Question

Write the equilibrium expression for each of the following heterogeneous equilibria. a. \(2 \mathrm{H}_{2}(g)+\mathrm{CO}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(l)\) b. \(2 \mathrm{K}(s)+\mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{KH}(s)\) c. \(\operatorname{MgO}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{MgCO}_{3}(s)\)

Short answer

The equilibrium expressions for the given heterogeneous equilibria are: a. \(K_a = \frac{1}{[H_2]^2 [CO]}\) b. \(K_b = \frac{1}{[H_2]}\) c. \(K_c = \frac{1}{[CO_2]}\)

Step-by-step solution

a. Equilibrium expression for \(2H_2(g) + CO(g) \rightleftharpoons CH_3OH(l)\)

First, let's identify the states of the substances: - \(H_2(g)\): Gas - \(CO(g)\): Gas - \(CH_3OH(l)\): Liquid Now, we will write the equilibrium expression: \[K_a = \frac{[CH_3OH]}{[H_2]^2 [CO]}\] However, since \(CH_3OH\) is a liquid, it will not appear in the equilibrium expression. So, the equilibrium expression becomes: \[K_a = \frac{1}{[H_2]^2 [CO]}\]

b. Equilibrium expression for \(2K(s) + H_2(g) \rightleftharpoons 2KH(s)\)

First, let's identify the states of the substances: - \(K(s)\): Solid - \(H_2(g)\): Gas - \(KH(s)\): Solid Now, we will write the equilibrium expression: \[K_b = \frac{[KH]^2}{[K]^2 [H_2]}\] However, since \(K\) and \(KH\) are solids, they will not appear in the equilibrium expression. So, the equilibrium expression becomes: \[K_b = \frac{1}{[H_2]}\]

c. Equilibrium expression for \(MgO(s) + CO_2(g) \rightleftharpoons MgCO_3(s)\)

First, let's identify the states of the substances: - \(MgO(s)\): Solid - \(CO_2(g)\): Gas - \(MgCO_3(s)\): Solid Now, we will write the equilibrium expression: \[K_c = \frac{[MgCO_3]}{[MgO][CO_2]}\] However, since \(MgO\) and \(MgCO_3\) are solids, they will not appear in the equilibrium expression. So, the equilibrium expression becomes: \[K_c = \frac{1}{[CO_2]}\]

Key concepts

Heterogeneous Equilibria

Understanding heterogeneous equilibria is crucial to grasp how reactions involving different phases reach a state of balance. A typical heterogeneous equilibrium involves substances in various states of matter—gases, liquids, and solids—interacting within a single reaction.

Let's focus on how we express the equilibrium for reactions with solids and liquids. Solids and liquids are considered to have constant concentrations because their densities do not change significantly with typical reaction conditions. As a result, they do not appear in the equilibrium expression, which is formulated only in terms of the concentrations of gases and the concentrations of substances dissolved in solution.

For instance, in the equilibrium process involving hydrogen gas and carbon monoxide reacting to produce methanol in a liquid state, the concentration of liquid methanol doesn't appear in the final expression. Similarly, solid reactants and products are omitted from equilibrium constants for reactions b and c from our exercise examples, simplifying the expressions significantly and focusing only on the concentration of gases.

Chemical Equilibrium

At chemical equilibrium, the rate of the forward reaction equals the rate of the reverse reaction, leading to no net change in the concentrations of reactants and products over time. This state is dynamic, meaning that the reactions are still occurring, but are balanced.

The equilibrium constant (\(K\)) is vital here, as it quantifies the equilibrium state depending on the concentrations of the reactants and products. For gas and aqueous phases, concentrations are typically used, while pressures can be used for gases in terms of partial pressures. Keep in mind, however, the constant is specific for a given reaction at a specific temperature; changing the temperature would change the value of \(K\).

An interesting fact to consider is that the equilibrium constant provides a glimpse into the reaction's favorability. A large \(K\) suggests the reaction heavily favors products at equilibrium, while a small \(K\) indicates a reactant-favored system. This insight is essential when predicting the outcome of chemical processes and for understanding the underpinnings of many industrial and biological systems.

Reaction Quotient

The reaction quotient (\(Q\)), often confused with the equilibrium constant, is another important concept that tells us whether a reaction mixture is at equilibrium and, if not, in which direction the reaction will proceed to reach equilibrium.

\(Q\) is calculated using the same formula as the equilibrium constant but with the initial concentrations or partial pressures of the reactants and products. Comparing \(Q\) to the equilibrium constant, \(K\), provides valuable directional information:

• If \(Q < K\), the forward reaction will proceed to produce more products and reach equilibrium.
• If \(Q > K\), the reaction will proceed in the reverse direction to produce more reactants and reach equilibrium.
• If \(Q = K\), the system is already at equilibrium, and no net change will occur.

Through repetition and practice, such as solving the problems provided in our exercise, students will gain a more robust understanding of how to apply these principles to real-world chemical reactions.