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)

For any equilibrium involving gases, the equilibrium constant in terms of pressure, $$K_p$$, is written by taking the partial pressure of every gaseous species raised to the power of its stoichiometric coefficient and dividing the product of such terms for products by that for reactants.

The balanced reaction supplied is

$$A(s) \;\rightleftharpoons\; M(s) + \tfrac{1}{2}\,O_2(g)$$

Here, both $$A(s)$$ and $$M(s)$$ are solids. According to the law of mass action, the activity (or “effective concentration”) of a pure solid is taken as unity, so solids do not appear explicitly in the expression for $$K_p$$. Only the gaseous component, $$O_2(g)$$, contributes.

Writing the expression, we have

$$K_p \;=\; \bigl(P_{O_2}\bigr)^{\,\tfrac{1}{2}}$$

because the stoichiometric coefficient of $$O_2(g)$$ is $$\tfrac12$$, and the pressure term must be raised to that power. All other species are solids and hence are omitted (equivalently, treated as 1).

The numerical value of the equilibrium constant is given as $$K_p = 4$$. Substituting this value, we get

$$4 \;=\; \bigl(P_{O_2}\bigr)^{\tfrac{1}{2}}$$

To isolate $$P_{O_2}$$, we remove the square root (i.e., we square both sides). Stating the algebraic operation explicitly:

$$\bigl(P_{O_2}\bigr)^{\tfrac{1}{2}} \;=\; 4 \;\;\Longrightarrow\;\; \left[\bigl(P_{O_2}\bigr)^{\tfrac{1}{2}}\right]^{2} \;=\; 4^{2}$$

Since $$\bigl(x^{1/2}\bigr)^{2} = x$$, the left-hand side simplifies directly to $$P_{O_2}$$. Performing the squaring on the right-hand side, we obtain

$$P_{O_2} \;=\; 16$$

The unit carried by a partial pressure in a $$K_p$$ calculation is atmospheres (atm). Therefore, the equilibrium partial pressure of oxygen is

$$P_{O_2} = 16\ \text{atm}$$

Because the question instructs us to round to the nearest integer, and 16 is already an integer, no further adjustment is necessary.

So, the answer is $$16$$.