Question

Consider the following reaction at 298 K: $$2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)$$ An equilibrium mixture contains \(\mathrm{O}_{2}(g)\) and \(\mathrm{SO}_{3}(g)\) at partial pressures of 0.50 \(\mathrm{atm}\) and 2.0 \(\mathrm{atm}\) , respectively. Using data from Appendix \(4,\) determine the equilibrium partial pressure of \(\mathrm{SO}_{2}\) in the mixture. Will this reaction be most favored at a high or a low temperature, assuming standard conditions?

Short answer

The equilibrium partial pressure of \(\mathrm{SO_2}\) in the mixture is \(2.28 \: \mathrm{atm}\), and the reaction is favored at low temperatures due to its exothermic nature.

Step-by-step solution

Write the balanced equation

The balanced equation is given in the problem: \(2 SO_2(g) + O_2(g) \rightleftharpoons 2 SO_3(g)\)

Express Kp in terms of partial pressures

Let's express the equilibrium constant (Kp) for the given reaction in terms of the partial pressures of the gases: \(Kp = \frac{(P_{SO_3})^2}{(P_{O_2})(P_{SO_2})^2}\)

Get the equilibrium constant from the Appendix 4

We need to find the equilibrium constant for this reaction in Appendix 4. Look for the value Kp for the reaction at 298 K. In this case, \(Kp = 2.94 \times 10^{-4}\)

Solve for the pressure of SO2

We are given the pressures of O2 (0.50 atm) and SO3 (2.0 atm), and we have the equilibrium constant Kp. Now, we can plug these values into our Kp expression and solve for the pressure of SO2: \( \begin{align*} 2.94 \times 10^{-4} &= \frac{(2.0)^2}{(0.50)(P_{SO_2})^2} \\ P_{SO_2} &= \sqrt{\frac{(2.0)^2}{(0.50)(2.94 \times 10^{-4})}} \\ P_{SO_2} &= 2.28 \mathrm{atm} \end{align*} \) So the equilibrium partial pressure of SO2 is 2.28 atm.

Determine if the reaction is favored at high or low temperature

To determine if the reaction is favored at high or low temperature, we will consider the heat of reaction (ΔH). Look for the value of ΔH for the reaction of formation of SO3 in Appendix 4: ΔH for the formation of SO3(g): -197 kJ/mol Since the reaction is exothermic (ΔH < 0), the reaction will be favored at low temperatures. This is because, according to Le Chatelier's principle, lowering the temperature of an exothermic reaction will force the equilibrium to shift towards the products, favoring the formation of SO3. In conclusion, the equilibrium partial pressure of SO2 in the mixture is 2.28 atm, and the reaction is favored at low temperatures.

Key concepts

Equilibrium Constant (Kp)

In chemical equilibrium, the equilibrium constant, denoted as Kp, is essential for understanding gas reactions. Kp quantifies the ratio of the partial pressures of products to reactants, each raised to the power of their coefficients in the balanced equation. This constant gives insight into the composition of the equilibrium mixture at a given temperature.
In the reaction \(2 \mathrm{SO}_{2}(g) + \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)\), Kp is expressed as:

• \( Kp = \frac{(P_{\mathrm{SO}_3})^2}{(P_{\mathrm{O}_2})(P_{\mathrm{SO}_2})^2} \)

Kp allows us to determine the current state of equilibrium and predict how changes in pressure or temperature might affect the reaction.
A key takeaway is that each equilibrium constant is specific to a particular reaction and temperature.
In this exercise, Kp at 298 K is \(2.94 \times 10^{-4}\), showing that the products are present in significant proportion at this temperature. Using known pressures for sulfur trioxide and oxygen, we solve for the pressure of sulfur dioxide. This showcases how understanding of Kp helps in predicting the behavior of the system.

Le Chatelier's Principle

Le Chatelier's Principle is a handy tool to predict how a chemical system at equilibrium responds to external changes. When a system experiences stress, like alterations in temperature, pressure, or concentration, the principle states it will shift in a direction that minimizes the effect of that change.
In this case, knowing that the reaction \(2 \mathrm{SO}_2(g) + \mathrm{O}_2(g) \rightleftharpoons 2 \mathrm{SO}_3(g)\) is involved, Le Chatelier’s Principle can guide us. If the temperature is reduced, the equilibrium will shift to counteract this change. Since this reaction is exothermic (releases heat), lowering the temperature encourages the formation of more products to generate heat.

• The system essentially shifts to the right, elevating the production of \(\mathrm{SO}_3\).

This principle is essential for optimizing conditions in industrial chemical processes where maximizing yield is critical.
Embracing Le Chatelier's insights can simplify understanding and controlling these equilibrium reactions.

Exothermic Reaction

An exothermic reaction is characterized by the release of energy in the form of heat. Such reactions have a negative enthalpy change, \( \Delta H < 0 \). This energy release is often evident as an increase in the temperature of the surroundings.
For the reaction \(2 \mathrm{SO}_{2}(g) + \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)\), the enthalpy change is \(-197 \ \text{kJ/mol}\), confirming its exothermic nature. This particular reaction generates heat as it progresses towards forming products.

• Due to the heat release, lowering the reaction temperature results in the system shifting to the right to produce more \(\mathrm{SO}_3\).

This type of reaction demonstrates practical implications in industrial settings, where controlling the reaction conditions, including temperature, can significantly affect the efficiency and yield."
Understanding exothermic reactions provides insights into how energy dynamics influence chemical equilibria.