Question

Consider the following reaction at \(298 \mathrm{K}:\) $$2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g)$$ An equilibrium mixture contains \(\mathrm{O}_{2}(g)\) and \(\mathrm{SO}_{3}(g)\) at partial pressures of 0.50 atm and 2.0 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 SO2 in the mixture is approximately 0.019 atm. The reaction is most favored at low temperatures, assuming standard conditions, as increasing the temperature will shift the equilibrium towards the reactants due to Le Châtelier's principle.

Step-by-step solution

Write the equilibrium expression and find Kp

Using the balanced chemical equation, we can write the equilibrium expression for Kp. \(K_p = \frac{[SO_3]^2}{[SO_2]^2[O_2]}\) Next, let's find the value of Kp at 298 K. According to the Appendix 4: - ΔGº(SO2) = -300.19 kJ/mol - ΔGº(O2) = 0 kJ/mol - ΔGº(SO3) = -370.40 kJ/mol We can calculate ΔGº for the reaction and use it to find the Kp value: ΔGº(reaction) = 2 x ΔGº(SO3) - 2 x ΔGº(SO2) - ΔGº(O2) ΔGº(reaction) = 2 x (-370.40) - 2 x (-300.19) - 0 = -140.42 kJ/mol Now, we can use the formula: \(ΔGº = -RT \ln{K_p}\) where R is the ideal gas constant, \(8.314 J/(mol \cdot K)\), and T is the temperature in Kelvin. Solving for Kp: \(K_p = e^{\frac{-ΔGº}{RT}}\) \(K_p = e^{\frac{-(-140.42 \times 10^3)}{(8.314)(298)}}\) \(K_p \approx 1.04 \times 10^5\)

Solve for the partial pressure of SO2

Now that we have the Kp value, we can plug in the given partial pressures of O2 and SO3 and solve for the partial pressure of SO2: \((1.04 \times 10^5) = \frac{(2.0)^2}{[SO_2]^2(0.50)}\) To solve for [SO2]: \([SO_2] = \sqrt{\frac{(2.0)^2}{(1.04 \times 10^5)(0.50)}}\) \([SO_2] \approx 0.019\,atm\) The equilibrium partial pressure of SO2 in the mixture is approximately 0.019 atm.

Check if the reaction is favored at high or low temperature

Looking at the given reaction, we see that there are two moles of gaseous reactants and three moles of gaseous products. According to Le Châtelier's principle, increasing the temperature of the reaction will shift the reaction toward the side with fewer moles of gas. In this case, increasing the temperature will shift the equilibrium towards the reactants. Thus, the reaction is most favored at low temperatures, assuming standard conditions.

Key concepts

Equilibrium Expressions

In chemistry, equilibrium expressions are vital for understanding how reactions behave when reaching a state where the rate of the forward reaction equals the rate of the reverse reaction. This state is called dynamic equilibrium.
The equilibrium expression for a given reaction is derived from its balanced chemical equation. It allows us to calculate the equilibrium constant, known as either \( K_c \) when using concentrations or \( K_p \) when dealing with partial pressures. For example, for the reaction:

• \( 2\text{SO}_{2}(g) + \text{O}_{2}(g) \rightarrow 2\text{SO}_{3}(g) \)

The equilibrium expression for \( K_p \) is:

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

Here, each concentration or partial pressure is raised to the power of its coefficient in the balanced equation. Calculating these expressions helps predict how changes in conditions can shift reactants and products' concentrations.

Gibbs Free Energy

Gibbs Free Energy \( (\Delta G^\circ) \) is a thermodynamic property that allows us to predict whether a chemical reaction can occur spontaneously. The relationship between Gibbs Free Energy and equilibrium is given by the formula:

• \( \Delta G^\circ = -RT \ln{K} \)

where \( R \) is the gas constant, \( T \) is the temperature in Kelvin, and \( K \) is the equilibrium constant.
This formula tells us that when \( \Delta G^\circ \) is negative, the reaction is spontaneous, and \( K \) is greater than one. Conversely, if \( \Delta G^\circ \) is positive, the reaction is not favorable, and \( K \) is less than one. Applying this to our reaction, calculated \( \Delta G^\circ \) was -140.42 kJ/mol, suggesting the reaction proceeds in the forward direction under standard conditions. Hence, it is spontaneous.

Le Châtelier's Principle

Le Châtelier's Principle helps predict the direction of a reaction's shift when subjected to changes such as pressure, temperature, or concentration. This principle is easy to understand: if a system at equilibrium experiences a change, it will adjust to counteract that change and re-establish equilibrium.
In the given reaction from the exercise, increasing the temperature favors the endothermic direction. Given that two moles of SO₂ and one mole of O₂ form two moles of SO₃, there are fewer moles of gas on the product side. According to Le Châtelier's Principle, a decrease in temperature will shift the equilibrium towards the side with fewer moles, favoring reactions under low-temperature conditions. This understanding is essential for optimizing industrial chemical processes.

Partial Pressure

Partial pressure refers to the pressure a gas in a mixture would exert if it occupied the entire volume by itself. Each gas's contribution in a mixture is independent of others and is described by Dalton's Law of Partial Pressures.
In chemical equilibrium, partial pressures are used similarly to concentrations in equilibrium expressions, typically denoted as \( K_p \). In our example problem, the given partial pressures for O₂ and SO₃ are 0.50 atm and 2.0 atm, respectively. Using these, along with \( K_p \), we can solve for the unknown partial pressure of SO₂ and establish equilibrium conditions.
This concept is crucial for reactions involving gases, as it allows chemists to manipulate conditions, like pressure, to favor either the forward or reverse reactions, achieving desired outcomes in various settings.