Question

The degree of dissociation of $S{O}_3$ is $\alpha$ at equilibrium pressure P. $K_P$ for $S{O}_3\left(g\right)\rightleftharpoons S{O}_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)$

• A. $\frac{P^2{\alpha }^2}{2(1-\alpha {)}^3}$
• B. $\frac{P^{\frac{1}{2}}{\alpha }^{3/2}}{(1-\alpha )(2+\alpha {)}^{1/2}}$
• C. $\frac{P^2{\alpha }^2}{2(1-\alpha {)}^2}$
• D. None of these

Answer 1

The correct option is B $\frac{P^{\frac{1}{2}}{\alpha }^{3/2}}{(1-\alpha )(2+\alpha {)}^{1/2}}$

$S{O}_3\left(g\right)\rightleftharpoons S{O}_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)$
$\begin{array}{cccccc}1& & -& & -& \\ 1-\alpha & & \alpha & & & \frac{\alpha }{2}\end{array}$
Total no. of moles $=1+\frac{\alpha }{2}$
$P_{S{O}_3}=\frac{(1-\alpha )P}{1+\frac{\alpha }{2}}{P}_{S{O}_2}=\frac{\left(\alpha P\right)}{1+\frac{\alpha }{2}}{P}_{O_2}=\frac{\left(\frac{\alpha }{2}\right)P}{1+\frac{\alpha }{2}}$
$K_P=\frac{P_{S{O}_2}\times P_{O_2}^{\frac{1}{2}}}{P_{S{O}_3}}$

putting the respective values,

$K_P=\frac{P^{\frac{1}{2}}{\alpha }^{3/2}}{(1-\alpha )(2+\alpha {)}^{1/2}}$