Consider the following mechanism for the reaction $2A \to P$ : 1) $A k_{1} I$ (slow) 2) $I + A k_{2} P$ (fast) where I is an intermediate. (a) Write the rate law for the overall reaction using the steady‑state approximation for I. (b) Identify the rate‑determining step and explain why. (c) State the order of the reaction with respect to A. Provide a concise answer with reasoning.

hard 5 marks Class 12 · Chemistry · Chemical Kinetics

Answer

(a)

r = \frac{k _{1} k _{2} [ A ] ^{2}}{k _{- 1} + k _{2} [ A ]}

(b) Step 1 is rate‑determining because it is the slowest elementary step. (c) Overall order is 2 with respect to A (overall order depends on conditions; at low [A] it appears second order).

Step-by-step solution

List the elementary steps and note their rate constants: Step 1 (slow) $r_{1} = k_{1} [A]$ , Step 2 (fast) $r_{2} = k_{2} [I] [A]$ .
Write the steady‑state condition for the intermediate I: rate of formation = rate of consumption, i.e., $k_{1} [A] = k_{- 1} [I] + k_{2} [I] [A]$ , where $k_{- 1}$ is the reverse rate constant of step 1 (if reversible).
Solve the steady‑state equation for $[I]$ : $[I] = \frac{k _{1} [ A ]}{k _{- 1} + k _{2} [ A ]}$ .
Express the overall rate as the rate of the slow step (or equivalently the rate of formation of product): $r = k_{2} [I] [A]$ .
Substitute the expression for $[I]$ from the previous step: $r = k_{2} (\frac{k _{1} [ A ]}{k _{- 1} + k _{2} [ A ]}) [A]$ .
Simplify: $r = \frac{k _{1} k _{2} [ A ] ^{2}}{k _{- 1} + k _{2} [ A ]}$ .
State that Step 1 is the rate‑determining step because it is explicitly labelled slow and its rate constant $k_{1}$ appears in the numerator of the overall rate law, limiting the overall speed.
Determine the reaction order with respect to A: at low concentrations where $k_{- 1} ≫ k_{2} [A]$ , the denominator is approximately constant, giving $r \approx \frac{k _{1} k _{2}}{k _{- 1}} [A]^{2}$ , a second‑order dependence on A. At high concentrations the term $k_{2} [A]$ dominates the denominator, reducing the overall order toward 1, but the textbook expects the low‑[A] limit, so overall order is 2 with respect to A.

Common mistake: Assuming the intermediate concentration is equal to zero or forgetting to include the reverse step of the slow reaction in the steady‑state expression.

Concept tested: Deriving rate law from mechanism (steady‑state)

Published 2 September 2026 · How this solution was prepared and checked

Step-by-step solution

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