Question

A first-order reaction is 75.0% complete in 320s.

a. What are the first and second half-lives for this reaction?

b. How long does it take for 90.0% completion?

Short answer

1. The first and second half lives are 16 and 0.00216s.
2. The time taken for 90% conversion is 5325 s.

Step-by-step solution

Calculating the first and second half-lives

For first-half life,

$\begin{array}{l}\mathit{I}\mathit{n}\left[A\right]\mathbf{=}\mathit{I}\mathit{n}{\left[A\right]}_{\mathbf{0}}\mathbf{}\mathbf{-}\mathit{k}\mathit{t}\\ In\left(25\right)=In\left(100\right)-k\left(320\right)\\ In\left(0.25\right)=-320k\\ k=0.0433s^{-1}\end{array}$

Solve further as:

$\begin{array}{c}t\left(\frac{1}{2}\right)=\frac{0.693}{k}\\ =\frac{0.693}{0.0433}\\ =16\mathrm{s}\end{array}$

For second-half life,

$\begin{array}{l}In\left[A\right]=In{\left[A\right]}_0-k{t}_{\frac{1}{2}}\\ In\left(0.5\right)=In\left(1\right)--t_{\frac{1}{2}}\left(0.0433\right)\\ In\left(0.5\right)=-0.0433t_{\frac{1}{2}}\\ t_{\frac{1}{2}}=0.00216\mathrm{s}\end{array}$

The first and second half lives are 16 and 0.00216 s.

Calculating how long time taken for 90% conversion

Solve for the time as:

$\begin{array}{l}In\left[A\right]=In{\left[A\right]}_0-kt\\ In\left(10\right)-In\left(100\right)=-\left(0.0433\right)t\\ t=5325s\end{array}$

Hence, the time taken for 90% conversion is 5325 s.