Question

A sample of O2(g) is placed in an otherwise empty, rigid container at 4224 K at an initial pressure of 4.97 atm, where it decomposes to O(g) by the reaction below.

O2(g) ⇄ 2 O(g)

At equilibrium, the partial pressure of O2 is 0.28 atm. Calculate Kp for this reaction at 4224 K.

Answer 1

Answer:

The value of $K_p$ at 4224 K is 314.23.

Explanation:

$O_2(g)\rightleftharpoons 2O(g)$

Initially

4.97 atm            0

At equilibrium

4.97 - p               2p

At initial stage, the partial pressure of oxygen gas = =4.97 atm

At equilibrium, the partial pressure of oxygen gas = $p_{O_2}=0.28 atm$

So, 4.97 - p = 0.28 atm

p = 4.69 atm

At equilibrium, the partial pressure of O gas = $p_{O}=2p=2\times 4.69 atm=9.38 atm$

The expression of $K_p$ is given as :

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

$K_p=\frac{(9.38 atm)^2}{0.28 atm}=314.23$

The value of $K_p$ at 4224 K is 314.23.