Question

' $a$ ' moles of $PCl _{5}$ are heated in a closed container to equilibriate $PCl _{5}(g) \rightleftharpoons PCl _{3}(g)+ Cl _{2}(g)$ at a pressure of $p$ atm. If $x$ moles of $PCl _{5}$ dissociate at equilibrium, then

• A. $\frac{x}{a}=\left(\frac{K_{p}}{p}\right)^{1 / 2}$
• B. $\frac{x}{a}=\frac{K_{p}}{K_{p}+p}$
• C. $\frac{x}{a}=\left(\frac{K_{p}}{K_{p}+p}\right)^{1 / 2}$
• D. $\frac{x}{a}=\left(\frac{K_{p}+p}{K_{p}}\right)^{1 / 2}$

Answer 1

Answer: (C)

$\frac{x}{a}=\left(\frac{K_{p}}{K_{p}+p}\right)^{1 / 2}$

Answer 2

$\underset{1 - \alpha}{PCl _{5}(g)} \rightleftharpoons \underset{\alpha}{PCl _{3}}+ \underset{\alpha}{Cl _{2}(g)}$ (where, $\alpha=$ degree of dissociation) Total moles $=(1-\alpha)+\alpha+\alpha=1+\alpha$ $K_{p}=\frac{p_{ PCl _{3}} \times p_{ Cl _{2}}}{p_{ PCl _{5}}}$ $=\frac{\left[\left(\frac{\alpha}{1+\alpha}\right) p\right]\left[\left(\frac{\alpha}{1+\alpha}\right) p\right]}{\left[\left(\frac{1-\alpha}{1+\alpha}\right)\right]}$ $K_{p}=\frac{\alpha^{2} p}{1-\alpha^{2}}$ or $K_{p}-\alpha^{2} K_{p}=\alpha^{2} p$ or $\alpha^{2}\left(K_{p}+p\right)=K_{p}$ or $\alpha=\left(\frac{K_{p}}{K_{p}+p}\right)^{1 / 2}$