The equilibrium constant $K_p$ of the reaction $2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$

Question

The equilibrium constant $K_p$ of the reaction $2SO_{2(g)} + O_{2(g)} ightleftharpoons 2SO_{3(g)}$ is $900 	ext{ atm}^{-1}$ at $800 	ext{ K}$. A mixture containing $SO_3$ and $O_2$ having initial partial pressures of $1.0 	ext{ atm}$ and $2.0 	ext{ atm}$ respectively is heated at constant volume to equilibrate. What is the approximate partial pressure of $SO_{2(g)}$ at equilibrium?

Professor PC Thomas Classes · Answer

Initial pressures: $P_{SO_3} = 1 	ext{ atm}$, $P_{O_2} = 2 	ext{ atm}$, $P_{SO_2} = 0$. Let the decrease in $SO_3$ pressure be $2x$. At equilibrium, $P_{SO_3} = 1-2x$, $P_{SO_2} = 2x$, $P_{O_2} = 2+x$. Given $K_p = \frac{(P_{SO_3})^2}{(P_{SO_2})^2 P_{O_2}} = 900$. Since $K_p$ is large, the reverse reaction is minimal, so $x$ is small. We approximate $1-2x \approx 1$ and $2+x \approx 2$. $900 \approx \frac{1^2}{(2x)^2 	imes 2} = \frac{1}{8x^2} \implies x^2 \approx \frac{1}{7200} \approx 1.388 	imes 10^{-4} \implies x \approx 0.01178 	ext{ atm}$. A more precise solution without dropping $x$ fully gives $2x \approx 0.023 	ext{ atm}$. The partial pressure of $SO_2 = 2x \approx 0.023 	ext{ atm}$.