Question

The equilibrium constant Kp of the reaction, 2$S{O}_2$(g) + $O_2$(g) \(\leftrightharpoons\) 2$S{O}_3$(g) is 900 atm. at 800 K. A mixture containing $S{O}_3$and $O_2$ having initial pressure of 1 and 2 atm respectively is heated at constant volume to equilibrate. Calculate the partial pressure of $S{O}_2$ gas at 800 K.

• A. 0.0236 atm
• B. 0.05 atm
• C. 1.03 atm
• D. 0.442 atm

Answer 1

The correct option is A

0.0236 atm

Consider the reaction,

2$S{O}_3$ \(\leftrightharpoons\) 2$S{O}_2$ + $O_2{K}_p$ = $\frac{1}{900}$ atm.

Initial pressure 1 0 2

at equibrium (1- x) 0 (2 + $\frac{x}{2}$)

$K_p$ = $\frac{{(P_{S{O}_2)}}^2{\left(P_{S_2}\right)}^2}{{\left(P_{S{O}_2}\right)}^2}$

$\frac{1}{900}$ = $\frac{X^2(2+\frac{x}{2})}{{(1-X)}^2}$

Since $K_p$ of this reaction is small and thus,

x<1

$\frac{1}{900}$ = $\frac{X^2\left(2\right)}{{(1-x)}^2}$ (2 + $\frac{X}{2}$ = 2)

=$\frac{{2x}^2}{{(1-x)}^2}$

$\frac{1}{30}$ = $\frac{\sqrt{2x}}{(1-x)}$

X= 0.0236