Question

The rate law for the decomposition of phosphine $\left({\mathrm{PH}}_3\right)$ is $\mathbf{Rate}\mathbf{=}\mathbf{d}\left[{\mathrm{PH}}_3\right]\mathbf{/}\mathbf{dt}\mathbf{=}\mathbf{k}\left[{\mathrm{PH}}_3\right]$. It takes 120s for the concentration of 1.00 M PH₃ to decrease to 0.250M. How much time is required for 2.00 M$\mathbf{\left(}{\mathbf{PH}}_{\mathbf{3}}\mathbf{\right)}$ to decrease to a concentration of 0.350 M?

Short answer

The time required for 2.00M$\left({\mathrm{PH}}_3\right)$ to decrease the concentration of 0.350M is 150.83s.

Step-by-step solution

Describing the calculation of the rate law and rate constant

$\mathrm{Rate}=\frac{\mathrm{d}\left[\mathrm{PH}3\right]}{\left[\mathrm{PH}3\right]}=\mathrm{k}\left[{\mathrm{PH}}_3\right]$

${\int }_{\left[\mathrm{PH}3\right]0}^{\left[\mathrm{PH}3\right]}\frac{\mathrm{d}\left[\mathrm{PH}3\right]}{\left[\mathrm{PH}3\right]}=-\mathrm{k}{\int }_0^{\mathrm{t}}\mathrm{t}$

By solving,

$\begin{array}{rcl}\mathrm{In}\left[\mathrm{PH}3\right]& =& \mathrm{In}{\left[\mathrm{PH}3\right]}_0-\mathrm{kt}\\ \mathrm{In}\left(0.250\right)& =& \mathrm{In}\left(1\right)--\mathrm{k}\left(120\right)\\ -1.38& =& 0--120\mathrm{k}\\ \mathrm{k}& =& 0.01155{\mathrm{s}}^{-1}\end{array}$

Describing the calculation of the required time

Using the first order concentration time equation,

$\begin{array}{rcl}\mathrm{In}\left[\mathrm{PH}3\right]& =& \mathrm{In}[\mathrm{PH}3{]}_0-\mathrm{kt}\\ \mathrm{In}\left(0.350\right)& =& \mathrm{In}\left(2\right)--0.01155\mathrm{t}\\ -1.049& =& 0.693--0.01155\mathrm{t}\\ 1.74& =& 0.01155\mathrm{t}\\ \mathrm{t}& =& 150.83\mathrm{s}\end{array}$

Therefore, the time required for 2.00M PH₃ to decrease the concentration of 0.350M is 150.83 s.