# Question 3d1a8

Aug 24, 2016

The equation of the gaseuos decomposition raction of $P C {l}_{5}$ with ICE table

$P C {l}_{5} \left(g\right) \text{ "rightleftharpoons" "PCl_3(g)" "+" } C {l}_{2} \left(g\right)$

$I \text{ "1" "mol" "0" "mol" "0" } m o l$

$C \text{ "alpha" "mol" "alpha" "mol" "alpha" } m o l$

$E \text{ "1-alpha" "mol" "alpha" "mol" "alpha" } m o l$

color(red)("where "alpha" degree of dissociation"=10%=0.1#

Total no. of moles in the reaction mixture at equilibrium is given by

$n = 1 - \alpha + \alpha + \alpha = 1 + \alpha$

Mole fractions of the components at equilibrium

${\chi}_{P C {l}_{5}} = \frac{1 - \alpha}{1 + \alpha}$

${\chi}_{P C {l}_{3}} = \frac{\alpha}{1 + \alpha}$

${\chi}_{C {l}_{2}} = \frac{\alpha}{1 + \alpha}$

If atequilibrium the total pressure of the reaction mixture is P then the partial pressures of the components in the mixture will be

${p}_{P C {l}_{5}} = \frac{\left(1 - \alpha\right) P}{1 + \alpha}$

${p}_{P C {l}_{3}} = \frac{\alpha P}{1 + \alpha}$

${p}_{C {l}_{2}} = \frac{\alpha P}{1 + \alpha}$

$\text{The equilibrium constant in respect of preesure}$

${K}_{p} = \frac{{p}_{P C {l}_{3}} \times {p}_{C {l}_{2}}}{p} _ \left(P C {l}_{5}\right)$

Substituting respective values

${K}_{p} = {\left(\frac{\alpha P}{1 + \alpha}\right)}^{2} / \left(\frac{\left(1 - \alpha\right) P}{1 + \alpha}\right)$

$= \frac{{\alpha}^{2} P}{1 - {\alpha}^{2}}$

Now it is given $\alpha = 0.1 \mathmr{and} P = 5 \text{ atm}$

So

${K}_{p} = \frac{{\left(0.1\right)}^{2} \cdot 5}{1 - {\left(0.1\right)}^{2}} \approx 0.05 a t m$