Question

Write the equilibrium constant expressions for the following reactions. (a) \(\mathrm{S}_{2}(g)+2 \mathrm{H}_{2}(g) \rightleftarrows 2 \mathrm{H}_{2} \mathrm{~S}(g)\) (b) \(\mathrm{H}_{2} \mathrm{~S}(a q)+\mathrm{Cl}_{2}(a q) \rightleftarrows \mathrm{S}(s)+2 \mathrm{HCl}(a q)\) (c) \(\mathrm{Br}_{2}(g)+\mathrm{Cl}_{2}(g) \rightleftarrows 2 \mathrm{BrCl}(g)\) (d) \(\mathrm{C}(s)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftarrows \mathrm{CO}(g)+\mathrm{H}_{2}(g)\)

Short answer

(a) \( K_c = \frac{[\mathrm{H}_{2}\mathrm{S}]^2}{[\mathrm{S}_{2}][\mathrm{H}_{2}]^2} \); (b) \( K_c = \frac{[\mathrm{HCl}]^2}{[\mathrm{H}_{2}\mathrm{S}][\mathrm{Cl}_{2}]} \); (c) \( K_c = \frac{[\mathrm{BrCl}]^2}{[\mathrm{Br}_{2}][\mathrm{Cl}_{2}]} \); (d) \( K_c = \frac{[\mathrm{CO}][\mathrm{H}_{2}]}{[\mathrm{H}_{2}\mathrm{O}]} \).

Step-by-step solution

Identify the Reaction Type

For each reaction, identify whether the reactants and products are in gaseous, liquid, aqueous, or solid states. The equilibrium expression only includes species in the gas and aqueous states; solids and liquids are omitted.

Write the Equilibrium Expression for Reaction (a)

For the reaction \(\mathrm{S}_{2}(g)+2 \mathrm{H}_{2}(g) \rightleftarrows 2 \mathrm{H}_{2} \mathrm{~S}(g)\), the equilibrium constant expression \(K_c\) is based on the concentrations of the gases:\[ K_c = \frac{[\mathrm{H}_{2}\mathrm{S}]^2}{[\mathrm{S}_{2}][\mathrm{H}_{2}]^2} \]

Write the Equilibrium Expression for Reaction (b)

For the reaction \(\mathrm{H}_{2} \mathrm{~S}(a q)+\mathrm{Cl}_{2}(a q) \rightleftarrows \mathrm{S}(s)+2 \mathrm{HCl}(a q)\), the equilibrium constant expression \(K_c\) involves only the aqueous species since \(\mathrm{S}(s)\) is a solid and is omitted:\[ K_c = \frac{[\mathrm{HCl}]^2}{[\mathrm{H}_{2}\mathrm{S}][\mathrm{Cl}_{2}]} \]

Write the Equilibrium Expression for Reaction (c)

For the reaction \(\mathrm{Br}_{2}(g)+\mathrm{Cl}_{2}(g) \rightleftarrows 2 \mathrm{BrCl}(g)\), the equilibrium constant expression \(K_c\) is:\[ K_c = \frac{[\mathrm{BrCl}]^2}{[\mathrm{Br}_{2}][\mathrm{Cl}_{2}]} \]

Write the Equilibrium Expression for Reaction (d)

For the reaction \(\mathrm{C}(s)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftarrows \mathrm{CO}(g)+\mathrm{H}_{2}(g)\), the equilibrium constant expression \(K_c\) does not include \(\mathrm{C}(s)\) because it is a solid:\[ K_c = \frac{[\mathrm{CO}][\mathrm{H}_{2}]}{[\mathrm{H}_{2}\mathrm{O}]} \]

Key concepts

Reaction Types

The type of reaction plays a vital role in determining how to set up an equilibrium expression. Chemical reactions can involve substances in different states, including gaseous, aqueous, solid, or liquid. In the context of equilibrium constant expressions, we focus primarily on reactions involving gaseous and aqueous states. This is because equilibrium expressions do not include substances in solid or liquid states, as their concentrations do not change.
- **Gaseous Reactions**: These involve substances that are entirely in the gas phase, making it relevant to include all reactants and products when composing the equilibrium expression.
- **Aqueous Reactions**: Only the species dissolved in water are considered, while solids that might form as precipitates are excluded.
Understanding the state of each component in a reaction helps us write accurate equilibrium expressions by determining which species to include or ignore.

Equilibrium Expression

The equilibrium expression, represented as the equilibrium constant (\(K_c\)), is a mathematical way to describe the ratio of the concentrations of the products to the reactants at equilibrium. When writing out these expressions, the concentrations of gaseous and aqueous reactants and products are the only ones considered.

• Identify each reactant and product's state.
• Include substances only in gaseous (g) and aqueous (aq) states.
• Simplify the expression by excluding solids and liquids.

For instance, in Reaction (a), \( ext{S}_2(g) + 2 ext{H}_2(g) ightleftarrows 2 ext{H}_2 ext{S}(g)\), the equilibrium expression is:\[ K_c = \frac{[ ext{H}_2 ext{S}]^2}{[ ext{S}_2][ ext{H}_2]^2}\]It includes only gases, following the rule to exclude solids and liquids.

Gas and Aqueous States

Reactions that involve gas or aqueous states are crucial in equilibrium expressions because these are the states in which concentration can change and thus affect the reaction equilibrium. Understanding these states is key to setting up equilibrium constants correctly.
- **Gas State**: For reactions involving gases, concentrations can be expressed in terms of partial pressures or molarity. Equilibrium constant expressions can also sometimes be represented as \(K_p\) if pressures are used.
- **Aqueous State**: These refer to dissolved substances in a solution. Their concentrations can fluctuate based on solubility and dissolution dynamics, making them necessary for inclusion in \(K_c\) expressions.
A firm grasp of the different roles these states play ensures precision when forming equilibrium expressions by focusing on how dynamically these concentration phases impact the overall equilibrium.

Concentration

The concept of concentration is central to the formulation of equilibrium expressions. Concentrations refer to the amount of a substance within a given volume and are often expressed in moles per liter (Molarity, \(M\)). In equilibrium contexts, concentration measurements help predict and calculate the point at which the reaction system remains stable over time.
- **Concentration Changes**: The strength of concentration in \(K_c\) expressions is depicted as the power to which the concentration is raised, determined by the coefficients in a balanced chemical equation.
For example, in Reaction (c), \( ext{Br}_2(g) + ext{Cl}_2(g) ightleftarrows 2 ext{BrCl}(g)\),\[ K_c = \frac{[ ext{BrCl}]^2}{[ ext{Br}_2][ ext{Cl}_2]}\], the doubling coefficient of \( ext{BrCl}\) in the reaction equation becomes a square in the equilibrium expression. Comprehending how concentration influences equilibrium lays the foundation for predicting how changes in conditions may influence the overall reaction.