Question

Write the equilibrium expression for each of the following heterogeneous equilibria. a. \(\mathrm{C}(s)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{CO}(g)\) b. \(\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(g)\) c. \(4 \mathrm{~B}(s)+3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{~B}_{2} \mathrm{O}_{3}(s)\)

Short answer

The short answer for the equilibrium expressions for each heterogeneous equilibria are: a. \(K = \frac{[\mathrm{H}_{2}][\mathrm{CO}]}{[\mathrm{H}_{2}\mathrm{O}]}\) b. \(K = [\mathrm{H}_{2}\mathrm{O}(g)]\) c. \(K = \frac{1}{[\mathrm{O}_{2}]^3}\)

Step-by-step solution

Identify the reactants and products in the gas phase

In this reaction, \(\mathrm{H}_{2} \mathrm{O}(g)\) is a reactant and both \(\mathrm{H}_{2}(g)\) and \(\mathrm{CO}(g)\) are products in the gas phase.

Write the equilibrium expression

Exclude the solid \(\mathrm{C}(s)\) from the expression since we do not include solids. The equilibrium expression is given by \[K = \frac{[\mathrm{H}_{2}][\mathrm{CO}]}{[\mathrm{H}_{2}\mathrm{O}]}\] b. Reaction: \(\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(g)\)

Identify the reactants and products in the gas phase

In this reaction, the gas phase product is \(\mathrm{H}_{2}\mathrm{O}(g)\).

Write the equilibrium expression

Exclude the liquid \(\mathrm{H}_{2}\mathrm{O}(l)\) from the expression. The equilibrium expression is given by \[K = [\mathrm{H}_{2}\mathrm{O}(g)]\] c. Reaction: \(4 \mathrm{B}(s)+3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{B}_{2}\mathrm{O}_{3}(s)\)

Identify the reactants and products in the gas phase

In this reaction, \(\mathrm{O}_{2}(g)\) is the only reactant in the gas phase.

Write the equilibrium expression

Exclude the solids \(\mathrm{B}(s)\) and \(\mathrm{B}_{2}\mathrm{O}_{3}(s)\) from the expression. The equilibrium expression is given by \[K = \frac{1}{[\mathrm{O}_{2}]^3}\]

Key concepts

Heterogeneous Equilibria

When understanding chemical reactions, it's essential to recognize that not all reactions occur with elements and compounds in the same physical state. Reactants and products can exist as solids, liquids, or gases, and when they are in different phases, the equilibrium they establish is known as heterogeneous equilibria. In the provided exercise examples, each involves reactants and products in various states, such as solids (s) and gases (g).

For instance, in \(\mathrm{C}(s)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{CO}(g)\), carbon is solid, and the other reactants and products are gases. It is crucial to note that when writing equilibrium expressions for heterogeneous equilibria, the concentrations of pure solids and liquids are not included. This is because their concentrations do not change during the reaction, as they have fixed densities and are thus thought to have constant activity. Therefore, equilibrium expressions only include the concentrations of gaseous and aqueous species, which can change and hence, affect the equilibrium state.

Chemical Equilibrium

At the heart of these examples lies the principle of chemical equilibrium, a state in a chemical reaction where the rate of the forward reaction equals the rate of the reverse reaction. As a result, there is no net change in the concentrations of reactants and products, even though both reactions continue to occur. This dynamic balance is essential to chemistry and is described quantitatively by the equilibrium constant \(K\).

The equilibrium constant offers insight into the extent of a reaction, the position of equilibrium, and how it's affected by different conditions. Returning to our example \(\mathrm{C}(s)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{CO}(g)\), the equilibrium constant \(K\) allows us to predict how much \(\mathrm{H}_{2}(g)\) and \(\mathrm{CO}(g)\) will be present at equilibrium given a certain amount of \(\mathrm{H}_{2} \mathrm{O}(g)\). This predictive ability is a powerful tool for chemists analyzing reaction systems.

Reaction Quotient

In the journey to reach equilibrium, chemical systems pass through states that are not at equilibrium. The reaction quotient \(Q\) helps determine the direction a non-equilibrium system will shift to reach equilibrium. It is calculated using the same formula as \(K\), the equilibrium constant, but with the initial concentrations of the reactants and products.

If \(Q\) is less than \(K\), the forward reaction is favored to form more products. If \(Q\) is greater than \(K\), the reaction needs to shift in the reverse direction to reach equilibrium. In the context of the provided examples, comparing \(Q\) and \(K\) would inform you which direction the reaction would proceed to reach equilibrium given initial gas concentrations. Learning to calculate and interpret \(Q\) is a vital skill for students to predict how a reaction mixture will behave over time.