Question

After 42.0 min, 26.0% of a compound has decomposed. What is the half-life of this reaction assuming first-order kinetics?

_(answer)____ min

Answer 1

Concepts and reason

The concept used to solve this problem is first-order reaction.

The rate for a first-order reaction with respect to reactant ${\rm{A}}$ is equal to product of rate constant and the concentration of reactant ${\rm{A}}$ . The half-life of a first order reaction is independent of the concentration of reactant. It is calculated by dividing 0.693 by the rate constant of the reaction.

Fundamentals

The integrated rate law for the first order kinetics is:

$\left[ {\rm{A}} \right] = {\left[ {\rm{A}} \right]_{\rm{o}}}{e^{ - kt}}$

Here, $\left[ {\rm{A}} \right]$ is the concentration of reactant ${\rm{A}}$ at time $t$ , ${\left[ {\rm{A}} \right]_0}$ is the concentration of reactant ${\rm{A}}$ at time 0, $k$ is the first-order rate constant.

The half-life $\left( {{t_{1/2}}} \right)$ of a first order reaction is calculated as:

${t_{1/2}} = \frac{{0.693}}{k}$

Here, $k$ is the first-order rate constant.

Let the initial amount of the $\left( {{{\left[ {\rm{A}} \right]}_0}} \right)$ compound be 100%.

The amount of the compound decomposed is 26%.

The remaining amount $\left( {\left[ {\rm{A}} \right]} \right)$ of the sample is:

$\begin{array}{c}\\\left[ {\rm{A}} \right] = \left( {100 - 26} \right)\% \\\\ = 74\% \\\end{array}$

The integrated rate law for the first order kinetics is:

$\left[ {\rm{A}} \right] = {\left[ {\rm{A}} \right]_{\rm{o}}}{e^{ - kt}}$

Rearrange the above equation for $k$ .

$k = - \frac{1}{t}\ln \frac{{\left[ {\rm{A}} \right]}}{{{{\left[ {\rm{A}} \right]}_{\rm{o}}}}}$

Substitute $42.0{\rm{ min}}$ for $t$ , $74\%$ for $\left[ {\rm{A}} \right]$ and 100% for ${\left[ {\rm{A}} \right]_{\rm{o}}}$ .

$\begin{array}{c}\\k = - \frac{1}{{42.0{\rm{ min}}}}\ln \left( {\frac{{74}}{{100}}} \right)\\\\ = - \frac{1}{{42.0{\rm{ min}}}}\left( { - 0.3011} \right)\\\\ = 7.17 \times {10^{ - 3}}{\rm{ mi}}{{\rm{n}}^{ - 1}}\\\end{array}$

Therefore, the rate constant from the first-order rate equation is $7.17 \times {10^{ - 3}}{\rm{ mi}}{{\rm{n}}^{ - 1}}$ .

The half-life $\left( {{t_{1/2}}} \right)$ of a first order reaction is calculated as:

${t_{1/2}} = \frac{{0.693}}{k}$

Substitute $7.17 \times {10^{ - 3}}{\rm{ mi}}{{\rm{n}}^{ - 1}}$ for $k$ in the above equation.

$\begin{array}{c}\\{t_{1/2}} = \frac{{0.693}}{{7.17 \times {{10}^{ - 3}}{\rm{ mi}}{{\rm{n}}^{ - 1}}}}\\\\ = 96.65{\rm{ min}}\\\end{array}$

Therefore, the half-life of the first order reaction is $96.65{\rm{ min}}$ .

Ans:

The half-life of the reaction assuming first-order kinetics is $96.65{\rm{ min}}$ .