Question

$K_{p}$ for the reaction $PCl_{5} (g \rightleftharpoons PCl_{3}(g)+Cl_{2}(g)$ at $\text{250}^{\text{o}} \text{C}$ is $\text{0} \text{.82}$ . Calculate the degree of dissociation at given temperature under a pressure of $\text{5} \, \text{atm}$ . What will be the degree of dissociation if the equilibrium pressure is $\text{10} \, \text{atm}$ at same temperature.

• A. 27.5%
• B. 23%
• C. 35.5%
• D. 40%

Answer 1

Answer: (A)

$K_{P}=\frac{\left(\frac{x}{1 + x}\right) P \times \left(\frac{x}{1 + x}\right) P}{\left(\frac{1 - x}{1 + x}\right) P}$

$K_{P}=\frac{x^{2}}{1 - x^{2}}P=0.82$

$x^{2}=\sqrt{\frac{0.82}{5.82}}$

$x=0.375$

Now new pressure is 10

$K_{P}=\frac{y^{2} \times P}{1 - y^{2}}$ , $0.82=\frac{y^{2}}{1 - y^{2}}\times 10$

$y=0.275$ , $y=27.5\%$