Question

The decomposition of acetaldehyde is a second-order reaction with a rate constant of \({\bf{4}}{\bf{.71 \times 1}}{{\bf{0}}^{{\bf{ - 8 }}}}{\bf{L mo}}{{\bf{l}}^{{\bf{ - 1}}}}{\bf{ s}}{{\bf{ }}^{{\bf{ - 1}}}}\). What is the instantaneous rate of decomposition of acetaldehyde in a solution with a concentration of \({\bf{5}}{\bf{.55 \times 1}}{{\bf{0}}^{{\bf{ - 4}}}}{\bf{M}}\)?

Short answer

The instantaneous rate of decomposition of aldehyde is \({\bf{1}}{\bf{.45}}\,{\bf{ \times 1}}{{\bf{0}}^{{\bf{ - 13}}}}\,{\bf{mol}}{{\bf{L}}^{{\bf{ - 1}}}}{{\bf{s}}^{{\bf{ - 1}}}}\).

Step-by-step solution

Definitions

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. It can also be a second-order reaction.

An instantaneous rate of change is the change in the rate at a particular instant.

Rate of reaction

The decomposition of aldehyde is a second-order reaction.

The rate law of reaction can be represented as

\({\bf{rate}}\;{\bf{of}}\;{\bf{reaction = k(acetaldehyde}}{{\bf{)}}^{\bf{2}}}\)

Given the value of the rate constant is\({\bf{4}}{\bf{.71 \times 1}}{{\bf{0}}^{{\bf{ - 8 }}}}{\bf{L mo}}{{\bf{l}}^{{\bf{ - 1}}}}{\bf{ s}}{{\bf{ }}^{{\bf{ - 1}}}}\)and the concentration of aldehyde \({\bf{5}}{\bf{.55 \times 1}}{{\bf{0}}^{{\bf{ - 4}}}}{\bf{M}}\).

The instantaneous rate of decomposition is equal to the rate of reaction.

Hence the rate of reaction can be calculated as follow

\(\begin{aligned}{}{\bf{Rate}}\,\,{\bf{ = k(Acetaldehyde}}{{\bf{)}}^{\bf{2}}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,{\bf{ = 4}}{\bf{.71 \times 1}}{{\bf{0}}^{{\bf{ - 7}}}}{\bf{Lmo}}{{\bf{l}}^{{\bf{ - 1}}}}{{\bf{s}}^{{\bf{ - 1}}}}{{\bf{(5}}{\bf{.55 \times 1}}{{\bf{0}}^{{\bf{ - 4}}}}{\bf{mol}}{{\bf{L}}^{{\bf{ - 1}}}}{\bf{)}}^{\bf{2}}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,{\bf{ = 1}}{\bf{.45}}\,{\bf{ \times 1}}{{\bf{0}}^{{\bf{ - 13}}}}\,{\bf{mol}}{{\bf{L}}^{{\bf{ - 1}}}}{{\bf{s}}^{{\bf{ - 1}}}}\end{aligned}\)

Thus, the instantaneous rate of decomposition is \({\bf{1}}{\bf{.45}}\,{\bf{ \times 1}}{{\bf{0}}^{{\bf{ - 13}}}}\,{\bf{mol}}{{\bf{L}}^{{\bf{ - 1}}}}{{\bf{s}}^{{\bf{ - 1}}}}\).