For equilibrium
$3\mathrm{Fe}(\mathrm{s})+4{\mathrm{H}}_2\mathrm{O}(\mathrm{g})\rightleftharpoons {\mathrm{Fe}}_3{\mathrm{O}}_4+4{\mathrm{H}}_2(\mathrm{g})$
Pressure exerted by water vapour was 40% of total pressure at equilibrium. Find equilibrium constant.
(1) 1.5
(2) 2.25
(3) 3.37
(4) 5.06​

To find the equilibrium constant for the given reaction, we need to determine the expression for the equilibrium constant using the partial pressures of the reactants and products.

The balanced equation for the reaction is:
$3\mathrm{Fe}(\mathrm{s})+4{\mathrm{H}}_2\mathrm{O}(\mathrm{g})\rightleftharpoons {\mathrm{Fe}}_3{\mathrm{O}}_4+4{\mathrm{H}}_2(\mathrm{g})$

Let's assume the total pressure at equilibrium is P.
According to the given information, the pressure exerted by water vapor (H2O) is 40% of the total pressure. So, the partial pressure of water vapor is 0.4P.

The partial pressure of hydrogen gas (H2) is:
${\mathrm{P}}_{\text{Total }}={\mathrm{P}}_{{\mathrm{H}}_2\mathrm{O}}+{\mathrm{P}}_{{\mathrm{H}}_2}=0.4P+{\mathrm{PH}}_2\mathrm{O}$
${\mathrm{P}}_{{\mathrm{H}}_2}=P-O.4P=0.6P$
The equilibrium constant expression (Kp) for the reaction can be written as:
Kp = $(P_{H2}{)}^4/(P_{H2O}{)}^4$

Substituting the values, we get:
Kp = (0.6P)^4 / (0.4P)^4
Kp = 5.06

Therefore, the equilibrium constant (Kp) for the given reaction is 5.06.