Consider the mechanism for the reaction between X, Y and Z: 1. $X + Y ⇌ I$ (fast equilibrium, equilibrium constant $K$ ) 2. $I + Z x r i g h t a r r o w k_{2} P$ (slow)

medium 4 marks Class 12 · Chemistry · Chemical Kinetics

Answer

Rate law:

r = k_{2} K [X] [Y] [Z]

. Overall order: first order in X, Y and Z (total order 3).

The correct answer is Rate law: $r = k_{2} K [X] [Y] [Z]$ . Overall order: first order in X, Y and Z (total order 3). — From the rate law, the reaction is first order in X, in Y and in Z; total order = 3.

Step-by-step solution

Write the equilibrium expression for step 1: $K = \frac{[ I ]}{[ X ] [ Y ]}$ .
Express the concentration of the intermediate: $d i s pl a y s t y l e [I] = K [X] [Y]$ .
The rate of the slow step 2 is $r = k_{2} [I] [Z]$ .
Substitute $[I]$ from step 1: $r = k_{2} K [X] [Y] [Z]$ .
Thus the rate law is $r = k_{2} K [X] [Y] [Z]$ .
From the rate law, the reaction is first order in X, in Y and in Z; total order = 3.

Common mistake: Omitting the equilibrium constant

K

or mis‑assigning the order

Concept tested: Pre‑equilibrium rate law

Published 9 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked

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Consider the mechanism for the reaction between X, Y and Z: 1. X+Y⇌I (fast equilibrium, equilibrium constant K) 2. I+Zxrightarrowk2​P (slow)

Step-by-step solution

Consider the mechanism for the reaction between X, Y and Z: 1. $X + Y ⇌ I$ (fast equilibrium, equilibrium constant $K$ ) 2. $I + Z x r i g h t a r r o w k_{2} P$ (slow)