Question

At 900.0 K, the equilibrium constant (Kp) for the following reaction is 0.345. 2SO2(g)+O2(g)→2SO3(g) At equilibrium, the partial pressure of SO2 is 30.6 atm and that of O2 is 13.9 atm. The partial pressure of SO3 is __________ atm.

Answer 1

Answer : The partial pressure of $SO_3$ is, 67.009 atm

Solution :  Given,

Partial pressure of $SO_2$ at equilibrium = 30.6 atm

Partial pressure of $O_2$ at equilibrium = 13.9 atm

Equilibrium constant = $K_p=0.345$

The given balanced equilibrium reaction is,

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

The expression of $K_p$ will be,

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

Now put all the values of partial pressure, we get

$0.345=\frac{(p_{SO_3})^2}{(30.6)^2\times (13.9)}$

$p_{SO_3}=67.009atm$

Therefore, the partial pressure of $SO_3$ is, 67.009 atm