Nitrogen monoxide, NO, reacts with hydrogen, H₂, according to the following equation.
[tex]2NO + 2H_2 \rightarrow N_2 + 2H_2O[/tex]
What would the rate law be if the mechanism for this reaction were as follows?
(Rate expressions take the general form: rate = k . [A]a . [B]b.)
[tex]2 NO + H_2 \rightarrow N_2 + H_2O_2[/tex] (slow)
[tex]H_2O_2 + H_2 \rightarrow 2 H_2O[/tex] (fast)

Rate law : It is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

As we are given the mechanism for the reaction :

Step 1 : [tex]2NO+H_2\rightarrow N_2+H_2O_2[/tex] (slow)

Step 2 : [tex]H_2O_2+H_2\rightarrow 2H_2O[/tex] (fast)

Overall reaction : [tex]2NO+2H_2\rightarrow N_2+2H_2O[/tex]

The rate law expression for overall reaction should be in terms of [tex]NO\text{ and }H_2[/tex].

As we know that the slow step is the rate determining step. So,

The slow step reaction is,

[tex]2NO+H_2\rightarrow N_2+H_2O_2[/tex]

The expression of rate law for this reaction will be,

[tex]Rate=k[NO]^2[H_2][/tex]

Hence, the rate law for the overall reaction is, [tex]Rate=k[NO]^2[H_2][/tex]