Question

In a study of the reaction below at 1181 K, it was observed that when the equilibrium partial pressure of water vapor is 53.7 torr, the total pressure at equilibrium is 69.4 torr. Calculate Kp for this reaction at 1181 K. 3 Fe(s) + 4 H2O(g) equilibrium reaction arrow Fe3O4(s) + 4 H2(g)

Answer 1

Answer:

The value $K_p$ for this reaction at 1181 K is $7.306\times 10^{-3}$.

Explanation:

$Fe(s) + 4 H_2O(g)\rightleftharpoons Fe_3O_4(s) + 4 H_2(g)$

The expression of an equilibrium constant in terms of partial pressure can be given as:

$K_p=\frac{(p_{H_2})^4}{(p_{H_2O})^4}$

Total pressure of the gases at equilibrium = P = 69.4 Torr

Partial pressure of hydrogen gas = $p_{H_2}=x$

Partial pressure of water vapor = $p_{H_2O}=53.7 Torr$

$p_{H_2O}+p_{H_2}=P$ (Dalton's law of partial pressure)

$53.7 Torr+x=69.4 Torr$

x = 69.4 Torr - 53.7 Torr = 15.7 Torr

$K_p=\frac{(p_{H_2})^4}{(p_{H_2O})^4}$

$K_p=\frac{(15.7 Torr)^4}{(53.7 Torr)^4}=7.306\times 10^{-3}$

The value $K_p$ for this reaction at 1181 K is $7.306\times 10^{-3}$.