The equilibrium constant for the reaction A(s) M(s) + 1/2 O2(g) is K_p = 4. At equilibrium, the

Q: Why don't we include the solids $A$ and $M$ in the $K_p$ equation?

The concentration of a pure solid is determined by its density and molar mass, both of which are constant at a given temperature. Therefore, their active mass is considered to be $1$ and they are omitted from the equilibrium expression.

Q: Where does the power of $1/2$ come from?

In the equilibrium constant expression, the partial pressure (or concentration) of each species is raised to the power of its stoichiometric coefficient from the balanced chemical equation. Since $O_2$ has a coefficient of $1/2$, its partial pressure is raised to $1/2$.

Q: If I add more solid $A$ to the container, will the pressure of $O_2$ increase?

No. Adding or removing pure solids does not shift the equilibrium position. The partial pressure of $O_2$ will remain exactly $16 \text{ atm}$ as long as the temperature is constant and there is at least some solid $A$ and $M$ present.

Q: What if the reaction was written as $2A(s) \rightleftharpoons 2M(s) + O_2(g)$?

If you multiply the entire reaction by $2$, the new equilibrium constant $K_p'$ would be the square of the original $K_p$ (so $K_p' = 4^2 = 16$). The expression would be $K_p' = p_{O_2}$. Solving this gives $p_{O_2} = 16 \text{ atm}$, which is exactly the same physical result!

Q: Does the volume of the container affect the final answer?

In this specific case, no. The equilibrium pressure of $O_2$ is fixed by the equilibrium constant $K_p$, which only depends on temperature. If you increase the volume, more solid $A$ will decompose to maintain the $O_2$ pressure at $16 \text{ atm}$.

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)

Sigma Percentile · Accepted Answer

The problem presents us with a classic case of heterogeneous chemical equilibrium. We are given the reaction:

$$A(s) ightleftharpoons M(s) + \frac{1}{2}O_2(g)$$

And we are told that the equilibrium constant in terms of partial pressures, $K_p$, is $4$. Our goal is to find the partial pressure of the oxygen gas, $O_2$, at equilibrium.

### Analyzing the Setup

Imagine a closed flask where this reaction is taking place. We have a solid reactant, $A$, decomposing to form another solid product, $M$, and a gaseous product, $O_2$.

The most crucial concept to remember here is the **golden rule of heterogeneous equilibrium**: the active mass (or partial pressure) of pure solids and pure liquids is always taken as unity ($1$). This is because their concentration (density divided by molar mass) remains constant regardless of how much of the substance is present.

### The Master Equation

Because $A$ and $M$ are pure solids, they will not appear in our equilibrium constant expression. The equilibrium constant $K_p$ will depend solely on the gaseous species present in the reaction, which in this case is only oxygen.

According to the law of mass action, the expression for $K_p$ is written by taking the partial pressures of the products raised to their stoichiometric coefficients, divided by the partial pressures of the reactants raised to their stoichiometric coefficients.

For our specific reaction, the expression simplifies beautifully to:

$$K_p = (p_{O_2})^{\frac{1}{2}}$$

Notice the power of $\frac{1}{2}$. This comes directly from the stoichiometric coefficient of $O_2$ in the balanced chemical equation.

### Final Calculation

We are given that $K_p = 4$. Substituting this value into our master equation, we get:

$$4 = (p_{O_2})^{\frac{1}{2}}$$

To isolate the partial pressure of oxygen, $p_{O_2}$, we simply need to square both sides of the equation. This is a straightforward algebraic step:

$$(4)^2 = \left((p_{O_2})^{\frac{1}{2}}ight)^2$$

$$16 = p_{O_2}$$

Therefore, the partial pressure of oxygen gas at equilibrium is exactly **$16 	ext{ atm}$**.

This problem elegantly demonstrates how pure solids are excluded from equilibrium expressions, making what might seem like a complex reaction into a very simple calculation!

Animated Solution for Chemistry - Chemical Equilibrium: The equilibrium constant for the reaction $A (s) ⇌ M (s) + \frac{1}{2} O_{2} (g)$ is $K_{p} = 4$ . At equilibrium, the partial pressure of $O_{2}$ is ......... atm. (Round off to the nearest integer)

Enter Numerical Value:

Visualized Solution

The Reaction Setup

Heterogeneous Equilibrium

The $K_{p}$ Expression

Substituting the Value

Calculating Partial Pressure

Final Answer

Food for Thought

The Sigma Insight: Law of Mass Action

Analyzing the Setup

The Master Equation

Final Calculation

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