Question

The reaction mechanism for the reaction $P\to R$ is as follows :
$P\underset{K_2}{\overset{K_1}{\to }}2Q\left(fast\right);2Q+P\stackrel{K_3}{\to }R\left(slow\right)$the rate law for the main reaction ($P\to R$) is:

• A. $K_1\left[P\right]\left[Q\right]$
• B. $K_1{K}_2\left[P\right]$
• C. $\frac{K_1{K}_3[P{]}^2}{K_2}$
• D. $K_1{K}_2\left[a\right]$

Answer 1

The correct option is C $\frac{K_1{K}_3[P{]}^2}{K_2}$

From the slow step, which is rate determining step,
rate $r=K_3\left[Q{]}^2\right[P]$......(1)

From the fast step, which is the equilibrium step,
$\frac{K_1}{K_2}=\frac{[Q{]}^2}{\left[P\right]}$ or

$[Q{]}^2=\frac{K_1}{K_2}[P]$......(2)

Substitute equation (2) in equation (1)
$r=K_3\frac{K_1}{K_2}\left[P\right]\times \left[P\right]$

Hence, the rate $r=\frac{K_1{K}_3[P{]}^2}{K_2}$