Question

Iodine-131 is a radioactive isotope that is used to diagnose and treat some forms of thyroid cancer. Iodine-131 decays to xenon-131 according to the equation:I-131⟶Xe-131 + electron. The decay is first-order with a rate constant of 0.138 d−1. All radioactive decay is first order. How many days will it take for 90% of the iodine−131 in a 0.500 M solution of this substance to decay to Xe-131?

Short answer

It will take 16.7 days will take for 90% of the iodine−131 in a 0.500 M solution of this substance to decay to Xe-131.

Step-by-step solution

Rate of a Reaction

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

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

Value of r            \(\)

Equation of following order:

\({\bf{ln }}{\left( {\bf{A}} \right)_{\bf{0}}}{\bf{/\;}}\left( {\bf{A}} \right){\bf{ = kt}}{\bf{.}}\)

In this case we know (A)₀, (A), and k, and need to find t. The initial concentration of Xe-131, (A)0, is not provided, but the provision that 90.0% of the sample has decomposed is enough information to solve this problem. Let x be the initial concentration, in which case the concentration after 90.0% decomposition is 10.0% of x or 0.010x. \(\)

\(\begin{array}{}\begin{array}{{}{}}{{\bf{t = }}\left( {{\bf{ln }}\left( {\bf{x}} \right){\bf{ / }}\left( {{\bf{0}}{\bf{.010x}}} \right)} \right){\bf{/ k}}}\\{{\bf{t = 2}}{\bf{.303}}\left( {{\bf{log 0}}{\bf{.100 mol }}{{\bf{L}}^{{\bf{ - 1 /}}}}{\bf{0}}{\bf{.010 mol }}{{\bf{L}}^{{\bf{ - 1}}}}} \right){\bf{ / 0}}{\bf{.139 }}{{\bf{d}}^{{\bf{ - 1}}}}}\\{{\bf{t = 2}}{\bf{.303 \times 0}}{\bf{.139 }}{{\bf{d}}^{{\bf{ - 1}}}}}\end{array}\\{\bf{t = 16}}{\bf{.7 days}}\end{array}\)