Question

For the reaction $2H_2+2NO\to N_2+2H_{2}O$, the following mechanism has been suggested :
$2NO\rightleftharpoons N_2{O}_2$ equilibrium constant $k_1$ (fast)
$N_2{O}_2+H_2\stackrel{k_2}{\to }{N}_{2}O+H_{2}O\left(slow\right)$
$N_{2}O+H_2\stackrel{K_3}{\to }{N}_2+H_{2}O\left(fast\right)$
Establish the rate law for given reaction.

• A. $r=K\left[NO{]}^2\right[H_2],$ where $K=K_2\times K_1$
• B. $r=K\left[NO{]}^{-2}\right[H_2],$ where $K=2K_2\times K_1$
• C. $r=K\left[NO{]}^{-2}\right[H_2],$ where $K=K_2\times 2K_1$
• D. None of these

Answer 1

The correct option is A $r=K\left[NO{]}^2\right[H_2],$ where $K=K_2\times K_1$

From the slow step, which is rate determining step, the rate $r=k_2\left[N_2{O}_2\right]\left[H_2\right]$.......(1)

From the equilibrium step, $k_1=\frac{\left[N_2{O}_2\right]}{[NO{]}^2}$ or $\left[N_2{O}_2\right]=k_1\times [NO{]}^2$......(2)

Substitute equation (2) in equation (1)
$r=k_2\times k_1\left[NO{]}^2\right[H_2],$

Substitute $k=k_2\times k_1$ in the above equation
$r=k\left[NO{]}^2\right[H_2],$