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].

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;

$K_p=\frac{(p_2)^4}{(p_1)^4}$

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

$K_p=\frac{(21.3 Torr)^4}{(15.0 Torr)^4}$

$K_p=4.066$

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