Question

At 900.0 K, the equilibrium constant (Kp) for the following reaction is 0.345. 2SO2 + O2 (g) →2 SO3 (g) At equilibrium, the partial pressure of SO2 is 36.9 atm and that of O2 is 16.8 atm. The partial pressure of SO3 is ________ atm.

Answer 1

Answer : The partial pressure of $SO_3$ is, 88.84 atm

Solution :  

The given equilibrium reaction is,

$2SO_2(g)+O_2(g)\rightleftharpoons 2SO_3(g)$

$\text{Partial pressure of }SO_2$ = 36.9 atm

$\text{Partial pressure of }O_2$ = 16.8 atm

The expression of $K_p$ will be,

$K_p=\frac{(p_{SO_3})^2}{(p_{O_2})(p_{SO_2})^2}$

Now put all the values in this expression, we get

$0.345=\frac{(p_{SO_3})^2}{(16.8)\times (36.9)^2}$

$(p_{SO_3})=88.84atm$

Therefore, the partial pressure of $SO_3$ is, 88.84 atm