Question

If $\alpha$ is the degree of dissociation of the reaction $2H_{2}S\left(g\right)\leftrightharpoons 2H_2\left(g\right)+S_2\left(g\right)$ and p is total equilibrium pressure, then $K_p$ of the reaction is given by

• A. $K_p=\frac{{\alpha }^{3}p}{(1+\alpha )(1-\alpha {)}^2}$
• B. $K_p=\frac{(2+\alpha )p}{(1+\alpha )(1-\alpha {)}^2}$
• C. $K_p=\frac{{\alpha }^{3}p}{(2+\alpha )(1-\alpha {)}^2}$
• D. $K_p=\frac{(1+\alpha )p}{{\alpha }^2(1-\alpha {)}^2}$

Answer 1

The correct option is C

$K_p=\frac{{\alpha }^{3}p}{(2+\alpha )(1-\alpha {)}^2}$

Given that α is the degree of dissociation and the equilibrium total pressure is p. Let there initially be n moles of $H_{2}S$ (g)
$\underset{n(1-\alpha )}{2H_{2}S\left(g\right)}\leftrightharpoons \underset{n\alpha }{2H_2\left(g\right)}+\underset{n\alpha /2}{S_2\left(g\right)}$
Total amount of gases = $n(1+\alpha /2)=n(2+\alpha )/2$
$p\left(H_{2}S\right)=\frac{2(1-\alpha )}{2+\alpha }p;p\left(H_2\right)=\frac{2\alpha }{2+\alpha }p;p\left(S_2\right)=\frac{\alpha }{2+\alpha }p$
$K_p=\frac{\left\{2\right(\alpha \left)p/\right(2+\alpha \left){\}}^2\right\{\alpha p/(2+\alpha )\}}{\left\{2\right(1-\alpha \left)p/\right(2+\alpha ){\}}^2}=\frac{{\alpha }^{3}p}{(2+\alpha )(1-\alpha {)}^2}$