol
(d) zero
Show Answer & Explanation

Answer: (c) 10 kcal/mol

Explanation:
1. E_a(reverse) = E_a(forward) − ΔH for the forward reaction = 15 − 5 = 10 kcal/mol.

Teacher's Note:
1. Reverse barrier = forward barrier minus ΔH.

Question 18: The rate of a chemical reaction doubles for every 10°C rise in temperature. If the temperature is increased by 60°C, the rate of the reaction increases by :
(a) 20 times
(b) 32 times
(c) 64 times
(d) 128 times
Show Answer & Explanation

Answer: (c) 64 times

Explanation:
1. Number of 10°C steps = 6.
2. Factor = 2⁶ = 64.

Teacher's Note:
1. Raise 2 to the number of steps.

Question 19: According to the Arrhenius equation, a straight line is obtained by plotting the logarithm of the rate constant (log k) against :
(a) T
(b) log T
(c) 1/T
(d) log (1/T)
Show Answer & Explanation

Answer: (c) 1/T

Explanation:
1. \( \log k = \log A - \dfrac{E_a}{2.303RT} \) is linear in 1/T.

Teacher's Note:
1. The slope is −Ea/2.303R.

Question 20: Which plot will give the value of the activation energy?
(a) k against T
(b) 1/k against T
(c) ln k against T
(d) ln k against 1/T
Show Answer & Explanation

Answer: (d) ln k against 1/T

Explanation:
1. \( \ln k = \ln A - E_a/RT \) is a straight line in 1/T with slope −Ea/R.

Teacher's Note:
1. Use the Arrhenius plot.

Question 21: Given that k is the rate constant at temperature T, the value of \( \lim_{T \to \infty} \log k \) is :
(a) \( A/2.303 \)
(b) \( A \)
(c) \( 2.303A \)
(d) \( \log A \)
Show Answer & Explanation

Answer: (d) \( \log A \)

Explanation:
1. \( \log k = \log A - E_a/(2.303RT) \).
2. As T → ∞ the second term goes to 0, leaving log A.

Teacher's Note:
1. k approaches A at infinite temperature.

Question 22: From the following data, the activation energy (cal/mol) for H₂ + I₂ → 2HI is : T = 769 K, 1/T = 1.3 × 10⁻³ K⁻¹, log k = 2.9; T = 667 K, 1/T = 1.5 × 10⁻³ K⁻¹, log k = 1.1.
(a) \( 4 \times 10^{4} \)
(b) \( 2 \times 10^{4} \)
(c) \( 8 \times 10^{4} \)
(d) \( 3 \times 10^{4} \)
Show Answer & Explanation

Answer: (a) \( 4 \times 10^{4} \)

Explanation:
1. Slope = (2.9 − 1.1)/(1.3 × 10⁻³ − 1.5 × 10⁻³) = −9000 K = −Ea/(2.303R).
2. \( E_a = 9000 \times 2.303 \times 1.987 \approx 4.1 \times 10^{4} \) cal/mol.

Teacher's Note:
1. Slope of log k against 1/T is −Ea/2.303R.

Question 23: The rate constant, the activation energy and the Arrhenius parameter of a reaction at 25°C are \( 3 \times 10^{-4} \) s⁻¹, 104.4 kJ mol⁻¹ and \( 6.0 \times 10^{14} \) s⁻¹. The value of the rate constant as T → ∞ is :
(a) \( 2 \times 10^{8} \) s⁻¹
(b) \( 6 \times 10^{14} \) s⁻¹
(c) infinity
(d) \( 3.6 \times 10^{30} \) s⁻¹
Show Answer & Explanation

Answer: (b) \( 6 \times 10^{14} \) s⁻¹

Explanation:
1. As T → ∞, \( e^{-E_a/RT} \to 1\), so k → A = \( 6.0 \times 10^{14} \) s⁻¹.

Teacher's Note:
1. The limit of k at infinite temperature equals A.

Question 24: For an endothermic reaction, where ΔH represents the enthalpy of the reaction in kJ/mol, the minimum value for the energy of activation will be :
(a) less than ΔH
(b) zero
(c) more than ΔH
(d) equal to ΔH
Show Answer & Explanation

Answer: (c) more than ΔH

Explanation:
1. E_a(forward) = ΔH + E_a(reverse), and E_a(reverse) is greater than zero.
2. So E_a is more than ΔH.

Teacher's Note:
1. The reverse barrier cannot be negative.

Question 25: If the concentration of the reactants is increased by x, then the rate constant k becomes :
(a) ln (k/x)
(b) k/x
(c) k + x
(d) k
Show Answer & Explanation

Answer: (d) k

Explanation:
1. The rate constant is independent of concentration.
2. It stays k.

Teacher's Note:
1. k depends on temperature only.

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Chemical Kinetics Objective Questions & Solutions for JEE Chemistry

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