Question

A sample of SO3 is introduced into an evacuated sealed container and heated to 600 K. The following equilibrium is established: 2 SO3( g) ∆ 2 SO2( g) + O2( g) The total pressure in the system is 3.0 atm and the mole fraction of O2 is 0.12. Find Kp

Answer 1

Answer: The value of $K_p$ is 0.050.

Explanation:

According to Raoult's law, the vapor pressure of a component at a given temperature is equal to the mole fraction of that component multiplied by the vapor pressure of that component in the pure state.

$p_x=x\times P$

As we know the mole fraction of $O_2$ is 0.12

The partial pressure of $O_2=0.12\times 3.0atm=0.36atm$

The partial pressure of $SO_2=2\times 0.36atm=0.72atm$Thus the partial pressure of $SO_3$ is = [3 - (0.36+0.720)] atm = 1.92 atm

$p_{SO3}$= 1.92 atm

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

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

$K_p=\frac{0.36\times (0.72)^2}{(1.92)^2}$

$K_p=0.050$

The value of $K_p$ is 0.050.