Question

The volume percentage of $Cl_{2}$ at equilibrium in the dissociation of $PCl_{5}$ under a total pressure of $1.5atm$ is $\left(\right.Kp=0.202\left.\right)$ ,

• A. 74.5
• B. 36.5
• C. 63.5
• D. 26.6

Answer 1

Answer: (D)

$PCl_{5}\rightleftharpoons PCl_{3}+Cl_{2}$

$1-x$ $x$ $x$

Total moles $=1-x+x+x=1+x$

$\mathrm{k}_{\mathrm{p}}=\frac{\mathrm{P}_{\mathrm{PCl}_3} \cdot \mathrm{PCl}_2}{\mathrm{P}_{\mathrm{PCl}_5}}=\frac{\left(\frac{\mathrm{X}}{1+\mathrm{x}}\right) \mathrm{P}}{\left(\frac{1-\mathrm{X}}{1+\mathrm{x}}\right) \mathrm{P}}\left(\frac{\mathrm{x}}{1+\mathrm{x}}\right) \mathrm{P}$

$k_{P}=\frac{x^{2} P}{1 - x^{2}}$ $1-x^{2}=1$

$k_{P}=x^{2}P$

$x=\sqrt{\frac{k p}{P}}=\sqrt{\frac{0.202}{1.5}}=\sqrt{0.134}=0.366$

Mole $ \propto $ volume at constant $P\&T$

Volume percentage $=\frac{m o l e \, o f \, C l_{2}}{t o t a l m o l e s}\times 100$

$=\frac{0.366}{1.366}\times 100=26.86\%$