Question

Consider these two equilibria involving \(\mathrm{SO}_{2}(\mathrm{~g})\) and their corresponding equilibrium constants. $$ \begin{array}{cl} \mathrm{SO}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_{3}(\mathrm{~g}) & K_{\mathrm{c}_{1}} \\ 2 \mathrm{SO}_{3}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) & K_{\mathrm{c}_{2}} \end{array} $$ Which of these expressions correctly relates \(K_{c_{1}}\) to \(K_{c_{2}} ?\) (a) \(K_{\mathrm{c}_{2}}=K_{\mathrm{c}_{1}}^{2}\) (b) \(K_{\mathrm{c}_{2}}^{2}=K_{\mathrm{c}_{1}}\) (c) \(K_{\mathrm{c}_{2}}=1 / K_{\mathrm{c}_{1}}\) (d) \(K_{\mathrm{c}_{2}}=K_{\mathrm{c}_{1}}\) (e) \(K_{\mathrm{c},}=1 / K_{\mathrm{c}}^{2}\)

Short answer

The correct relation is (e) \(K_{c_2} = \frac{1}{K_{c_1}^2}\).

Step-by-step solution

Write Equilibrium Expressions

For the first equilibrium: \( \text{SO}_{2}(\text{g}) + \frac{1}{2} \text{O}_{2}(\text{g}) \rightleftharpoons \text{SO}_{3}(\text{g}) \), the equilibrium constant is given by \( K_{c_1} = \frac{[\text{SO}_{3}]}{[\text{SO}_{2}][\text{O}_{2}]^{1/2}} \). For the second equilibrium: \( 2 \text{SO}_{3}(\text{g}) \rightleftharpoons 2 \text{SO}_{2}(\text{g}) + \text{O}_{2}(\text{g}) \), the equilibrium constant is given by \( K_{c_2} = \frac{[\text{SO}_{2}]^{2}[\text{O}_{2}]}{[\text{SO}_{3}]^{2}} \).

Relate the Equilibrium Expressions

Notice that the second reaction is the reverse of twice the first reaction. By properties of equilibrium constants, if a reaction is reversed and then multiplied by a coefficient, we use the reciprocal of the equilibrium constant raised to the power of that coefficient. The second reaction can be considered the reverse of the first multiplied by 2, therefore: \( K_{c_2} = \left(\frac{1}{K_{c_1}}\right)^2 \). Thus, \( K_{c_2} = \frac{1}{K_{c_1}^2} \).

Verify and Match with Given Expressions

We derived that \( K_{c_2} = \frac{1}{K_{c_1}^2} \). Comparing to the options given: (a) \( K_{c_2} = K_{c_1}^2 \) - Incorrect. (b) \( K_{c_2}^2 = K_{c_1} \) - Incorrect. (c) \( K_{c_2} = \frac{1}{K_{c_1}} \) - Incorrect. (d) \( K_{c_2} = K_{c_1} \) - Incorrect. (e) \( K_{c_2} = \frac{1}{K_{c_1}^2} \) - Correct.

Key concepts

Equilibrium Constants

In chemical equilibrium, reactions do not go to completion but reach a state where the rates of the forward and reverse reactions are equal. At this point, the concentrations of reactants and products remain constant over time. The equilibrium constant, denoted by \( K_c \), is a numerical value that represents the ratio of the concentrations of the products to the reactants at equilibrium. Each species' concentration is raised to the power of its stoichiometric coefficient from the balanced chemical equation.
For the reaction \( aA + bB \rightleftharpoons cC + dD \), the equilibrium constant expression is:

• \( K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b} \)

The value of \( K_c \) indicates the extent of the reaction at equilibrium. A large \( K_c \) means products are favored, while a small \( K_c \) suggests reactants are more prevalent at equilibrium. Equilibrium constants are unique to specific reactions and depend on temperature but remain constant under set conditions.

Reversible Reactions

Reversible reactions are chemical reactions where the conversion of reactants to products and the conversion of products back to reactants occur simultaneously. They can proceed in both the forward and backward directions. These reactions do not go to completion, meaning they never fully convert all reactants to products.
In a reversible reaction, such as \( A + B \rightleftharpoons C + D \), the double arrow signifies that the reaction can proceed in both directions. Over time, a dynamic equilibrium is achieved in a closed system where the rate of the forward reaction equals the rate of the reverse reaction. The concentrations of the reactants and products stabilize and do not change, unless external conditions such as pressure or temperature are altered. Understanding reversible reactions helps in predicting how changes in conditions can shift the equilibrium position and alter the balance between reactants and products. This is crucial for industrial processes where conditions are adjusted to maximize product yield.

Reaction Quotients

The reaction quotient, \( Q_c \), is similar to the equilibrium constant, \( K_c \), but it is used to determine the direction a reaction will proceed to reach equilibrium. The expression for \( Q_c \) is the same as that of \( K_c \):

• \( Q_c = \frac{[C]^c[D]^d}{[A]^a[B]^b} \)

The key difference is that \( Q_c \) is calculated using the concentrations of reactants and products at any point in time, not just at equilibrium. By comparing \( Q_c \) to \( K_c \), you can predict the shift required for the reaction to reach equilibrium:

• If \( Q_c < K_c \), the reaction will proceed in the forward direction to form more products.
• If \( Q_c > K_c \), the reaction will shift in the reverse direction to form more reactants.
• If \( Q_c = K_c \), the reaction is already at equilibrium.

Knowing how to calculate and interpret \( Q_c \) is crucial for predicting the behavior of reactions under non-equilibrium conditions and for adjusting processes to achieve desired outcomes in chemical manufacturing and research.