Question

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

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

Answer 1

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

${SO}_3\rightleftharpoons {2SO}_2+O_2$

$1$ $-$ $-$

$1-\alpha$ $2\alpha$ $\alpha$

Total no. of moles $=1+2\alpha$

$P_{{SO}_3}=\frac{(1-\alpha )P}{1+2\alpha }{P}_{{SO}_2}=\frac{\left(2\alpha \right)P}{1+2\alpha }{P}_{O_2}=\frac{\left(\alpha \right)P}{1+2\alpha }$

$K_P=\frac{{P_{{SO}_2}}^2+P_{O_2}}{P_{{SO}_3}}=\frac{{\left(\frac{\left(2\alpha \right)P}{1+2\alpha }\right)}^2\left(\frac{\left(\alpha \right)P}{1+2\alpha }\right)}{\left(\frac{(1-\alpha )P}{1+2\alpha }\right)}$

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