Question

For the reaction $\mathit{A}\mathbf{}\mathbf{\to }\mathit{p}\mathit{r}\mathit{o}\mathit{d}\mathit{u}\mathit{c}\mathit{t}\mathit{s}$, successive half-lives are

observed to be 10.0, 20.0, and 40.0 min for an experiment

in which ${\left[A\right]}_{\mathbf{0}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{10}\mathbf{M}$. Calculate the concentration

of A at the following times.

a. 80.0 min b. 30.0 min

Short answer

1. The concentration at 89 min is $\left[\mathrm{A}\right]=1.1\times {10}^{-2}\mathrm{MS}$
2. The concentration at 30 min is 0.025

Step-by-step solution

(a) Calculating the concentration of A at 80 min

Given data

$\begin{array}{l}{\mathrm{t}}_{\frac{1}{2}}=10,20\mathrm{and}40\min ,\\ {\left[\mathrm{A}\right]}_0=0.10\mathrm{M},\end{array}$

The reaction is bimolecular.

For t ½,

$\begin{array}{l}{\mathbf{t}}_{\frac{\mathbf{1}}{\mathbf{2}}}\mathbf{=}\mathbf{1}\mathbf{/}{\left[A\right]}_{\mathbf{0}}\mathbf{k}\\ {\mathrm{t}}_{\frac{1}{2}}=1/0.1\mathrm{k}\\ \mathrm{k}=1/\left(0.1\right)\left(10\right)\\ \mathrm{k}=1{\mathrm{M}}^{-1}{\min }^{-1}\end{array}$

Solve for concentration:

$\begin{array}{l}1/\left[\mathrm{A}\right]=1/{\left[\mathrm{A}\right]}_0+\mathrm{kt}\\ 1/\left[\mathrm{A}\right]=1/.1+1.0\left(80.0\right)\\ 1/\left[\mathrm{A}\right]=90\\ \left[\mathrm{A}\right]=1.1\times {10}^{-2}\mathrm{MS}\end{array}$

(b)Calculating the concentration of A at 30 min

$\begin{array}{l}1/\left[\mathrm{A}\right]=1/{\left[\mathrm{A}\right]}_0+\mathrm{kt}\\ 1/\left[\mathrm{A}\right]=1/.1+1.0\left(30.0\right)\\ 1/\left[\mathrm{A}\right]=40\\ \left[\mathrm{A}\right]=0.025\mathrm{M}\end{array}$

The concentration at 30 min is 0.025