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

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

Step-by-step 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* 4.69 atm=9.38 atm`

The expression of
`K_p` is given as :

`K_p=((p_(O))^2)/((p_(O_2)))`

`K_p=((9.38 atm)^2)/(0.28 atm)=314.23`

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