Question

The equilibrium constant for the reaction
A(s) $\rightleftharpoons$ M(s) + $\frac{1}{2}$O$_2$(g)
is K$_p$=4. At equilibrium, the partial pressure of O$_2$ is ________ atm. (Round off to the Nearest Integer).

Hint:

When writing expressions for equilibrium constants (K$_p$ or K$_c$), remember that the concentrations or partial pressures of pure solids and pure liquids are considered constant and are omitted from the expression (or treated as having an activity of 1).

Answer 1

Correct Answer: 16

The given reaction at equilibrium is:
A(s) $\rightleftharpoons$ M(s) + $\frac{1}{2}$O$_2$(g)
The expression for the equilibrium constant in terms of partial pressures, K$_p$, is written based on the products and reactants.
$K_p = \frac{[\text{Products}]}{[\text{Reactants}]}$
For this reaction, the expression is:
$K_p = \frac{ P_M \cdot (P_{O_2})^{1/2} }{ P_A }$
However, for pure solids (like A(s) and M(s)), their activities are considered to be constant and equal to 1. Therefore, they are not included in the K$_p$ expression.
The expression simplifies to:
$K_p = (P_{O_2})^{1/2}$.
We are given that $K_p = 4$.
$4 = (P_{O_2})^{1/2}$.
To find the partial pressure of O$_2$, we need to square both sides of the equation.
$(4)^2 = ((P_{O_2})^{1/2})^2$.
$16 = P_{O_2}$.
Therefore, the partial pressure of O$_2$ at equilibrium is 16 atm.