Question

The degree of dissociation of $N_2{O}_4$ into $N{O}_2$ at 1.5 atmosphere and ${40}^{\circ }C$ is 0.25. Calculate its $K_p$ at ${40}^{\circ }C$.

• A. $K_p=0.4$
• B. $K_p=0.8$
• C. $K_p=0.2$
• D. $K_p=0.6$

Answer 1

Answer: (A)

$K_p=0.4$

$N_2{O}_4\to 2N{O}_2$

$N_a$ $0$

At eqb. $N_a(1-\alpha )$ $2N_a\alpha$

Total number of moles= $N_a(1+\alpha )$

Mole fraction of $N_2{O}_4=\frac{(1-\alpha )}{(1+\alpha )}$

Mole fraction of $N{O}_2=\frac{2\alpha }{(1+\alpha )}$

Given: $\alpha =$ 0.25

$K_p=P_t\frac{\left(4{\alpha }^2\right)}{(1-{\alpha }^2)}$

$=1.5\frac{({0.25}^2\times 4)}{(1-(0.25{)}^2)}$

$=0.4$