Question

Nitrogen (II) oxide, \({\rm{NO}}\), reacts with hydrogen, \({{\rm{H}}_{\rm{2}}}\), according to the following equation: What would the rate law be if the mechanism for this reaction were:

\({\rm{2NO + 2}}{{\rm{H}}_{\rm{2}}}{\rm{ }} \to {\rm{ }}{{\rm{N}}_{\rm{2}}}{\rm{ + 2}}{{\rm{H}}_{\rm{2}}}{\rm{O}}\)

What would the rate law be if the mechanism for this reaction were: \(\begin{array}{\rm{2NO + }}{{\rm{H}}_{\rm{2}}}{\rm{ }} \to {\rm{ }}{{\rm{N}}_{\rm{2}}}{\rm{ + }}{{\rm{H}}_{\rm{2}}}{{\rm{O}}_{\rm{2}}}{\rm{ (slow)}}\\{{\rm{H}}_{\rm{2}}}{{\rm{O}}_{\rm{2}}}{\rm{ + }}{{\rm{H}}_{\rm{2}}}{\rm{ }} \to {\rm{ 2}}{{\rm{H}}_{\rm{2}}}{\rm{O (fast)}}\end{array}\)

Short answer

The rate law for the first reaction mechanism would be,

\({\bf{rate = k(NO}}{{\bf{)}}^{\bf{2}}}{{\bf{(}}{{\bf{H}}_{\bf{2}}}{\bf{)}}^{\bf{2}}}\)

and

The rate law for the second reaction mechanism would be,

\({\rm{rate = }}k{{\rm{(NO)}}^2}{\rm{(}}{{\rm{H}}_{\rm{2}}}{\rm{)}}\)

Step-by-step solution

Definition of Rate Equation (Rate Law)

The rate law (or rate equation) is the mathematical expression which explains the the relationship between the rate of a chemical reaction and the concentration of its reactants.

\({\rm{rate = }}k{{\rm{(A)}}^x}{{\rm{(B)}}^y}{{\rm{(C)}}^z}.....\)

Where,

(A), (B), and (C) denotes the molar concentrations of reactants.

kis the rate constant.

Exponents m, n, and pare generally positive integers.

Rate Law for the First Reaction \({\rm{2NO   +  2}}{{\rm{H}}_{\rm{2}}}{\rm{ }} \to {\rm{ }}{{\rm{N}}_{\rm{2}}}{\rm{   +  2}}{{\rm{H}}_{\rm{2}}}{\rm{O}}\) 

For the first reaction with mechanism,

\({\rm{2NO + 2}}{{\rm{H}}_{\rm{2}}}{\rm{ }} \to {\rm{ }}{{\rm{N}}_{\rm{2}}}{\rm{ + 2}}{{\rm{H}}_{\rm{2}}}{\rm{O}}\)

The rate law will be,

\({\bf{rate = k(NO}}{{\bf{)}}^{\bf{2}}}{{\bf{(}}{{\bf{H}}_{\bf{2}}}{\bf{)}}^{\bf{2}}}\)

Rate Law for the Second Reactions \(\begin{array}{\rm{2NO  +   }}{{\rm{H}}_{\rm{2}}}{\rm{ }} \to {\rm{ }}{{\rm{N}}_{\rm{2}}}{\rm{  +   }}{{\rm{H}}_{\rm{2}}}{{\rm{O}}_{\rm{2}}}{\rm{ (slow)}}\\{{\rm{H}}_{\rm{2}}}{{\rm{O}}_{\rm{2}}}{\rm{  +   }}{{\rm{H}}_{\rm{2}}}{\rm{ }} \to {\rm{ 2}}{{\rm{H}}_{\rm{2}}}{\rm{O (fast)}}\end{array}\) 

For the second reaction with mechanism,

\(\begin{align}{\rm{2NO + }}{{\rm{H}}_{\rm{2}}}{\rm{ }} \to {\rm{ }}{{\rm{N}}_{\rm{2}}}{\rm{ + }}{{\rm{H}}_{\rm{2}}}{{\rm{O}}_{\rm{2}}}{\rm{ (slow)}}\\{{\rm{H}}_{\rm{2}}}{{\rm{O}}_{\rm{2}}}{\rm{ + }}{{\rm{H}}_{\rm{2}}}{\rm{ }} \to {\rm{ 2}}{{\rm{H}}_{\rm{2}}}{\rm{O (fast)}}\end{align}\)

The rate law is determined by depending on its slow reaction. The slow step in the reaction mechanism is considered as rate determining step. Thus, the rate law will be

\({\bf{rate = k(NO}}{{\bf{)}}^{\bf{2}}}{\bf{(}}{{\bf{H}}_{\bf{2}}}{\bf{)}}\)

Therefore, the rate law will be\({\bf{rate = k(NO}}{{\bf{)}}^{\bf{2}}}{{\bf{(}}{{\bf{H}}_{\bf{2}}}{\bf{)}}^{\bf{2}}}\)and\({\bf{rate = k(NO}}{{\bf{)}}^{\bf{2}}}{\bf{(}}{{\bf{H}}_{\bf{2}}}{\bf{)}}\)for both the reactions with different mechanisms.