Question

The degree of dissociation of $S{O}_2$ is $\alpha$ at equilibrium pressure $P_2$.
$K_p$ for $2S{O}_3\left(g\right)\rightleftharpoons 2S{O}_2\left(g\right)+O_2\left(g\right)$ is

• A. $\left[\right(P_2{\alpha }^3\left)/2\right(1-\alpha {)}^2]$
• B. $\left[\right(P_3{\alpha }^2\left)/\right(2+\alpha \left)\right(1-\alpha {)}^2]$
• C. $\left[\right(P_a{\alpha }^2\left)/2\right(1-\alpha {)}^2]$
• D. $Noneofthese$

Answer 1

The correct option is A $\left[\right(P_2{\alpha }^3\left)/2\right(1-\alpha {)}^2]$

$\begin{array}{cc}{P}_2=2P-2P\alpha +2P\alpha +P\alpha & Kp=\frac{P_{SO}{{}_2}^2{P}_{O_2}}{{P_{S{O}_3}}^2}\\ P_2=2P+P\alpha & Kp=\frac{{\left(2P\alpha \right)}^{2}P\alpha }{{\left(2P\left(1-\alpha \right)\right)}^2}\Rightarrow \frac{4P^2{\alpha }^{2}P\alpha }{4P^2{\left(1-\alpha \right)}^2}\\ P_2=P\left(2+\alpha \right)& Kp=\frac{P{\alpha }^3}{{\left(1-\alpha \right)}^2}\Rightarrow \frac{P_2{\alpha }^3}{\left(2+\alpha \right){\left(1-\alpha \right)}^2}\\ P=\frac{P_2}{2+\alpha }& \end{array}$

Option B is correct.