Question

The rate constant for the radioactive decay of 14C is \({\bf{1}}{\bf{.21 \times 1}}{{\bf{0}}^{{\bf{ - 4}}}}{\bf{ yea}}{{\bf{r}}^{{\bf{ - 1}}}}\). The products of the decay are nitrogen atoms and electrons (beta particles): \(\begin{aligned}{}_{\bf{6}}^{{\bf{14}}}{\bf{C}} \to _{\bf{6}}^{{\bf{14}}}{\bf{N + }}{{\bf{e}}^{\bf{ - }}}\\{\bf{rate = k(}}_{\bf{6}}^{{\bf{14}}}{\bf{C)}}\end{aligned}\).

What is the instantaneous rate of production of N atoms in a sample with a carbon-14 content of \({\bf{ 6}}{\bf{.5 \times 1}}{{\bf{0}}^{{\bf{ - 9 }}}}{\bf{M}}\)?

Short answer

The rate of production of N atoms is \({\bf{7}}{\bf{.865 \times 1}}{{\bf{0}}^{{\bf{ - 13}}}}{\bf{mol}}{{\bf{L}}^{{\bf{ - 1}}}}{\bf{Yea}}{{\bf{r}}^{{\bf{ - 1}}}}\).

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.

\(_{\bf{6}}^{{\bf{14}}}{\bf{C}} \to _{\bf{6}}^{{\bf{14}}}{\bf{N + }}{{\bf{e}}^{\bf{ - }}}\)

For the above reaction, the rate law is represented as

\({\bf{rate = k(}}_{\bf{6}}^{{\bf{14}}}{\bf{C)}}\)

The instantaneous rate of production of N-atoms

The instantaneous rate of production of N atoms is equal to the rate of reaction.

Hence the instantaneous rate can be calculated as;

\(\begin{aligned}{}{\bf{Rate}}\,{\bf{of}}\,{\bf{production}}\,{\bf{of}}\,{\bf{N}}\,{\bf{atoms = k(}}_{\bf{6}}^{{\bf{14}}}{\bf{C)}}\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{\bf{ = 1}}{\bf{.21 \times 1}}{{\bf{0}}^{{\bf{ - 4}}}}{\bf{yea}}{{\bf{r}}^{{\bf{ - 1}}}}{\bf{ \times 6}}{\bf{.5 \times 1}}{{\bf{0}}^{{\bf{ - 9}}}}{\bf{M}}\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{\bf{ = 7}}{\bf{.865 \times 1}}{{\bf{0}}^{{\bf{ - 13}}}}{\bf{mol}}{{\bf{L}}^{{\bf{ - 1}}}}{\bf{Yea}}{{\bf{r}}^{{\bf{ - 1}}}}\end{aligned}\)

Thus, the rate of production of N atoms is \({\bf{7}}{\bf{.865 \times 1}}{{\bf{0}}^{{\bf{ - 13}}}}{\bf{mol}}{{\bf{L}}^{{\bf{ - 1}}}}{\bf{Yea}}{{\bf{r}}^{{\bf{ - 1}}}}\).