Question

Indicate whether the stated equilibrium expression is correct for each of the following heterogeneous equilibria. For those expressions that are not correct, provide the correct expression along with an explanation. a. \(2 \mathrm{NBr}_{3}(s) \rightleftharpoons \mathrm{N}_{2}(g)+3 \mathrm{Br}_{2}(g), K=\frac{\left[\mathrm{N}_{2}\right]\left[\mathrm{Br}_{2}\right]^{3}}{\left[\mathrm{NBr}_{3}\right]^{2}}\) b. \(\mathrm{CuO}(s)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{Cu}(l)+\mathrm{H}_{2} \mathrm{O}(g), K=\frac{[\mathrm{Cu}]\left[\mathrm{H}_{2} \mathrm{O}\right]}{\left[\mathrm{H}_{2}\right]}\) c. \(4 \mathrm{Fe}(s)+3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{Fe}_{2} \mathrm{O}_{3}(s), K=\frac{1}{\left[\mathrm{O}_{2}\right]^{3}}\)

Short answer

The correct equilibrium expressions for the given heterogeneous equilibria are: a. \(K=\left[\mathrm{N}_{2}\right]\left[\mathrm{Br}_{2}\right]^{3}\) b. \(K=\frac{\left[\mathrm{H}_{2} \mathrm{O}\right]}{\left[\mathrm{H}_{2}\right]}\) c. \(K=\frac{1}{\left[\mathrm{O}_{2}\right]^{3}}\)

Step-by-step solution

a. Check given expression for 2NBr3(s) ↔ N2(g) + 3Br2(g)

The given equilibrium expression is: \[K=\frac{\left[\mathrm{N}_{2}\right]\left[\mathrm{Br}_{2}\right]^{3}}{\left[\mathrm{NBr}_{3}\right]^{2}}\] Since NBr3 is a solid, its concentration should not be in the equilibrium expression. The correct expression for this equilibrium is: \[K=\left[\mathrm{N}_{2}\right]\left[\mathrm{Br}_{2}\right]^{3}\]

b. Check given expression for CuO(s) + H2(g) ↔ Cu(l) + H2O(g)

The given equilibrium expression is: \[K=\frac{[\mathrm{Cu}]\left[\mathrm{H}_{2} \mathrm{O}\right]}{\left[\mathrm{H}_{2}\right]}\] Since CuO is a solid and Cu is a liquid, their concentrations should not be in the equilibrium expression. The correct expression for this equilibrium is: \[K=\frac{\left[\mathrm{H}_{2} \mathrm{O}\right]}{\left[\mathrm{H}_{2}\right]}\]

c. Check given expression for 4Fe(s) + 3O2(g) ↔ 2Fe2O3(s)

The given equilibrium expression is: \[K=\frac{1}{\left[\mathrm{O}_{2}\right]^{3}}\] Since Fe and Fe2O3 are both solids, their concentrations should not be in the equilibrium expression, making the given expression correct. The equilibrium expression for this equilibrium is: \[K=\frac{1}{\left[\mathrm{O}_{2}\right]^{3}}\]

Key concepts

Equilibrium Expression

In chemical reactions, especially in heterogeneous equilibria, it’s crucial to write the correct equilibrium expression. This expression describes the ratio of the concentrations of products to reactants, each raised to the power of their coefficients in the balanced chemical equation. However, it excludes solids and pure liquids, as their concentrations do not change during the reaction.

For example, consider the reaction of 2NBr3(s) ↔ N2(g) + 3Br2(g). The solid NBr3 is excluded from the equilibrium expression. Therefore, the correct equilibrium expression is:

• For gases: Only gaseous substances are included in the equilibrium constant expression.
• For solids and solvents: These do not appear in the expression because their concentrations can be considered constant.

Including solids in the equilibrium expression would inaccurately depict the reaction dynamics since their amount doesn't influence the equilibrium state.

Chemical Equilibrium

Chemical equilibrium is a state in a chemical reaction where the concentrations of reactants and products remain constant over time. Once a reaction reaches equilibrium, the forward and reverse reactions occur at the same rate, maintaining a consistent environment.

• Dynamic Process: Although no net change is observed at equilibrium, it is dynamic, with continuous forward and reverse reactions.
• Equilibrium Constant (K): A constant value that gives the ratio of concentrations of products to reactants at equilibrium.

The value of the equilibrium constant gives insight into the position of equilibrium—whether products or reactants are favored. Large K values signify product-favored equilibria, while small values indicate reactant-favored conditions.

Solid and Liquid Phases in Equilibria

In heterogeneous equilibria, solid and liquid phases play a specific role. Here, only gases and aqueous solutions are considered dynamic in the equilibrium constant calculation. Solids and pure liquids are often omitted because their concentrations are constant and do not affect the equilibrium position.

Take the reaction CuO(s) + H2(g) ↔ Cu(l) + H2O(g). Here, the solids CuO and Cu are omitted from the equilibrium expression. This rule helps focus only on the changing concentrations in the gaseous or aqueous phase, which can impact the reaction equilibrium.

• Solids: These have a fixed concentration and do not appear in the K expression.
• Liquids: Like solids, pure liquids do not fluctuate in concentration and thus are excluded.

Understanding this concept is crucial when deriving the correct equilibrium expression for heterogeneous reactions.