Question

At temperature $TK,PC{l}_5$ is $50\%$ dissociated at an equilibrium pressure of $4atm$. At what pressure it would dissociate to the extent of $80\%$ at the same temperature?

• A. $0.05atm$
• B. $0.60atm$
• C. $0.75atm$
• D. $2.50atm$

Answer 1

Answer: (C)

$0.75atm$

${PCl}_5\rightleftharpoons {PCl}_3+{Cl}_2$

(moles) Initial $1$ $0$ $0$

At eqilibrium $1-x$ $x$ $x$

Total no. of moles at eqm. $=1-x+x+x=1+x$

Partial pressure of a gas = mole fraction $\times$ total pressure

Partial pressure $\frac{1-x}{1+x}P\frac{x}{1+x}P\frac{x}{1+x}P$

$K_p=\frac{P_{{PCl}_3}\times P_{{Cl}_2}}{P_{{PCl}_5}}=\frac{\frac{x}{1+x}P\frac{x}{1+x}P}{\frac{1-x}{1+x}P}=\frac{x^{2}P}{1-x^2}$

$=\frac{{\left(0.5\right)}^2\times 4}{1-{\left(0.5\right)}^2}=1.333(\because x=50$% dissociation)

For $80$% dissociation,

$K_P=\frac{x^{2}P}{1-x^2}\Rightarrow 1.33=\frac{{\left(0.8\right)}^2\times P}{1-{\left(0.8\right)}^2}$

$\therefore P=0.75atm$

Hence, the correct option is C.