Problem 26 - Chemistry The Molecular Science

Question

At \(627{ }^{\circ} \mathrm{C}, K_{\mathrm{c}}=0.76\) for the reaction $$ 2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g}) $$ Calculate \(K_{\mathrm{c}}\) at \(627{ }^{\circ} \mathrm{C}\) for (a) synthesis of 1 mol sulfur trioxide gas. (b) decomposition of \(2 \mathrm{~mol} \mathrm{SO}_{3}\)

Step-by-Step Solution

Answer

(a) 0.872 (b) 1.316

Step 1: Write the expression for Kc of the original reaction

The equilibrium constant expression for the reaction \(2 \mathrm{SO}_{2}(\mathrm{g}) + \mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{g})\) is \[ K_{\mathrm{c}} = \frac{[\mathrm{SO}_{3}]^2}{[\mathrm{SO}_{2}]^2[\mathrm{O}_{2}]} = 0.76 \] at the given temperature of \(627^{\circ} \mathrm{C}\).

Step 2: Understand the change in reaction for 1 mol synthesis

When the synthesis of 1 mol of \(\mathrm{SO}_{3}\) is considered, the reaction becomes \(\mathrm{SO}_{2}(\mathrm{g}) + \frac{1}{2}\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{SO}_{3}(\mathrm{g})\).

Step 3: Calculate Kc for synthesis of 1 mol SO3

For the reaction \(\mathrm{SO}_{2} + \frac{1}{2}\mathrm{O}_{2} \rightleftharpoons \mathrm{SO}_{3}\), the equilibrium constant \(K'_{\mathrm{c}} = \sqrt{K_{\mathrm{c}}}\) because the new reaction is half of the original. \[ K'_{\mathrm{c}} = \sqrt{0.76} \approx 0.872 \]

Step 4: Consider the decomposition of 2 mol SO3

For the decomposition reaction of \(2\, \text{mol of } \mathrm{SO}_{3}\), it remains the same as the original reaction \(2\mathrm{SO}_{3}(\mathrm{g}) \rightleftharpoons 2\mathrm{SO}_{2}(\mathrm{g}) + \mathrm{O}_{2}(\mathrm{g})\).

Step 5: Calculate Kc for decomposition of 2 mol SO3

The equilibrium constant for the decomposition reaction is the inverse of the original \(K_{\mathrm{c}}\). Hence, \[ K_{\mathrm{decomp}} = \frac{1}{K_{\mathrm{c}}} = \frac{1}{0.76} \approx 1.316 \]

Key Concepts

Chemical Equilibrium

Chemical equilibrium is a state in a chemical reaction where the concentrations of reactants and products remain constant over time. This occurs in a closed system when forward and reverse reactions happen at the same rate. At equilibrium, there is no further change in the concentration of either reactants or products. Instead of stopping, the reactions continue to proceed—but, the rates of the forward and backward reactions are equal which keeps the concentrations steady.

In the context of our exercise, equilibrium is reached in the reaction involving sulfur dioxide (\(\mathrm{SO}_2\)), oxygen (\(\mathrm{O}_2\)), and sulfur trioxide (\(\mathrm{SO}_3\)). The equilibrium constant (\(K_c\)) for this reaction gives us insight into the ratio of product concentration to reactant concentration at equilibrium. For the given reaction:

• The equilibrium constant reflects the concentration of \(\mathrm{SO}_3\) produced from \(\mathrm{SO}_2\) and \(\mathrm{O}_2\) at \(627{ }^{\circ} \mathrm{C}\).
• A high \(K_c\) value implies that, at equilibrium, the products are favored over the reactants.

The equilibrium constant thus serves as a criterion for determining the extent of a chemical reaction.

Reaction Quotient

The reaction quotient, \(Q_c\), is a measure similar to the equilibrium constant, \(K_c\), but it applies to any point in the reaction, not just at equilibrium. \(Q_c\) is calculated using the same expression as \(K_c\) but with current concentrations of reactants and products. By comparing \(Q_c\) to \(K_c\), one can determine which direction the reaction will proceed to reach equilibrium.

• If \(Q_c < K_c\), the reaction must proceed in the forward direction to reach equilibrium as more products need to be formed.
• If \(Q_c > K_c\), then the reaction will shift backward, favoring the formation of reactants.
• If \(Q_c = K_c\), the system is already at equilibrium.

In our exercise, the values of \(Q_c\) can provide insights into the reaction's current status relative to its equilibrium position. Understanding the reaction quotient is crucial for predicting how a system that is not at equilibrium will behave.

Chemical Kinetics

Chemical kinetics involves the study of reaction rates and the steps that lead to the transformation of reactants to products. Although equilibrium constants do not directly provide information about the speed of a reaction, understanding kinetics helps us foresee how quickly equilibrium can be achieved.

Several factors influence chemical kinetics:

• Concentration: Higher concentrations generally increase reaction rates as there are more frequent collisions between reactant molecules.
• Temperature: Raising the temperature usually increases reaction rates by imparting more energy to the molecules, facilitating more collisions with sufficient energy to overcome the activation barrier.
• Catalysts: These substances increase reaction rates by lowering the activation energy needed for the reaction to occur.

In the context of our sulfur dioxide and oxygen reaction to form sulfur trioxide, the temperature at \(627{ }^{\circ} \mathrm{C}\) is critical because it affects both the rates of the forward and backward reactions. While the kinetics provide the speed, equilibrium concepts give us the position the reaction will gravitate towards over time.