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. 3 Fe(s) 4 H2O(g) equilibrium reaction arrow Fe3O4(s) 4 H2(g) Calculate the value of Kp for this reaction at 1200 K. Hint: Apply Dalton's law of partial pressures.

Answer 1

Answer:

The value of $K_p$ for this reaction at 1200 K is 4.066.

Explanation:

Partial pressure of water vapor at equilibrium = $p^o_{H_2O}=15.0 Torr$

Partial pressure of hydrogen gas at equilibrium = $p^o_{H_2}=?$

Total pressure of the system at equilibrium P = 36.3 Torr

Applying Dalton's law of partial pressure to determine the partial pressure of hydrogen gas at equilibrium:

$P=p^o_{H_2O}+p^o_{H_2}$

$p^o_{H_2}=P-p^o_{H_2O}=36.3 Torr- 15.0 Torr = 21.3 Torr$

$3 Fe(s) 4 H_2O(g)\rightleftharpoons Fe_3O_4(s) 4 H_2(g)$

The expression of $K_p$ is given by:

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

$K_p=\frac{(21.3 Torr)^4}{(15.0 Torr)^4}=4.066$

The value of $K_p$ for this reaction at 1200 K is 4.066.