Question

In a $500ml$ flask, the degree of dissociation of $PC{l}_5$ at equilibrium is $40\%$ and the initial amount is $5$ moles. The value of equilibrium constant in $mol.li{t}^{-1}$ for the decomposition of $PC{l}_5$ is :

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

Answer 1

Answer: (B)

$2.66$

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

5 0 0 moles initially

3 2 2 moles at equilibrium

$\therefore \left[PC{l}_5\right]=6M$

$\left[PC{l}_3\right]=4M$

$\left[C{l}_2\right]=4M$

Now according to law of mass action, the expression of $K_c$ is given as:

$K_c=\frac{4\times 4}{6}=2.66$