Question

The equilibrium constant \(K_{P}\) for the reaction: $$ 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftarrows 2 \mathrm{SO}_{3}(g) $$ is \(5.60 \times 10^{4}\) at \(350^{\circ} \mathrm{C}\). The initial pressures of \(\mathrm{SO}_{2}\) \(\mathrm{O}_{2},\) and \(\mathrm{SO}_{3}\) in a mixture are \(0.350,0.762,\) and \(0 \mathrm{~atm},\) respectively, at \(350^{\circ} \mathrm{C}\). When the mixture reaches equilibrium, is the total pressure less than or greater than the sum of the initial pressures?

Short answer

Total pressure decreases as the reaction shifts toward fewer gas moles (equilibrium).

Step-by-step solution

Write the Expression for Equilibrium Constant

For the reaction \(2 \mathrm{SO}_{2}(g) + \mathrm{O}_{2}(g) \rightleftarrows 2 \mathrm{SO}_{3}(g)\), the equilibrium constant in terms of partial pressures, \(K_{P}\), is expressed as: \[K_{P} = \frac{(P_{\mathrm{SO}_{3}})^2}{(P_{\mathrm{SO}_{2}})^2 \cdot P_{\mathrm{O}_{2}}}\] where \( P_{\mathrm{SO}_{2}}, P_{\mathrm{O}_{2}}\), and \( P_{\mathrm{SO}_{3}}\) are the partial pressures of \( \mathrm{SO}_{2}\), \( \mathrm{O}_{2}\), and \( \mathrm{SO}_{3}\) at equilibrium.

Use Initial Conditions to Setup ICE Table

An ICE (Initial, Change, Equilibrium) table is used to track changes in the concentrations or pressures of the species involved.- Initial pressures: \( P_{\mathrm{SO}_{2}} = 0.350 \, \mathrm{atm}\), \( P_{\mathrm{O}_{2}} = 0.762 \, \mathrm{atm}\), \( P_{\mathrm{SO}_{3}} = 0 \, \mathrm{atm}\).- Change in pressures: Let \( x \) be the change in pressure for \( \mathrm{SO}_{3} \), so \( +2x \) for \( \mathrm{SO}_{3} \), \( -2x \) for \( \mathrm{SO}_{2} \), and \( -x \) for \( \mathrm{O}_{2} \).- Equilibrium pressures: - \( P_{\mathrm{SO}_{2}} = 0.350 - 2x \) - \( P_{\mathrm{O}_{2}} = 0.762 - x \) - \( P_{\mathrm{SO}_{3}} = 2x \).

Substitute Equilibrium Pressures into \(K_{P}\) Expression

Substitute the equilibrium expressions into the \(K_{P}\) equation:\[K_{P} = \frac{(2x)^2}{(0.350 - 2x)^2 \cdot (0.762 - x)} = 5.60 \times 10^{4}\].

Solve for x

Solving for \( x \) requires iterative or numerical methods, as this is a polynomial equation that may not have a simple analytic solution. However, if directly solved or approximated, this will allow you to find the specific changes in pressure, leading to equilibrium.

Calculate Total Initial and Equilibrium Pressures

- Total initial pressure \( P_{\text{initial}} = 0.350 + 0.762 + 0 = 1.112 \, \mathrm{atm}\).- Calculate the total equilibrium pressure using: \[ P_{\text{equilibrium}} = (0.350 - 2x) + (0.762 - x) + 2x\]. After solving for \( x \) and substituting it back, the change in total pressure across reactants and products can be calculated.

Compare Initial and Equilibrium Total Pressures

Use the solved value of \( x \) to determine if the total equilibrium pressure exceeds or falls short of the initial total pressure. Due to the stoichiometry \(2:1:2\), and the expectation that \(x\) will increase \(P_{\mathrm{SO}_{3}}\) significantly while reducing \(P_{\mathrm{SO}_{2}}\) and \(P_{\mathrm{O}_{2}}\), you will then compare whether:- \( P_{\text{equilibrium}} < P_{\text{initial}} \).

Key concepts

Equilibrium Constant

The equilibrium constant, denoted as \(K_{P}\), is a critical concept in chemical equilibrium. It helps us predict the extent of a reaction under a given atmospheric pressure. For the reaction \(2 \mathrm{SO}_{2}(g) + \mathrm{O}_{2}(g) \rightleftarrows 2 \mathrm{SO}_{3}(g)\), the equilibrium constant is represented as the ratio of the partial pressures of the products to the reactants at equilibrium. The formula for \(K_{P}\) is: \[K_{P} = \frac{(P_{\mathrm{SO}_{3}})^2}{(P_{\mathrm{SO}_{2}})^2 \cdot P_{\mathrm{O}_{2}}}\] where \(P_{\mathrm{SO}_{2}}\), \(P_{\mathrm{O}_{2}}\), and \(P_{\mathrm{SO}_{3}}\) are the equilibrium pressures of \(\mathrm{SO}_{2}\), \(\mathrm{O}_{2}\), and \(\mathrm{SO}_{3}\), respectively. A high equilibrium constant, such as \(5.60 \times 10^{4}\) for this reaction, indicates that the reaction favors the formation of products at equilibrium.

Partial Pressures

Partial pressures refer to the pressure exerted by each individual gas in a mixture. Each gas component in a mix has its own partial pressure and the sum of all these gives us the total pressure of the system. In the problem provided, gases \(\mathrm{SO}_{2}\), \(\mathrm{O}_{2}\), and \(\mathrm{SO}_{3}\) have their own initial partial pressures of \(0.350\), \(0.762\), and \(0 \mathrm{~atm}\), respectively. These initial values help us set up the initial conditions in an ICE table to determine how the system evolves to equilibrium. Partial pressures play a vital role in calculating the equilibrium constant \(K_{P}\) and understanding the changes that occur when a reaction approaches equilibrium.

ICE Table

An ICE table, which stands for Initial, Change, Equilibrium, is a valuable tool used to visualize and calculate the changes in concentrations or pressures for the components involved in a chemical reaction. It allows us to systematically track the transition from initial conditions through change to the final equilibrium state. For the reaction \(2 \mathrm{SO}_{2}(g) + \mathrm{O}_{2}(g) \rightleftarrows 2 \mathrm{SO}_{3}(g)\), the ICE table starts with the initial pressures:

• \(P_{\mathrm{SO}_{2}} = 0.350\, \mathrm{atm}\)
• \(P_{\mathrm{O}_{2}} = 0.762\, \mathrm{atm}\)
• \(P_{\mathrm{SO}_{3}} = 0\, \mathrm{atm}\)

The changes in pressure for reaching equilibrium involve setting a variable \(x\), representing units of change. This results in:

• \(+2x\) for \(\mathrm{SO}_{3}\)
• \(-2x\) for \(\mathrm{SO}_{2}\)
• \(-x\) for \(\mathrm{O}_{2}\)

This way, the final equilibrium pressures become:

• \(P_{\mathrm{SO}_{2}} = 0.350 - 2x\)
• \(P_{\mathrm{O}_{2}} = 0.762 - x\)
• \(P_{\mathrm{SO}_{3}} = 2x\)

Equilibrium Pressures

Equilibrium pressures refer to the pressures of the reactants and products when the chemical reaction has reached its state of equilibrium. At this point, the rate of the forward reaction equals the rate of the reverse reaction, and the concentrations of reactants and products remain constant over time. To find these pressures, we substitute our change from the ICE table into our equilibrium expressions.In the given exercise, the equilibrium pressures are expressed as:

• \(P_{\mathrm{SO}_{2}} = 0.350 - 2x\)
• \(P_{\mathrm{O}_{2}} = 0.762 - x\)
• \(P_{\mathrm{SO}_{3}} = 2x\)

By solving the equilibrium expression \(K_{P} = \frac{(2x)^2}{(0.350 - 2x)^2 \cdot (0.762 - x)} = 5.60 \times 10^{4}\), the exact value of \(x\) can be computed either analytically or through numerical methods. The resulting equilibrium pressures can then be compared with initial pressures to understand whether the total pressure
at equilibrium is greater or less than that at the start.