Question

Write the equilibrium constant expressions for these reactions: (a) \(3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{O}_{3}(g)\) (b) \(\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(l)\) (c) \(\mathrm{MgCO}_{3}(s) \rightleftharpoons \mathrm{MgO}(s)+\mathrm{CO}_{2}(g)\) (d) \(2 \mathrm{Bi}^{3+}(a q)+3 \mathrm{H}_{2} \mathrm{~S}(a q) \rightleftharpoons \mathrm{Bi}_{2} \mathrm{~S}_{3}(s)+6 \mathrm{H}^{+}(a q)\)

Short answer

\(K_p = \frac{[O_3(g)]^2}{[O_2(g)]^3}\), \(K = [H_2O(g)]\), \(K = [CO_2(g)]\), \(K_{sp} = [Bi^{3+}(aq)]^2[H^{+}(aq)]^6\).

Step-by-step solution

- Writing Equilibrium Constant for Reaction (a)

For the reaction involving gases, the equilibrium constant expression is based on the partial pressures of the gases. For the reaction 3 O2(g) \(\rightleftharpoons\) 2 O3(g), the equilibrium constant expression (Kp) is written as: \(K_p = \frac{[O_3(g)]^2}{[O_2(g)]^3}\), where the concentrations [ ] are replaced by the partial pressures P, and the exponents correspond to the coefficients in the balanced equation.

- Writing the Equilibrium Constant for Reaction (b)

Since the reaction involves a pure liquid, the concentration of H2O(l) is constant and thus omitted from the equilibrium expression. For the reaction H2O(g) \(\rightleftharpoons\) H2O(l), the equilibrium constant expression is simply: \(K = \frac{1}{[H_2O(g)]}\) or more commonly written in terms of the reverse reaction as the pressure of the vapor above the liquid is equal to the equilibrium vapor pressure (Pvap) at a given temperature: \(K = [H_2O(g)]\).

- Writing Equilibrium Constant for Reaction (c)

In the reaction involving a solid and a gas, the concentration of solids is constant and not included in the equilibrium expression. For the reaction MgCO3(s) \(\rightleftharpoons\) MgO(s) + CO2(g), the equilibrium constant expression is: \(K = [CO_2(g)]\).

- Writing the Equilibrium Constant for Reaction (d)

For reactions involving ions in aqueous solution and the formation of a solid (precipitate), the solids are omitted from the equilibrium expression. In the reaction 2 Bi^{3+}(aq) + 3 H2S(aq) \(\rightleftharpoons\) Bi2S3(s) + 6 H+(aq), the equilibrium constant expression, known as the solubility product constant (Ksp), is: \(K_{sp} = [Bi^{3+}(aq)]^2[H^{+}(aq)]^6\). Note that the solid Bi2S3 is not included in the expression.

Key concepts

Chemical Equilibrium

In the world of chemistry, the concept of chemical equilibrium is like a dance between reactants and products where neither side completely takes over. At equilibrium, the rate of the forward reaction, where reactants turn into products, equals the rate of the reverse reaction, where products revert back into reactants. This balance doesn't mean the amounts of reactants and products are equal, but rather that their rates of formation are constant over time.

When writing equilibrium constant expressions, we're somewhat like accountants for molecules, tallying up the reactants and products, but instead of dollars, we use the concentrations or partial pressures of the substances involved. These expressions provide a numerical value, known as the equilibrium constant (K), which tells us the ratio of product concentrations to reactant concentrations at equilibrium, each raised to the power of their coefficients from the balanced equation. If K is large, it suggests products are favored; a small K indicates a preference for reactants.

Partial Pressures in Gas Reactions

When gases are involved in a chemical reaction and we're at a disco with high energy — I mean, high temperature — we often focus on partial pressures when talking about equilibrium. Partial pressure is just like it sounds: the individual pressure a gas would exert if it were alone at the party in the container. Naturally, when we successfully mix multiple gases, each brings its own partial pressure to the mix.

In our equilibrium expression, we swap out concentrations for the partial pressures when dealing with gases. Remember, like a good conversation, pressure also involves listening — or in this case, coefficients, as they tell us how much to 'raise the volume' of each gas' partial pressure.

Vapor Pressure of Liquids

Next up, we've got the vapor pressure of liquids, which is a bit like a magic act. Picture a closed container with a liquid inside. Some of the liquid molecules will escape into the air above it, forming a vapor. Now imagine those vapor molecules frequently disappearing and then reappearing as liquid — this is essentially the magic of vapor pressure.

At a certain point, the rate of molecules leaving the liquid equals the rate of vapor molecules returning to the liquid; this is the liquid's vapor pressure at a set temperature. In other words, the liquid and its vapor are at equilibrium. And if anything disrupts the act — if we change the temperature — the vapor pressure will adjust until a new equilibrium is reached.

Solubility Product Constant

Last but not least, we have the solubility product constant (Ksp), which is like the judge in a talent show, with ions auditioning to form a solid in water. Some solids are shy (low Ksp) and don't want to dissolve. Others are more outgoing (high Ksp) and dissolve easily. When a salt is just about to form a solid (precipitate) in water — when it's on the edge of its seat, so to speak — that's the solubility limit.

The Ksp gives us the scoop on how much solid can hang out in the water before it starts to form a solid. It quantifies the product of the concentrations of the ions, each raised to their star power - or stoichiometric coefficients. No solids or liquids in these expressions; they're constant and reliable, like a good backup dancer who doesn't compete with the star ions for attention.