Question

For which of the following reaction the degree of dissociation $\left(\alpha \right)$ and equilibrium constant $\left(K_p\right)$ are related as $K_p=\frac{{4\alpha }^{2}P}{(1-{\alpha }^2)}$?

• A. $N_2{O}_4\left(g\right)\rightleftharpoons 2{NO}_2\left(g\right)$
• B. $H_2\left(g\right)+I_2\left(g\right)\rightleftharpoons 2HI\left(g\right)$
• C. $N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2N{H}_3\left(g\right)$
• D. $PC{l}_3\left(g\right)+{Cl}_2\left(g\right)\rightleftharpoons PC{l}_5\left(g\right)$

Answer 1

Answer: (A)

$N_2{O}_4\left(g\right)\rightleftharpoons 2{NO}_2\left(g\right)$

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

initial $1$ $0$

- $\alpha$ $2\alpha$

$\overline{1-\alpha 2\alpha }$

total moles at $e{q}^m$

$=1-\alpha +2\alpha$

$1+\alpha$

$P$ is total pressure $=molefraction\times P_{total}$

$P_{N_2}{O}_4=\frac{(1-\alpha )}{1+\alpha }\times P_t$

$P_N{O}_2=\frac{2\alpha }{1+\alpha }\times P_t$

$K_p=\frac{(PN{O}_2{)}^2}{P_{N_2}{O}_4}$

$=\frac{\frac{(2\alpha {)}^2}{(1+\alpha {)}^2}}{\frac{1-\alpha }{1+\alpha }}{P}_t$

$=\frac{4{\alpha }^{2}p}{1-{\alpha }^2}$