Problem 46 - Chemistry The Central Science In Si Units

Question

At \(850 \mathrm{~K},\) the following reaction has \(K_{p}=0.0035:\) $$ 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g) $$ In an equilibrium mixture the partial pressures of \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) are \(18.24 \mathrm{kPa}\) and \(50.66 \mathrm{kPa}\), respectively. What is the equilibrium partial pressure of \(\mathrm{SO}_{3}\) in the mixture?

Step-by-Step Solution

Answer

The equilibrium partial pressure of \( \text{SO}_3 \) is approximately 7.686 kPa.

Step 1: Write the Expression for Kp

The equilibrium constant, \( K_p \), for the reaction \( 2 \text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2 \text{SO}_3(g) \) is given by the expression: \[ K_p = \frac{{P_{\text{SO}_3}^2}}{{P_{\text{SO}_2}^2 \cdot P_{\text{O}_2}}} \] Here, \( P_{\text{SO}_3} \), \( P_{\text{SO}_2} \), and \( P_{\text{O}_2} \) are the equilibrium partial pressures of \( \text{SO}_3 \), \( \text{SO}_2 \), and \( \text{O}_2 \) respectively. The given \( K_p \) is 0.0035.

Step 2: Substitute Known Values

We know that \( P_{\text{SO}_2} = 18.24 \text{kPa} \), \( P_{\text{O}_2} = 50.66 \text{kPa} \), and \( K_p = 0.0035 \). Substitute these values into the expression for \( K_p \): \[ 0.0035 = \frac{{P_{\text{SO}_3}^2}}{{(18.24)^2 \cdot 50.66}} \]

Step 3: Solve for P_SO3

To find \( P_{\text{SO}_3} \), rearrange the equation to solve for it:\[ P_{\text{SO}_3}^2 = 0.0035 \times (18.24)^2 \times 50.66 \] Calculate the right-hand side:\[ P_{\text{SO}_3}^2 = 0.0035 \times 332.6976 \times 50.66 \] \[ P_{\text{SO}_3}^2 = 59.0511 \] Now take the square root to find \( P_{\text{SO}_3} \):\[ P_{\text{SO}_3} = \sqrt{59.0511} \approx 7.686 \text{kPa} \]

Key Concepts

Equilibrium Constant

The equilibrium constant, denoted as \( K_p \) for reactions involving gases, represents the ratio of product pressures to reactant pressures when a reaction is at equilibrium. When dealing with chemical equilibrium in gases, \( K_p \) is calculated using the partial pressures of the gases involved.For the given chemical reaction \( 2 \text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2 \text{SO}_3(g) \):

• \( K_p = \frac{{P_{\text{SO}_3}^2}}{{P_{\text{SO}_2}^2 \cdot P_{\text{O}_2}}} \)

This equation shows that the products' pressures are divided by the reactants' pressures, adjusted for their coefficients in the balanced chemical equation. Each pressure term is raised to a power equal to its respective coefficient in the balanced equation.In our problem, \( K_p = 0.0035 \) at a temperature of \( 850 \mathrm{~K} \). This value indicates the degree to which the reactants are converted into products at equilibrium.A small \( K_p \) value, like in our scenario, suggests that, at equilibrium, the reactants are more prevalent than the products.

Partial Pressure

Partial pressure refers to the pressure that a single gas in a mixture contributes to the total pressure.Each gas in a mixture behaves as if it occupies the entire volume independently, contributing its own pressure, without affecting the other gases. For example, in the reaction \(2 \text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2 \text{SO}_3(g)\),the partial pressures of \( \text{SO}_2 \) and \( \text{O}_2 \) were given as \( 18.24 \text{kPa} \) and \( 50.66 \text{kPa} \) respectively.Knowing the partial pressures of all other components in the reaction allows us to solve for the unknown \( \text{SO}_3 \) partial pressure using the given \( K_p \) equation.

Le Châtelier's Principle

Le Châtelier's Principle helps us predict how changes in conditions can shift the equilibrium position of a chemical reaction.This principle states that if a dynamic equilibrium is disturbed by changing the conditions,

• The system will adjust itself to partially counteract the effect of the change and a new equilibrium will be established.

Consider changes such as temperature, pressure, or concentration when thinking about the gas reaction:\(2 \text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2 \text{SO}_3(g)\).If the pressure is increased, the system will shift to reduce this change by favoring the side of the reaction with fewer gas molecules.In our equation, increasing pressure would favor the production of \( \text{SO}_3 \), which has fewer moles of gas on the product side compared to its reactants.Thus, understanding Le Châtelier's Principle helps explain how external changes might affect the equilibrium conditions and outcomes.