Question

Regular flights of supersonic aircraft in the stratosphere are of concern because such aircraft produce nitric oxide, NO, as a by-product in the exhaust of their engines. Nitric oxide reacts with ozone, and it has been suggested that this could contribute to depletion of the ozone layer. The reaction \({\bf{NO + }}{{\bf{O}}_{\bf{3}}} \to {\bf{N}}{{\bf{O}}_{\bf{2}}}{\bf{ + }}{{\bf{O}}_{\bf{2}}}\) is first order with respect to both NO and \({{\bf{O}}_{\bf{3}}}\) with a rate constant of \({\bf{2}}{\bf{.20 \times 1}}{{\bf{0}}^{\bf{7}}}{\bf{mol}}{{\bf{L}}^{{\bf{ - 1}}}}{{\bf{s}}^{{\bf{ - 1}}}}\). What is the instantaneous rate of disappearance of NO when \(\left( {{\bf{NO}}} \right){\bf{ = 3}}{\bf{.3 \times 1}}{{\bf{0}}^{{\bf{ - 6}}}}{\bf{ M}}\) and \({\bf{(}}{{\bf{O}}_{\bf{3}}}{\bf{) = 5}}{\bf{.9 \times 1}}{{\bf{0}}^{{\bf{ - 7}}}}{\bf{ M}}\)?

Short answer

The rate of disappearance of NO is \({\bf{4}}{\bf{.3 \times 1}}{{\bf{0}}^{{\bf{ - 5}}}}{\bf{Mol}}{{\bf{L}}^{{\bf{ - 1}}}}{{\bf{s}}^{{\bf{ - 1}}}}\).

Step-by-step solution

rate law 

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.

Given reaction

\({\bf{NO + }}{{\bf{O}}_{\bf{3}}} \to {\bf{N}}{{\bf{O}}_{\bf{2}}}{\bf{ + }}{{\bf{O}}_{\bf{2}}}\)

The rate law or rate of disappearance of NO can be represented as

\({\bf{rate = k(NO)(}}{{\bf{O}}_{\bf{3}}}{\bf{)}}\)

Rate of reaction

Given the concentration value of NO is \({\bf{3}}{\bf{.3 \times 1}}{{\bf{0}}^{{\bf{ - 6}}}}\)M and \({{\bf{O}}_{\bf{3}}}\)is \({\bf{5}}{\bf{.9 \times 1}}{{\bf{0}}^{{\bf{ - 7}}}}{\bf{M}}\). The value of rate constant k is \({\bf{2}}{\bf{.20 \times 1}}{{\bf{0}}^{\bf{7}}}{\bf{M}}\) .

The rate of disappearance of NO can be calculated as;

\(\begin{aligned}{}{\bf{Rate}}\,{\bf{ = k(NO)(O}}{}_{\bf{3}}{\bf{)}}\\{\bf{ = 2}}{\bf{.2 \times 1}}{{\bf{0}}^{\bf{7}}}{\bf{LMo}}{{\bf{l}}^{{\bf{ - 1}}}}{{\bf{s}}^{{\bf{ - 1}}}}{\bf{(3}}{\bf{.3 \times 1}}{{\bf{0}}^{{\bf{ - 6}}}}{\bf{M)(5}}{\bf{.9 \times 1}}{{\bf{0}}^{{\bf{ - 7}}}}{\bf{M)}}\\{\bf{ = 4}}{\bf{.3 \times 1}}{{\bf{0}}^{{\bf{ - 5}}}}{\bf{Mol}}{{\bf{L}}^{{\bf{ - 1}}}}{{\bf{s}}^{{\bf{ - 1}}}}\end{aligned}\)