Question

The reaction mechanism for the reaction $P\to R$ is as follows :-
$P\stackrel{K_1}{\rightleftharpoons }2Q\left(fast\right);2Q+P\stackrel{K_2}{\to }R\left(slow\right)$

the rate law for the main reaction $(P\to R)$ is [where $K_1$ is an equilbrium constant]

• A. $K_1\left[P\right]\left[Q\right]$
• B. $K_1{K}_2\left[Q\right]$
• C. $K_1{K}_2[P{]}^2$
• D. $K_1{K}_2\left[P\right]$

Answer 1

Answer: (C)

$K_1{K}_2[P{]}^2$

$P\stackrel{K_1}{\rightleftharpoons }2Q$ (fast)

$k_1=\frac{[Q{]}^2}{\left[P\right]}\Rightarrow [Q{]}^2=k_1[P_1]$

rate is determined by slow step

$2Q+P\stackrel{K_2}{\to }R$ $rate=K_2\left[P\right][Q{]}^2$

Replacing $[Q{]}^2$

$rate=K_1{K}_2[P{]}^2$