Question

Consider the reaction :
$2H_2\left(g\right)+2NO\left(g\right)\to N_2\left(g\right)+2H_{2}O\left(g\right)$
The rate law for this reaction is :
$Rate=k\left[H_2\right][NO{]}^2$
Under what conditions could these steps represent the mechanism?
Step 1 : $2NO\left(g\right)\rightleftharpoons N_2{O}_2\left(g\right)$
Step 2 : $N_2{O}_2+H_2\to N_{2}O+H_{2}O$
Step 3 : $N_{2}O+H_2\to H_{2}O+N_2$

• A. These steps can never satisfy the rate law
• B. Step 1 should be the slowest step
• C. Step 2 should be the slowest step
• D. Step 3 should be the slowest step

Answer 1

Answer: (A)

These steps can never satisfy the rate law

The given reaction is:-

$2H_2\left(g\right)+2NO\left(g\right)⟶N_2\left(g\right)+2H_{2}O\left(g\right)$

The given rate law is:-

$Rate=K\left[H_2\right][NO{]}^2$

The rate of the chemical reaction is determined by the slowest step. So, in the slowest step we should have $2$ molecules of $NO$ and $1$ molecule of $H_2$ because the rate of the reaction is determined by that.

So, I. $2NO\left(g\right)+H_2\left(g\right)⟶N_2\left(g\right)+H_2{O}_2$ (slow)

II. $H_2{O}_2+H_2\left(g\right)⟶2H_{2}O\left(g\right)$ (fast)

This could be the mechanism of the reaction as given by rate law.