Question

Calculate the half-life, in years, for the reaction$\mathbf{2}\mathbf{X}\mathbf{\to }\mathbf{Y}$ when the starting concentration of $\mathbf{X}$ is$\mathbf{6}\mathbf{\mu M}$ and the rate constant is $\mathbf{3}\mathbf{.}\mathbf{6}\mathbf{\times }{\mathbf{10}}^{\mathbf{‐}\mathbf{3}}{\mathbf{M}}^{\mathbf{‐}\mathbf{1}}{\mathbf{s}}^{\mathbf{‐}\mathbf{1}}$.

Short answer

The half-life in years for the reaction $2\mathrm{X}\to \mathrm{Y}$is $1.47$years.

Step-by-step solution

Introduction

Thehalf-life period is the time it takes for a substance's concentration to decline to half of its initial value. The half-life of a reaction is unaffected by the reaction's initial concentration.

Calculate the half-life

The initial concentration of $\mathrm{X}$ is $6\mathrm{\mu M}$.

Rate constant$=3.6\times {10}^{‐3}{\mathrm{M}}^{‐1}{\mathrm{s}}^{‐1}$

The half-lifeof a bimolecular reaction is

$$\begin{gathered} {\mathrm{t}}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{1}{\mathrm{k}\left[\mathrm{A}\right]} \\ {\mathrm{t}}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{1}{\left(3.6\times {10}^{‐3}{\mathrm{M}}^{‐1}{\mathrm{s}}^{‐1}\right)\times \left(6\times {10}^{‐6}\mathrm{M}\right)} \\ =4.63\times {10}^9\mathrm{s} \end{gathered}$$

Calculate the half-life in years

The half-life period in years is

$$\begin{gathered} =4.63\times {10}^7\times \frac{1}{3600}\times \frac{1}{24}\times \frac{1}{365} \\ =1.47\mathrm{years} \end{gathered}$$

Thus, the half-life for the reaction of $2\mathrm{X}\to \mathrm{Y}$is $1.47$years.