Question

The degree of dissociation of $S{O}_3$ is $\alpha$ at equilibrium pressure $p^o$.
$K_p$ for $2S{O}_3\left(g\right)\rightleftharpoons 2S{O}_2\left(g\right)+O_2\left(g\right)$ is :

• A. $\frac{p^o{\alpha }^3}{2(1-\alpha {)}^3}$
• B. $\frac{p^o{\alpha }^3}{(2+\alpha )(1-\alpha {)}^2}$
• C. $\frac{p^o{\alpha }^2}{2(1-\alpha {)}^2}$
• D. None

Answer 1

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

Given, degree of dissociation of ${SO}_3$ is $\alpha$

equilibrium pressure $=P_0$

${2SO}_3\left(g\right)\rightleftharpoons {2SO}_2\left(g\right)+O_2\left(g\right)$

at eqn $2-2\alpha$ $2\alpha$ $\alpha$

total pressure $=(2-2\alpha )+\left(2\alpha \right)+\left(\alpha \right)=2+\alpha$

$P_{{SO}_3}=\left(\frac{2-2\alpha }{2+\alpha }\right)P_0$

$P_{{So}_2}=\left(\frac{2\alpha }{2+\alpha }\right)P_0$

$P_{O_2}=\left(\frac{\alpha }{2+\alpha }\right)P_0$

$K_P$ for the reaction given is

$K_P=\frac{{\left[P_{{SO}_2}\right]}^2\left[P_{O_2}\right]}{{\left[P_{{SO}_3}\right]}^2}$

$\Rightarrow K_P=\frac{{\left(\left(\frac{2\alpha }{2+\alpha }\right)P_0\right)}^2\left[\left(\frac{\alpha }{2+\alpha }\right)P_0\right]}{{\left[\left(\frac{2-2\alpha }{2+\alpha }\right)P_0\right]}^2}=\frac{P^0{\alpha }^3}{(2+\alpha ){(1-\alpha )}^2}$

$K_P$ for the reaction is $K_P=\frac{P^0{\alpha }^3}{(2+\alpha ){(1-\alpha )}^2}$