JEE Chemistry — MCQ Practice Questions & Answers

Question 1

For a reaction with mechanism: A ⇌ B (fast equilibrium), B + C → D (slow), the rate law is:

Accepted Answer

Correct answer: C. rate = k[A]^(1/2)[C] — From fast equilibrium: K = [B]/[A], so [B] = K[A]. The slow step rate law is rate = k'[B][C] = k'K[A][C] = k[A]^(1/2)[C] where k combines constants.

Answer

A. rate = k[B][C]

Answer

B. rate = k[A][C]

Answer

D. rate = k[A][B][C]

Question 2

The rate constant for a reaction increases 4 times when temperature increases from 27°C to 47°C. What is the activation energy? (R = 8.314 J mol⁻¹ K⁻¹)

Accepted Answer

Correct answer: B. 51.2 kJ/mol — Using ln(k₂/k₁) = (Eₐ/R)(1/T₁ - 1/T₂): ln(4) = (Eₐ/8.314)(1/300 - 1/320), solving gives Eₐ ≈ 51.2 kJ/mol.

Answer

A. 25.6 kJ/mol

Answer

C. 102.4 kJ/mol

Answer

D. 12.8 kJ/mol

Question 3

In a reaction, the rate increases by a factor of 8 when [A] doubles and by a factor of 2 when [B] doubles. What is the overall order of the reaction?

Accepted Answer

Correct answer: C. 4 — When [A] doubles, rate increases by 8 = 2³, so order w.r.t. A = 3. When [B] doubles, rate increases by 2 = 2¹, so order w.r.t. B = 1. Overall order = 3 + 1 = 4.

Answer

A. 2

Answer

B. 3

Answer

D. 5

Question 4

For the consecutive reaction A → B → C, if the rate constants are k₁ = 0.1 s⁻¹ and k₂ = 0.05 s⁻¹, and k₁ > k₂, which statement is true?

Accepted Answer

Correct answer: B. B accumulates and then decreases — Since k₁ > k₂, A converts to B faster than B converts to C, so B accumulates initially and then decreases as it slowly converts to C.

Answer

A. B is consumed faster than it is formed

Answer

C. B is immediately consumed after formation

Answer

D. No B is formed in the reaction

Question 5

In the Lindemann mechanism for unimolecular reactions, A* represents an activated molecule. The rate-determining step is:

Accepted Answer

Correct answer: B. A* → products (slow) — In the Lindemann mechanism: Step 1 (fast equilibrium): A + A ⇌ A* + A, Step 2 (slow): A* → products. The slow step is rate-determining.

Answer

A. A + A ⇌ A* + A (fast equilibrium)

Answer

C. A → A* (slow)

Answer

D. A* + A → products (slow)

Question 6

For a pseudo-first-order reaction where [B]₀ >> [A]₀, the rate law simplifies to first-order even though the actual order is higher. This is because:

Accepted Answer

Correct answer: A. [B] remains essentially constant throughout the reaction — When [B]₀ >> [A]₀, the concentration of B doesn't change significantly during the reaction, so it can be incorporated into the rate constant, making the reaction appear first-order in A only.

Answer

B. [A] is in large excess

Answer

C. The reaction temperature is constant

Answer

D. The rate constant becomes independent of [B]

Question 7

The rate constant for a reaction at 298 K is 2 × 10⁻⁵ s⁻¹ with Ea = 80 kJ/mol. What is the frequency factor (A) if rate = Ae^(-Ea/RT)?

Accepted Answer

Correct answer: D. 5.6 × 10¹³ s⁻¹ — A = k/e^(-Ea/RT) = (2 × 10⁻⁵)/e^(-80000/2.478) ≈ 5.6 × 10¹³ s⁻¹

Answer

A. 1.5 × 10⁻² s⁻¹

Answer

B. 8.0 × 10⁸ s⁻¹

Answer

C. 3.2 × 10¹¹ s⁻¹

Question 8

In the reaction mechanism: (1) A + B ⇌ AB (fast), (2) AB + C → ABC (slow), the order predicted from this mechanism is:

Accepted Answer

Correct answer: C. Third order overall — Rate = k₂[AB][C]. From fast equilibrium: [AB] = K₁[A][B]. So Rate = k₂K₁[A][B][C], making it third-order overall

Answer

A. First order overall

Answer

B. Second order overall

Answer

D. Fourth order overall

JEE Chemistry - MCQ Practice Questions