If the initial concentration of butadiene is 0.0200 M, what is the concentration remaining after 20.0 min?

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.

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