91Ó°ÊÓ

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 \(0.005\) \(M\).

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