Chapter 5, Problem 5.46E

(a)

Interpretation Introduction

Interpretation:

The equilibrium partial pressures of $H_{2}andP{H}_3$ has to be calculated.

Explanation of Solution

The given reaction is shown below,

  ${\text{2 PH}}_{\text{3}}\text{ (g)}\to \text{2P(s) }\text{ +}3{\text{H}}_{\text{2}}\text{(g) }$

The reaction can be expressed as follow,

The reaction tells that two molecule of phosphine gives two molecule of phosphorus and three molecules of water.

ICE table:

Pressure     ${\text{2 PH}}_{\text{3}}\text{ (g)}\to \text{2P(s) }\text{ +}3{\text{H}}_{\text{2}}\text{(g) }$

Initial concentration	$0.64$	0	0
Change	$-2\alpha$	$+2\alpha$	$+3\alpha$
At equilibrium	$0.64-2\alpha$	$+2\alpha$	$+3\alpha$

  $\begin{array}{l}{\text{2 PH}}_{\text{3}}\text{ (g)}\to \text{2P(s) }\text{ +}3{\text{H}}_{\text{2}}\text{(g) }\\ \text{K =}\frac{{\left[{\text{H}}_{\text{2}}\right]}^3}{{\left[{\text{PH}}_{\text{3}}\right]}^2}\end{array}$

Pressure,

  $\begin{array}{l}\text{P = 3}\alpha \text{ + 0}\text{.64}\text{- 2}\alpha =0.93\\ \text{P = }\alpha \text{ + 0}\text{.64}\text{=}\text{0}\text{.93}\\ \alpha \text{ =}\text{0}\text{.29}\end{array}$

The equilibrium partial pressures of $H_{2}andP{H}_3$ is given below,

  $\begin{array}{l}\left[{\text{PH}}_{\text{3}}\right]\text{ =}0.64-2\alpha \\ \left[{\text{PH}}_{\text{3}}\right]\text{ =}0.64-2(0.29)\\ \left[{\text{PH}}_{\text{3}}\right]\text{ =}0.06\end{array}$

  $\begin{array}{l}\left[{\text{H}}_2\right]\text{ =}3\alpha \\ \left[{\text{H}}_2\right]\text{ =}3(0.29)\\ \left[{\text{H}}_2\right]\text{ =}0.87\end{array}$

The equilibrium partial pressures of $H_2\text{ is }0.87andP{H}_{3}is0.06$.

(b)

Interpretation Introduction

Interpretation:

Once equilibrium is reached, the mass of phosphorus produced has be calculated.

Explanation of Solution

The equilibrium partial pressures of $P_2$ is given below,

  $\begin{array}{l}\left[\text{P}\right]\text{ =}2\alpha \\ \left[\text{P}\right]\text{ =}2(0.29\\ \left[\text{P}\right]\text{ =}0.58\end{array}$

Number of moles of phosphorous is calculated as follows,

  $\begin{array}{l}\text{Number}\text{of}\text{moles}\text{of}\text{phosphorous}\\ \text{PV}\text{=}\text{nRT}\\ \text{n}\text{=}\frac{\text{PV}}{\text{RT}}\\ \text{n}\text{=}\frac{\text{0}\text{.58}\text{bar×1}\text{L}}{\left(\text{0}\text{.0821}\text{L}\cdot \text{atm}\cdot {\text{mol}}^{\text{-1}}\cdot {\text{K}}^{\text{-1}}\text{×}\text{298}\text{K}\right)\text{×}\left(\frac{\text{1}\text{bar}}{{\text{1×10}}^{\text{5}}\text{pa}}\right)\text{×}\left(\frac{\text{101325}\text{pa}}{\text{1}\text{atm}}\right)}\\ \text{n}\text{=}\text{0}\text{.023396}\text{mol}\end{array}$

Mass in grams of phosphorous is calculated as follows,

  $\begin{array}{l}\text{Mass}\text{=}\text{Number}\text{of}\text{moles}\text{×}\text{Molar}\text{mass}\\ \text{Mass}\text{=}\text{0}\text{.0233965}\text{mol×}\text{30}\text{.973762 g/mol}\\ \text{Mass}\text{=}\text{0}\text{.72}\text{g}\end{array}$

Mass in grams of phosphorous is $\text{0}\text{.72}\text{g}$.

(c)

Interpretation Introduction

Interpretation:

Equilibrium constant K has be calculated.

Explanation of Solution

The given reaction is shown below,

  ${\text{2 PH}}_{\text{3}}\text{ (g)}\to \text{2P(s) }\text{ +}3{\text{H}}_{\text{2}}\text{(g) }$

The reaction can be expressed as follow,

The reaction tells that two molecule of phosphine gives two molecule of phosphorus and three molecules of water.

ICE table:

Pressure            ${\text{2 PH}}_{\text{3}}\text{ (g)}\to \text{2P(s) }\text{ +}3{\text{H}}_{\text{2}}\text{(g) }$

Initial concentration	$0.64$	0	0
Change	$-2\alpha$	$+2\alpha$	$+3\alpha$
At equilibrium	$0.64-2\alpha$	$+2\alpha$	$+3\alpha$

  $\begin{array}{l}{\text{2 PH}}_{\text{3}}\text{ (g)}\to \text{2P(s) }\text{ +}3{\text{H}}_{\text{2}}\text{(g) }\\ \text{K =}\frac{{\left[{\text{H}}_{\text{2}}\right]}^3}{{\left[{\text{PH}}_{\text{3}}\right]}^2}\end{array}$

Pressure,

  $\begin{array}{l}\text{P = 3}\alpha \text{ + 0}\text{.64}\text{- 2}\alpha =0.93\\ \text{P = }\alpha \text{ + 0}\text{.64}\text{=}\text{0}\text{.93}\\ \alpha \text{ =}\text{0}\text{.29}\end{array}$

The equilibrium partial pressures of $H_{2}andP{H}_3$ is given below,

  $\begin{array}{l}\left[{\text{PH}}_{\text{3}}\right]\text{ =}0.64-2\alpha \\ \left[{\text{PH}}_{\text{3}}\right]\text{ =}0.64-2(0.29)\\ \left[{\text{PH}}_{\text{3}}\right]\text{ =}0.06\end{array}$

  $\begin{array}{l}\left[{\text{H}}_2\right]\text{ =}3\alpha \\ \left[{\text{H}}_2\right]\text{ =}3(0.29)\\ \left[{\text{H}}_2\right]\text{ =}0.87\end{array}$

The equilibrium partial pressures of $H_2\text{ is }0.87andP{H}_{3}is0.06$.

Equilibrium constant K is calculated as follows,

  $\begin{array}{l}\text{K =}\frac{{\left[{\text{H}}_{\text{2}}\right]}^3}{{\left[{\text{PH}}_{\text{3}}\right]}^2}\\ \text{K =}\frac{{\left(0.87\right)}^3}{{\left(0.06\right)}^2}=182.9\\ \text{K =}183\end{array}$

Equilibrium constant K is 183.