Experiments were conducted to study the rate of the reaction represented by this equation.⁽²⁾\({\rm{2NO(g) + 2}}{{\rm{H}}_{\rm{2}}}{\rm{(g) }} \to {\rm{ }}{{\rm{N}}_{\rm{2}}}{\rm{(g) + 2}}{{\rm{H}}_{\rm{2}}}{\rm{O(g)}}\)Initial concentrations and rates of reaction are given here.

Experiment	Initial Concentration \(\left( {{\bf{NO}}} \right){\rm{ }}\left( {{\bf{mol}}/{\bf{L}}} \right)\)	Initial Concentration, \(\left( {{{\bf{H}}_{\bf{2}}}} \right){\rm{ }}\left( {{\bf{mol}}/{\bf{L}}} \right)\)	Initial Rate of Formation of \({{\bf{N}}_{\bf{2}}}{\rm{ }}\left( {{\bf{mol}}/{\bf{L}}{\rm{ }}{\bf{min}}} \right)\)
1	\({\bf{0}}.{\bf{0060}}\)	\({\bf{0}}.{\bf{00}}1{\bf{0}}\)	\({\bf{1}}.{\bf{8}} \times {\bf{1}}{{\bf{0}}^{ - {\bf{4}}}}\)
2	\({\bf{0}}.{\bf{0060}}\)	\({\bf{0}}.{\bf{00}}2{\bf{0}}\)	\({\bf{3}}.{\bf{6}} \times {\bf{1}}{{\bf{0}}^{ - {\bf{4}}}}\)
3	\({\bf{0}}.{\bf{00}}1{\bf{0}}\)	\({\bf{0}}.{\bf{0060}}\)	\({\bf{0}}.{\bf{30}} \times {\bf{1}}{{\bf{0}}^{ - {\bf{4}}}}\)
4	\({\bf{0}}.{\bf{00}}2{\bf{0}}\)	\({\bf{0}}.{\bf{0060}}\)	\({\bf{1}}.{\bf{2}} \times {\bf{1}}{{\bf{0}}^{ - {\bf{4}}}}\)

Consider the following questions:(a) Determine the order for each of the reactants, \({\bf{NO}}\) and \({{\bf{H}}_{\bf{2}}}\), from the data given and show your reasoning.(b) Write the overall rate law for the reaction.(c) Calculate the value of the rate constant, k, for the reaction. Include units.(d) For experiment 2, calculate the concentration of \({\bf{NO}}\)remaining when exactly one-half of the original amount of \({{\bf{H}}_{\bf{2}}}\) had been consumed.(e) The following sequence of elementary steps is a proposed mechanism for the reaction.Step 1:Step 2:Step 3:Based on the data presented, which of these is the rate determining step? Show that the mechanism is consistent with the observed rate law for the reaction and the overall stoichiometry of the reaction.

The rate equation is the mathematical expression which explains thethe 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.

The order of each reactant\({\bf{(NO)}}\)and\(({{\bf{H}}_{\bf{2}}})\)from the data given will be 2^nd order and 1^st order. As per the given data, we can see that the concentration of\({\bf{(NO)}}\)remains the same in the first and second experiment and the concentration of\(({{\bf{H}}_{\bf{2}}})\)is doubled. Due to this rate becomes double here. So, the order of\(({{\bf{H}}_{\bf{2}}})\)is 1 here.In the third and fourth experiment, the concentration of\(({{\bf{H}}_{\bf{2}}})\)remains the same in the and the concentration of\({\bf{(NO)}}\)is doubled. Due to this rate becomes greater four times. Thus, the order of\({\bf{(NO)}}\)is 2 here.

In the first and second experiment the concentration of \({\bf{(NO)}}\) remains the same and the concentration of \(({{\bf{H}}_{\bf{2}}})\) is doubled. Due to this rate becomes double here. So, the order of \(({{\bf{H}}_{\bf{2}}})\) is 1 here. The rate has increased from \({\bf{1}}.{\bf{8}} \times {\bf{1}}{{\bf{0}}^{ - {\bf{4}}}}\) to \({\bf{3}}.{\bf{6}} \times {\bf{1}}{{\bf{0}}^{ - {\bf{4}}}}\). While, in the third and fourth experiment, the concentration of \(({{\bf{H}}_{\bf{2}}})\) remains the same in the and the concentration of \({\bf{(NO)}}\) is doubled. Due to this rate becomes greater four times. Thus, the order of \({\bf{(NO)}}\) is 2 here. The rate has increased from \({\bf{0}}.{\bf{30}} \times {\bf{1}}{{\bf{0}}^{ - {\bf{4}}}}\) to \({\bf{1}}.{\bf{2}} \times {\bf{1}}{{\bf{0}}^{ - {\bf{4}}}}\). Therefore, by combining both, the overall rate law for the reaction will be,\(Rate = k{(NO)^2}{({H_2})^1}\)

The rate law for the reaction will be,\(Rate = k{(NO)^2}{({H_2})^1}\)Let us substitute values from the experiment 1, to the value of rate constant.

\(\begin{aligned}{l}Rate = k{(NO)^2}{({H_2})^1}\\{\bf{1}}.{\bf{8}} \times {\bf{1}}{{\bf{0}}^{ - {\bf{4}}}} = k{({\bf{0}}.{\bf{0060}})^2}({\bf{0}}.{\bf{0010}})\\k = 0.5 \times {\bf{1}}{{\bf{0}}^4}mo{l^{ - 2}}{l^2}{\min ^{ - 1}}\end{aligned}\)

For the reaction, one-half of the original amount of \({{\bf{H}}_{\bf{2}}}\)will be \(0.001\). Thus, the concentration of \({\bf{(NO)}}\) when exactly one-half of the original amount of \({{\bf{H}}_{\bf{2}}}\) had been consumed will be \(0.005\) \(M\).

\(\begin{aligned}{\rm{ 2NO(g) + 2}}{{\rm{H}}_{\rm{2}}}{\rm{(g) }} \to {\rm{ }}{{\rm{N}}_{\rm{2}}}{\rm{(g) + 2}}{{\rm{H}}_{\rm{2}}}{\rm{O(g)}}\\\begin{aligned}{{20}{l}}{Before\;\;\;\;\;\;\;\;\;\;\;0.006\;\;\;\;\;0.002}\\{Change\;\;\;\;\;\;\; - 0.001\;\;\;\;\; - 0.001}\\{After\;\;\;\;\;\;\;\;\;\;\;\;\;\;0.005\;\;\;\;\;0.001}\end{aligned}\end{aligned}\)