Question

At ${227}^{\circ }C$ and $4atm$, $PC{l}_5$ is dissociated to an extent $50\%$. At same temperature, extent of dissociation of $PC{l}_5$ is $0.75$ at a pressure of:

[Hint: $\alpha \propto \frac{1}{\sqrt{P}}]$

• A. $1.77atm$
• B. $2atm$
• C. $4atm$
• D. $8atm$

Answer 1

Answer: (A)

$1.77atm$

$\text{Since},\alpha \propto \frac{1}{\sqrt{P}}$

$\frac{{\alpha }_1}{{\alpha }_2}=\sqrt{\frac{P_2}{P_1}}$

${\left(\frac{{\alpha }_1}{{\alpha }_2}\right)}^2=\frac{P_2}{P_1}$

$P_2=\frac{{0.5}^2\times 4}{{0.75}^2}=1.77atm$