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Answer 1 · 2016-08-24T11:09:02

The equation of the gaseuos decomposition raction of $P C l_{5}$ $PCl_5$ with ICE table

$P C l_{5} (g) ⇌ P C l_{3} (g) + C l_{2} (g)$ $PCl_5(g)" "rightleftharpoons" "PCl_3(g)" "+" "Cl_2(g)$

$I 1 m o l 0 m o l 0 m o l$ $I" "1" "mol" "0" "mol" "0" "mol$

$C α m o l α m o l α m o l$ $C" "alpha" "mol" "alpha" "mol" "alpha" "mol$

$E 1 - α m o l α m o l α m o l$ $E" "1-alpha" "mol" "alpha" "mol" "alpha" "mol$

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

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

$n = 1 - α + α + α = 1 + α$ $n=1-alpha+alpha+alpha=1+alpha$

Mole fractions of the components at equilibrium

$χ_{P C l_{5}} = \frac{1 - α}{1 + α}$ $chi_(PCl_5)=(1-alpha)/(1+alpha)$

$χ_{P C l_{3}} = \frac{α}{1 + α}$ $chi_(PCl_3)=alpha/(1+alpha)$

$χ_{C l_{2}} = \frac{α}{1 + α}$ $chi_(Cl_2)=alpha/(1+alpha)$

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

$p_{P C l_{5}} = \frac{(1 - α) P}{1 + α}$ $p_(PCl_5)=((1-alpha)P)/(1+alpha)$

$p_{P C l_{3}} = \frac{α P}{1 + α}$ $p_(PCl_3)=(alphaP)/(1+alpha)$

$p_{C l_{2}} = \frac{α P}{1 + α}$ $p_(Cl_2)=(alphaP)/(1+alpha)$

$The equilibrium constant in respect of preesure$ $"The equilibrium constant in respect of preesure"$

$K_{p} = \frac{p_{P C l_{3}} \times p_{C l_{2}}}{p_{P C l_{5}}}$ $K_p=(p_(PCl_3)xxp_(Cl_2))/p_(PCl_5)$

Substituting respective values

$K_{p} = \frac{{(\frac{α P}{1 + α})}^{2}}{\frac{(1 - α) P}{1 + α}}$ $K_p= ((alphaP)/(1+alpha))^2 / ((( 1- alpha)P)/(1+alpha))$

$= \frac{α^{2} P}{1 - α^{2}}$ $=(alpha^2P)/(1-alpha^2)$

Now it is given $α = 0.1 and P = 5 atm$ $alpha=0.1 and P=5" atm"$

So

$K_{p} = \frac{{(0.1)}^{2} \cdot 5}{1 - {(0.1)}^{2}} \approx 0.05 a t m$ $K_p=((0.1)^2*5)/(1-(0.1)^2)~~0.05atm$