Hydrogen reacts with nitrogen monoxide to form dinitrogen monoxide (laughing gas) according to the equation:\({{\bf{H}}_{\bf{2}}}{\bf{(g) + 2NO(g)}} \to {{\bf{N}}_{\bf{2}}}{\bf{O(g) + }}{{\bf{H}}_{\bf{2}}}{\bf{O}}\).Determine the rate law, the rate constant, and the orders with respect to each reactant from the following data:

The rate law for a chemical reaction is an expression that provides a relationship between the rate of the reaction and the concentration of the reactants participating in it.

General rate law for the given reaction is represented as

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

From the given table we can make the following equations

\(\begin{align}2.835 \times {10^{ - 3}}mol{L^{ - 1}}{s^{ - 1}} &= k{(0.30)^m}{(0.60)^n}\,\,\,\,\,\,......(1)\left( {0,1} \right)\\1.134 \times {10^{ - 2}}mol{L^{ - 1}}{s^{ - 1}} &= k{(0.35)^m}{(0.35)^n}\,\,\,\,\,\,......(2)\\2.268 \times {10^{ - 2}}mol{L^{ - 1}}{s^{ - 1}} &= k{(0.60)^m}{(0.70)^n}\,\,\,\,\,\,......(3)\end{align}\)

The order of reaction refers to the power dependence of the rate on the concentration of each reactant.

On dividing equation (2) by (1), we get

\(\frac{{Rate(R{}_2)}}{{Rate(R{}_1)}} = \frac{{0.01134}}{{0.002835}} = \frac{{k{{(0.60)}^m}{{(0.35)}^n}}}{{k{{(0.30)}^m}{{(0.35)}^n}}}\)

\(\begin{align}{(2)^m} = 4\\m = 2\end{align}\)

On dividing equation (3) by (2), we get

\(\frac{{Rate(R{}_3)}}{{Rate(R{}_2)}} = \frac{{0.02268}}{{0.01134}} = \frac{{k{{(0.60)}^m}{{(0.70)}^n}}}{{k{{(0.60)}^m}{{(0.35)}^n}}}\)

\(\begin{align}{(2)^n} = 2\\n = 1\end{align}\)

Hence the rate of given reaction is

\({\bf{rate = k(NO}}{{\bf{)}}^{\bf{2}}}{{\bf{(}}{{\bf{H}}_{\bf{2}}}{\bf{)}}^{\bf{1}}}\)

The overall order of reaction \({\bf{ = m + n = 2 + 1 = 3}}\)

The rate constant is the proportionality constant in the equation that expresses the relationship between the rate of a chemical reaction and concentration of reacting substances.

\({\bf{Rate = k}}\left( {{\bf{NO}}} \right){{\bf{\;}}^{\bf{2}}}{\bf{(}}{{\bf{H}}_{\bf{2}}}{\bf{)}}\)

\(\begin{align}2.835 \times {10^{ - 3}}mol{L^{ - 1}}{s^{ - 1}} &= K{(0.30M)^2}(0.35M)\\k &= \frac{{2.835 \times {{10}^{ - 3}}mol{L^{ - 1}}{s^{ - 1}}}}{{{{(0.30M)}^2}(0.35M)}}\\ &= 9.0 \times {10^{ - 2}}{L^2}Mo{l^{ - 2}}{s^{ - 1}}\end{align}\)