Question

In a $500mL$ flask, the degree of dissociation of $P{Cl}_5$ at equilibrium is $40$% and the initial amount is $5$ moles. The value of equilibrium constant in $mol{L}^{-1}$ for the decomposition of $P{Cl}_5$ is:

• A. $2.33$
• B. $2.66$
• C. $5.32$
• D. $4.66$

Answer 1

Answer: (B)

$2.66$

$PC{l}_5\rightleftharpoons PC{l}_3+C{l}_2$

At a o o

At equilibrium $a(1-\alpha )$ $a\alpha$ $a\alpha$

$\alpha =\frac{40}{100}=0.4$

$a=\frac{5}{0.5}=10mol{L}^{-1}$

$\therefore$ Equilibrium constant(k)= $ \frac{(a\alpha) x (a\alpha)}{a(1-\alpha )} $

$=\frac{\left(10x0.4\right)x\left(10x0.4\right)}{10(1-0.4)}$

$\therefore$ Equilibrium constant(k)= 2.66 mol L${}^{-1}$