# One mole of ${\mathrm{PCl}}_{5}$  is heated in a closed vessel at volume V litre.  If x moles of ${\mathrm{PCl}}_{5}$ dissociates into ${\mathrm{PCl}}_{3}$ and ${\mathrm{Cl}}_{2}$ at equilibrium, then the expressions for ${\mathrm{K}}_{\mathrm{c}}$  and ${\mathrm{K}}_{\mathrm{P}}$ respectively are (P is the total pressure at equilibrium).

1. A

$\frac{{x}^{2}}{\left(1+x\right)V},\frac{{x}^{2}P}{\left(1+x\right)}$

2. B

$\frac{{\mathrm{x}}^{2}}{\left(1-\mathrm{x}\right)\mathrm{V}},\frac{{\mathrm{x}}^{2}\mathrm{P}}{\left(1-{\mathrm{x}}^{2}\right)}$

3. C

$\frac{{x}^{2}}{\left(1+{x}^{2}\right)V},\frac{{x}^{2}P}{\left(1+{x}^{2}\right)}$

4. D

$\frac{{x}^{2}}{\left(1-{x}^{2}\right)V},\frac{{x}^{2}P}{\left(1+{x}^{2}\right)}$

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### Solution:

$PC{I}_{5}⇌PC{I}_{3}+C{I}_{2}$

1                0        0                 initially

$\frac{1-x}{v}$           $\frac{x}{v}$       $\frac{x}{v}$              at equilibrium, ${K}_{c}=\frac{\frac{x}{v},\frac{x}{v}}{\left(\frac{1-x}{v}\right)}=\frac{{x}^{2}}{\left(1-x\right)v}$

$PC{I}_{5}⇌PC{I}_{3}+C{I}_{2}$

1               0          0                 initially

1 – x              x           x                  at equilibrium

$\frac{1-x}{1+x}p$     $\frac{x}{1+x}p$   $\frac{x}{1+x}p$     Partial pressure    :   total number of moles

= 1 – x + x + x = 1 + x     Partial pressure = mole fraction x total pressure (p)

${K}_{p}=\frac{\frac{x}{1+x}p\frac{x}{1+x}p}{\left(\frac{1-x}{1+x}\right)p}$=${K}_{p}=\frac{\frac{x}{1+x}p\frac{x}{1+x}p}{\left(\frac{1-x}{1+x}\right)p}$

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