Chapter 5, Problem 5.41E

(a)

Interpretation Introduction

Interpretation:

When temperature K is $4.96$, and the total pressure is $0.50bar$, the fractional decomposed has to be calculated.

Explanation of Solution

The decomposition reaction is given below,

${\text{PCl}}_{\text{5}}\text{ (g)}\to {\text{PCl}}_3\text{ (g) }\text{ +}{\text{Cl}}_{\text{2}}\text{ (g)}$

The dissociation constant can be expressed as follow,

ICE table:

Pressure            ${\text{PCl}}_{\text{5}}\text{ (g)}\to {\text{PCl}}_3\text{ (g) }\text{ +}{\text{Cl}}_{\text{2}}\text{ (g)}$

Initial concentration	$n$	0	0
Change	$-n\alpha$	$+n\alpha$	$+n\alpha$
At equilibrium	$n(1-\alpha )$		$+n\alpha$

  $\begin{array}{l}{\text{PCl}}_{\text{5}}\text{ (g)}\to {\text{PCl}}_{\text{3}}\text{ (g) }\text{ +}{\text{Cl}}_{\text{2}}\text{ (g)}\\ \text{K =}\frac{\left(\text{nα}\right)\left(\text{nα}\right)}{\text{n(1-α)}}\\ \text{K =}\frac{{\text{nα}}^{\text{2}}}{\text{(1-α)}}\end{array}$

Pressure,

  $\begin{array}{l}\text{P = n(1 - }\alpha \text{) + n}\alpha \text{ + n}\alpha \text{ = n(1 + }\alpha \text{)}\\ \text{P = n(1 + }\alpha \text{)}\\ \text{therefore,}\\ \text{n}\text{=}\frac{\text{P}}{\text{(1 + }\alpha \text{)}}\end{array}$

Substitute the values,

  $\begin{array}{l}\text{K =}\frac{{\text{nα}}^{\text{2}}}{\text{(1-α)}}\\ \text{K =}\left(\frac{\text{P}}{\text{(1 + α)}}\right)\left(\frac{{\text{α}}^{\text{2}}}{\text{(1-α)}}\right)\\ \text{K =}\left(\frac{{\text{Pα}}^{\text{2}}}{{\text{(1 - α}}^{\text{2}}\text{)}}\right)\\ \\ \text{K}\text{=}{\text{Pα}}^{\text{2}}\text{+}{\text{Kα}}^{\text{2}}\\ \\ {\text{α}}^{\text{2}}\text{=}\frac{\text{K}}{\left(\text{P}\text{+}\text{K}\right)}\\ \text{α}\text{=}\sqrt{\frac{\text{K}}{\left(\text{P}\text{+}\text{K}\right)}}\end{array}$

Conditions at  $K=4.96andP=0.50bar$,

  $\begin{array}{l}\text{α}\text{=}\sqrt{\frac{\text{K}}{\left(\text{P}\text{+}\text{K}\right)}}\\ \text{α}\text{=}\sqrt{\frac{4.96}{\left(0.50\text{+}4.96\right)}}\\ \text{α}\text{=}0.953\end{array}$

(b)

Interpretation Introduction

Interpretation:

When temperature K is $4.96$, and the total pressure is $\text{1}\text{.00 }bar$, the fractional decomposed has to be calculated.

Explanation of Solution

The decomposition reaction is given below,

${\text{PCl}}_{\text{5}}\text{ (g)}\to {\text{PCl}}_3\text{ (g) }\text{ +}{\text{Cl}}_{\text{2}}\text{ (g)}$

The dissociation constant can be expressed as follow,

ICE table:

Pressure            ${\text{PCl}}_{\text{5}}\text{ (g)}\to {\text{PCl}}_3\text{ (g) }\text{ +}{\text{Cl}}_{\text{2}}\text{ (g)}$

Initial concentration	$n$	0	0
Change	$-n\alpha$	$+n\alpha$	$+n\alpha$
At equilibrium	$n(1-\alpha )$		$+n\alpha$

  $\begin{array}{l}{\text{PCl}}_{\text{5}}\text{ (g)}\to {\text{PCl}}_{\text{3}}\text{ (g) }\text{ +}{\text{Cl}}_{\text{2}}\text{ (g)}\\ \text{K =}\frac{\left(\text{nα}\right)\left(\text{nα}\right)}{\text{n(1-α)}}\\ \text{K =}\frac{{\text{nα}}^{\text{2}}}{\text{(1-α)}}\end{array}$

Pressure,

  $\begin{array}{l}\text{P = n(1 - }\alpha \text{) + n}\alpha \text{ + n}\alpha \text{ = n(1 + }\alpha \text{)}\\ \text{P = n(1 + }\alpha \text{)}\\ \text{therefore,}\\ \text{n}\text{=}\frac{\text{P}}{\text{(1 + }\alpha \text{)}}\end{array}$

Substitute the values,

  $\begin{array}{l}\text{K =}\frac{{\text{nα}}^{\text{2}}}{\text{(1-α)}}\\ \text{K =}\left(\frac{\text{P}}{\text{(1 + α)}}\right)\left(\frac{{\text{α}}^{\text{2}}}{\text{(1-α)}}\right)\\ \text{K =}\left(\frac{{\text{Pα}}^{\text{2}}}{{\text{(1 - α}}^{\text{2}}\text{)}}\right)\\ \\ \text{K}\text{=}{\text{Pα}}^{\text{2}}\text{+}{\text{Kα}}^{\text{2}}\\ \\ {\text{α}}^{\text{2}}\text{=}\frac{\text{K}}{\left(\text{P}\text{+}\text{K}\right)}\\ \text{α}\text{=}\sqrt{\frac{\text{K}}{\left(\text{P}\text{+}\text{K}\right)}}\end{array}$

Conditions at  $K=4.96andP=1.00bar$,

  $\begin{array}{l}\text{α}\text{=}\sqrt{\frac{\text{K}}{\left(\text{P}\text{+}\text{K}\right)}}\\ \text{α}\text{=}\sqrt{\frac{4.96}{\left(1.00\text{+}4.96\right)}}\\ \text{α}\text{=}0.912\end{array}$