Radioactive phosphorus is used in the study of biochemical reaction mechanisms because phosphorus atoms are components of many biochemical molecules. The location of the phosphorus (and the location of the molecule it is bound in) can be detected from the electrons (beta particles) it produces:

\(\begin{aligned}{l}_{{\bf{15}}}^{{\bf{32}}}{\bf{P}} \to _{{\bf{16}}}^{{\bf{32}}}{\bf{S + }}{{\bf{e}}^{\bf{ - }}}\\{\bf{rate = 4}}{\bf{.85 \times 1}}{{\bf{0}}^{{\bf{ - 2}}}}\,{\bf{da}}{{\bf{y}}^{{\bf{ - 1}}}}{{\bf{(}}^{{\bf{32}}}}{\bf{p)}}\end{aligned}\)

What is the instantaneous rate of production of electrons in a sample with a phosphorus concentration of \({\bf{0}}{\bf{.0033 M}}\)?

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.

The rate of reaction or rate law for the given reaction is represented as

\({\bf{Rate}}\,{\bf{of}}\;{\bf{reaction = k(concentration}}\;{\bf{of}}\;{\bf{phosphorous)}}\)

Where k is the rate constant. Value of rate constant is \({\bf{4}}{\bf{.85 \times 1}}{{\bf{0}}^{{\bf{ - 2}}}}\,{\bf{da}}{{\bf{y}}^{{\bf{ - 1}}}}\)

Instantaneous rate of production of electrons is equal to the rate of reaction.

Hence the rate of production of electrons can be calculated as;

\(\begin{aligned}{}{\bf{rate}}\;{\bf{of}}\,{\bf{production}}\;{\bf{of}}\;{\bf{electron = k(concentration}}\;{\bf{of}}\;{\bf{phosphorus)}}\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{\bf{ = 4}}{\bf{.85 \times 1}}{{\bf{0}}^{{\bf{ - 2}}}}{\bf{da}}{{\bf{y}}^{{\bf{ - 1}}}}{\bf{ \times 0}}{\bf{.0033M}}\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{\bf{ = 1}}{\bf{.6 \times 1}}{{\bf{0}}^{{\bf{ - 4}}}}{\bf{mol}}{{\bf{L}}^{{\bf{ - 1}}}}{{\bf{d}}^{{\bf{ - 1}}}}\end{aligned}\)