Question

In a study of the following reaction at 1200 K it was observed that when the equilibrium partial pressure of water vapor is 15.0 torr, the total pressure at equilibrium is 36.3 torr.

Answer 1

The question is incomplete , the complete question is ;

In a study of the following reaction :

`3Fe(s)+4H_2O(g)\rightleftharpoons Fe_2O_4(s)+4H_2(g)`

at 1200 K it was observed that when the equilibrium partial pressure of water vapor is 15.0 torr, the total pressure at equilibrium is 36.3 torr. Calculate the
`K_p` foe this reaction at 1200 K.

Answer:

The
`K_p` for this reaction at 1200 K is 4.066[/tex].

Step-by-step explanation:

Partial pressure of the water vapor at equilibrium =
`p_1=15.0 Torr`

Partial pressure of the hydrogen gas at equilibrium =
`p_2`

Total pressure at equilibrium =
`P=36.3 Torr`

`P_1+P_2`

`p_2=P-P_1=36.3 Torr-15.0 Torr=21.3 Torr`

`3Fe(s)+4H_2O(g)\rightleftharpoons Fe_2O_4(s)+4H_2(g)`

The expression if
`K_p` is given as;

`K_p=((p_2)^4)/((p_1)^4)`

(the partial pressure of the gas will be taken along with which partial pressure of the solids are taken as unity)

`K_p=((21.3 Torr)^4)/((15.0 Torr)^4)`

`K_p=4.066`

The
`K_p` for this reaction at 1200 K is 4.066[/tex].