91Ó°ÊÓ

The rate of reaction may be defined as the speed of the reactant reacting to obtain a product in a particular reaction at a particular time. The concentration of the reactant and product are represented into mole/L.

Rate\({\bf{ = }}\,{\bf{k}}{\left( {\bf{A}} \right)^{\bf{m}}}{\left( {\bf{B}} \right)^{\bf{n}}}\)

The second-order reaction depends on the one concentration having rate law,

\({\bf{Rate = k}}{\left( {\bf{A}} \right)^{\bf{2}}}\)

For second-order reactions, the rate equation is:

\({\bf{1 / }}\left( {\bf{A}} \right){\bf{ = kt + 1 / }}{\left( {\bf{A}} \right)_{\bf{0}}}\)

For a second-order reaction, we have:

\({\bf{1 / }}\left( {\bf{A}} \right){\bf{ = kt + 1 / }}{\left( {\bf{A}} \right)_{\bf{0}}}\)

\({\left( {\bf{A}} \right)_{\bf{0}}}\)= 0.020 mol/L,

k = \({\bf{5}}{\bf{.76 \times 1}}{{\bf{0}}^{{\bf{ - 2}}}}{\bf{L}}{\left( {{\bf{molmin}}} \right)^{{\bf{ - 1}}}}\),

And t = 20.0 min.

\(\begin{aligned}{{}{}}{{\bf{1 / }}\left( {\bf{A}} \right){\bf{ = }}\left( {{\bf{5}}{\bf{.76 \times 1}}{{\bf{0}}^{{\bf{ - 2}}}}{\bf{L}}{{\left( {{\bf{molmin}}} \right)}^{{\bf{ - 1}}}}} \right){\bf{ }}\left( {{\bf{20 min}}} \right){\bf{ + 1 /0}}{\bf{.020 mol/L}}}\\{{\bf{1 / }}\left( {\bf{A}} \right){\bf{ = 0}}{\bf{.1152 + 50}}{\bf{.00}}}\\{{\bf{1 / }}\left( {\bf{A}} \right){\bf{ = 50}}{\bf{.1152}}}\\{{\bf{\;}}\,\,\,\,\,\,\left( {\bf{A}} \right){\bf{ = 1 / 50}}{\bf{.1152}}}\\{\,\,\,\,\,\,\left( {\bf{A}} \right){\bf{ = 0}}{\bf{.0196 mol/L}}}\end{aligned}\)