Question

$K_p$ for the reaction $PC{l}_5(g\rightleftharpoons PC{l}_3(g)+C{l}_2(g)$ 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

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

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