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.

Answer 2

The value of  Kp is 0.050.

Raoult's law:

As per this law, the vapor pressure of a component at a given temperature should be equivalent to the mole fraction of that component and then it should be multiplied by the vapor pressure of that component in the pure state.

$p_x = x\times P$

Since mole fraction of oxygen is 0.12

Now the partial pressure should be = 0.12(3) = 0.36

The partial pressure of SO_2 is = 2(0.36) = 0.72

Now the partial pressure of SO_3 is  [3 - (0.36+0.720)] atm = 1.92 atm

Now Kp is

$= 0.36 \times (0.72)^3 \div (1.92)^3$

= 0.050

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