Question

Consider the reaction ${\mathbf{\text{4PH}}}_{\mathbf{3}}\mathbf{\left(}\mathbf{\text{g}}\mathbf{\right)}\mathbf{\to }{\mathbf{\text{P}}}_{\mathbf{4}}\mathbf{\left(}\mathbf{\text{g}}\mathbf{\right)}\mathbf{+}\mathbf{6}{\mathbf{\text{H}}}_{\mathbf{2}}\mathbf{\left(}\mathbf{\text{g}}\mathbf{\right)}$. If, in a certain experiment, over a specific time period,$\mathbf{0}\mathbf{.}\mathbf{0048}\mathbf{}\mathbf{mol}$of${\mathbf{PH}}_{\mathbf{3}}$is consumed in a 2 L container during each second of the reaction, what are the rates of production ofrole="math" localid="1663754874175" ${\mathbf{P}}_{\mathbf{4}}$and${\mathbf{H}}_{\mathbf{2}}$in this experiment?

Short answer

The rates of ${\mathrm{P}}_2$and ${\mathrm{H}}_2$are:

$\frac{d\left[P_4\right]}{dt}=0.0006{\mathrm{molL}}^{-1}{\mathrm{s}}^{-1}$

$\frac{d\left[H_2\right]}{dt}=0.0036{\mathrm{molL}}^{-1}{\mathrm{s}}^{-1}$

Step-by-step solution

Determine the rate of reactant [PH3].

The given chemical reaction is:

$4{\mathrm{PH}}_3\left(\mathrm{g}\right)\to {\mathrm{P}}_4\left(\mathrm{g}\right)+6{\mathrm{H}}_2\left(\mathrm{g}\right)$

$\begin{array}{rcl}\mathrm{\Delta }\left[{\mathrm{PH}}_3\right]& =& \frac{0.0048\mathrm{mol}}{2\mathrm{L}}\\ & =& 0.0024{\mathrm{molL}}^{-1}\end{array}$

And,

$\begin{array}{rcl}\mathrm{Rate}& =& \frac{\mathrm{\Delta }\left[{\mathrm{PH}}_3\right]}{\mathrm{\Delta t}}\\ & =& \frac{0.0024{\mathrm{molL}}^{-1}}{1\mathrm{s}}\\ & =& 0.0024{\mathrm{molL}}^{-1}{\mathrm{s}}^{-1}\end{array}$

Determine the rate of products [P4] and [H2].

Rate of product $\left[{\mathrm{P}}_4\right]$:

$\begin{array}{rcl}\frac{\mathrm{\Delta }\left[{\mathrm{P}}_4\right]}{\mathrm{\Delta t}}& =& \frac{1}{4}\times \frac{\mathrm{\Delta }\left[{\mathrm{PH}}_3\right]}{\mathrm{\Delta t}}\\ & =& \frac{1}{4}\times 0.0024\\ & =& 0.0006{\mathrm{molL}}^{-1}{\mathrm{s}}^{-1}\end{array}$

Rate of product $\left[{\mathrm{H}}_2\right]$:

$\begin{array}{rcl}\frac{\mathrm{\Delta }\left[{\mathrm{H}}_2\right]}{\mathrm{\Delta t}}& =& \frac{6}{4}\times \frac{\mathrm{\Delta }\left[{\mathrm{PH}}_3\right]}{\mathrm{\Delta t}}\\ & =& \frac{6}{4}\times 0.0024\\ & =& 0.0036{\mathrm{molL}}^{-1}{\mathrm{s}}^{-1}\end{array}$