Question

Consider the equilibrium \(\mathrm{Na}_{2} \mathrm{O}(s)+\mathrm{SO}_{2}(g) \rightleftharpoons\) \(\mathrm{Na}_{2} \mathrm{SO}_{3}(s) .(\mathbf{a})\) Write the equilibrium-constant expression for this reaction in terms of partial pressures. (b) All the compounds in this reaction are soluble in water. Rewrite the equilibrium-constant expression in terms of molarities for the aqueous reaction.

Short answer

In terms of partial pressures, the equilibrium constant expression for the given reaction is \(K_p = \frac{1}{P_{\mathrm{SO}_{2}}}\). When all compounds are dissolved in water, the equilibrium constant expression in terms of molarities is \(K_c = \frac{[\mathrm{Na}_2\mathrm{SO}_{3}]}{[\mathrm{Na}_2\mathrm{O}][\mathrm{SO}_{2}]}\).

Step-by-step solution

Write the balanced equilibrium reaction

The balanced chemical equation for the given reaction is: \[ \mathrm{Na}_2\mathrm{O}(s) + \mathrm{SO}_{2}(g) \rightleftharpoons \mathrm{Na}_2\mathrm{SO}_{3}(s) \]

Write the equilibrium constant expression in terms of partial pressures

For any reaction, the equilibrium constant expression in terms of partial pressures is given by: \[ K_p = \frac{(\text{partial pressures of products})^{coefficients}}{(\text{partial pressures of reactants})^{coefficients}} \] In the balanced chemical equation, only one reactant \(\mathrm{SO}_{2}\) is in the gaseous state. All the other reactants and products are solids. Thus, the equilibrium constant in terms of partial pressure is given by: \[ K_p = \frac{1}{P_{\mathrm{SO}_{2}}} \] #b. Writing the equilibrium constant expression in terms of molarities for the aqueous reaction#

Write the balanced aqueous reaction

When all compounds are dissolved in water, the balanced chemical equation for the given reaction is: \[ \mathrm{Na}_2\mathrm{O}(aq) + \mathrm{SO}_{2}(aq) \rightleftharpoons \mathrm{Na}_2\mathrm{SO}_{3}(aq) \]

Write the equilibrium constant expression in terms of molarities

For any reaction, the equilibrium constant expression in terms of molarities is given by: \[ K_c = \frac{(\text{concentrations of products})^{coefficients}}{(\text{concentrations of reactants})^{coefficients}} \] In the balanced aqueous chemical equation, all reactants and products are in the aqueous state. Thus, the equilibrium constant in terms of molarities is given by: \[ K_c = \frac{[\mathrm{Na}_2\mathrm{SO}_{3}]}{[\mathrm{Na}_2\mathrm{O}][\mathrm{SO}_{2}]} \]

Key concepts

Equilibrium Constant

The concept of an equilibrium constant, often symbolized as \( K \), is central in understanding chemical equilibria. It provides insight into how concentrations of reactants and products relate to one another at equilibrium.
The value of \( K \) helps us predict the direction and extent of a reaction.- **Equilibrium State**: At equilibrium, the rate of the forward reaction equals the rate of the reverse reaction. The concentrations of reactants and products do not change.- **Expression of \( K \)**: It is generally expressed as a fraction where: - **Numerator**: Product concentrations raised to their stoichiometric coefficients. - **Denominator**: Reactant concentrations raised to their stoichiometric coefficients.It's essential to be aware of the physical states of the chemicals involved, as solids and pure liquids like solvents do not appear in the equation.
This makes \( K \) specific to different types of reactions such as gas-phase ones (\( K_p \)) and reactions in solution (\( K_c \)).

Partial Pressure

Partial pressure is a crucial concept for gaseous reactions and relates to the contribution any single gas makes to the total pressure in a system.
It helps in finding the equilibrium constant in terms of partial pressures, denoted as \( K_p \).- **Definition**: The partial pressure of a gas is the pressure it would exert if it occupied the entire volume alone.- **Dalton's Law**: The total pressure of a mixture of gases is the sum of the partial pressures of each individual gas, i.e., \( P_{total} = P_1 + P_2 + ... + P_n \).In the given exercise:- The reaction involves a single gaseous reactant, \( \mathrm{SO}_2 \), affecting \( K_p \).Thus, the equilibrium expression at gas-phase equilibrium is influenced solely by the partial pressure of \( \mathrm{SO}_2 \).
In non-soluble reactions, the solids and liquids have no partial pressure contribution.

Molarity

Molarity is a widely used unit of concentration for solutions in chemistry.
It provides a straightforward way to express the concentration of solutes in a solution.- **Definition**: Molarity \( M \) is defined as the number of moles of solute per liter of solution.- **Formula**: \( M = \frac{n}{V} \), where \( n \) is moles of solute and \( V \) is the volume of solution in liters.For the aqueous reaction in the exercise:- Reaction shifts to solution phase, where \( K_c \) is used.The equilibrium constant expression using molarity, \( K_c \), considers concentrations of substances in solution.
The expression for the given reaction in terms of molarities is based on the molarity of the aqueous reactants and products involved.

Aqueous Solutions

Aqueous solutions play a pivotal role in many chemical reactions and involve dissolving compounds in water.
This changes the reaction dynamics considerably compared to gaseous or solid-state reactions.- **Water as a Solvent**: Water's ability as a solvent arises from its polarity, which allows it to dissolve various ionic and molecular substances.- **Equilibrium in Solutions**: When substances dissolve, they often dissociate into ions, impacting the equilibrium constant expression in terms of molar concentration \( K_c \).In the provided reaction, each component is soluble in water:- This shifts the focus from partial pressures to molarities.Understanding how substances interact and reach equilibrium in aqueous solutions helps predict behaviors and reactions in diverse contexts.
In chemistry, many reactions occur in solutions, making the study of aqueous equilibria highly significant.