For a reaction with mechanism: A ⇌ B (fast equilibrium), B + C → D (slow), the rate law is:
Answer: 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]^(21)[C] where k combines constants.
Q.2Hard
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)
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?
Answer: C
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.
Q.4Hard
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?
Answer: B
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.
Q.5Hard
In the Lindemann mechanism for unimolecular reactions, A* represents an activated molecule. The rate-determining step is:
Answer: B
In the Lindemann mechanism: Step 1 (fast equilibrium): A + A ⇌ A* + A, Step 2 (slow): A* → products. The slow step is rate-determining.
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Q.6Hard
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:
Answer: A
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.
Q.7Hard
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)?
