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 \(\mathrm{O}_{2}(\mathrm{~g})\) 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 is 0.046 atm. Moreover, since the standard enthalpy change (ΔH°) is negative, indicating an exothermic reaction, the reaction will be most favored at a low temperature.

Step-by-step solution

Calculate the equilibrium constant (Kp)

We need to find the equilibrium constant Kp using the standard Gibbs free energy change. For this reaction, we can use the relation: \( ΔG° = -RT \ln Kp \) Where R is the gas constant (8.314 J/(mol K)), T is the temperature in Kelvin and ΔG° refers to the standard Gibbs free energy change, which can be calculated using the standard Gibbs free energy change of formation (ΔGf°) values from Appendix 4. ΔG° can be found by: \( ΔG° = Σn(ΔGf°)_\text{products} - Σn(ΔGf°)_\text{reactants} \)

Calculate ΔG° for the reaction

According to Appendix 4, the standard Gibbs free energy change of formation values for the reactants and products are: \(ΔGf°_\text{SO2(g)} = -300.19 \,\mathrm{kJ/mol}\) \(ΔGf°_\text{O2(g)} = 0 \,\mathrm{kJ/mol}\) \(ΔGf°_\text{SO3(g)} = -370.41 \,\mathrm{kJ/mol}\) Now using these values, let's obtain ΔG° for the reaction: \( ΔG° = 2(-370.41 \,\mathrm{kJ/mol}) - [2(-300.19 \,\mathrm{kJ/mol}) + 0] \) \( ΔG° = -140.44 \,\mathrm{kJ/mol} \)

Solve for Kp

Now we can calculate the equilibrium constant Kp using the formula: \(Kp= e^{-\frac{ΔG°}{RT}}\) At standard temperature 298 K and gas constant R (8.314 J/(mol K)): \(Kp= e^{-\frac{-140.44\,\mathrm{kJ/mol}}{(8.314\,\mathrm{J/(mol\,K)})(298\,\mathrm{K})}} = 2.91\times 10^{4}\)

Set up the equilibrium expression

Now that we have Kp, let's set up the equilibrium expression involving the equilibrium partial pressure of SO2 (x): \( Kp = \frac{(2.0)^2}{(0.50) (x)^2} \)

Solve for equilibrium partial pressure of SO2 (x)

Now we can solve for the equilibrium partial pressure of SO2 (x): \( (2.91\times 10^{4}) = \frac{(2.0)^2}{(0.50) (x)^2} \) \( x = \sqrt{\frac{(0.50) (2.0)^2}{(2.91\times 10^{4})}} = 0.046\,\text{atm} \) So, the equilibrium partial pressure of SO2 is 0.046 atm.

Assess the effect of temperature

To determine if the reaction is favored at high or low temperatures, we need to consider the standard enthalpy change (ΔH°) of the reaction. From Appendix 4: \(ΔHf°_\text{SO2(g)} = -296.83 \,\mathrm{kJ/mol}\) \(ΔHf°_\text{O2(g)} = 0 \,\mathrm{kJ/mol}\) \(ΔHf°_\text{SO3(g)} = -395.73 \,\mathrm{kJ/mol}\) For this reaction: \( ΔH° = Σn(ΔHf°)_\text{products} - Σn(ΔHf°)_\text{reactants} \) \( ΔH° = 2(-395.73 \,\mathrm{kJ/mol}) - [2(-296.83 \,\mathrm{kJ/mol}) + 0] \) \( ΔH° = -197.8 \,\mathrm{kJ/mol} \) Since ΔH° is negative (exothermic reaction), increasing the temperature will shift the equilibrium towards the reactants and decreasing temperature will favor the products. Therefore, this reaction will be favored at a low temperature.

Key concepts

Equilibrium Constant

The equilibrium constant is a fundamental concept in the study of chemical equilibrium. It is a quantitative measure of the ratio of concentrations of products to reactants when a reaction has reached a state of balance, where the rate of the forward reaction equals the rate of the reverse reaction. For the gaseous reaction involving sulfur dioxide and oxygen to produce sulfur trioxide, the equilibrium constant (denoted as Kp when expressed in terms of partial pressures) can be determined using thermodynamic data, specifically the standard Gibbs free energy change (ΔG°).

Using the equation \(Kp= e^{-\frac{ΔG°}{RT}}\), where R is the ideal gas constant and T is the temperature in Kelvin, we can calculate Kp. A large Kp value indicates that, at equilibrium, the concentration of the products is higher than that of the reactants, meaning the reaction tends towards the formation of products. In this exercise, the calculated Kp is \(2.91 \times 10^{4}\), suggesting a strong tendency towards product formation under the given conditions.

Gibbs Free Energy

Gibbs free energy, represented by the symbol ΔG, is the maximum amount of work that can be performed by a closed system at constant temperature and pressure. This thermodynamic quantity helps predict the direction of a chemical reaction and determines if the process is spontaneous. A negative ΔG° indicates a spontaneous reaction under standard conditions. In the reaction of sulfur dioxide and oxygen to form sulfur trioxide, ΔG° can be calculated using standard Gibbs free energies of formation for the products and reactants.

Once ΔG° is known, this value is instrumental in finding the equilibrium constant, as demonstrated in the steps of the provided exercise. The relationship between ΔG° and Kp emphasizes the interdependence of thermodynamics and equilibrium: the more negative the ΔG° is, the higher the value of Kp, which implies a reaction that favours the formation of products at equilibrium.

Partial Pressure

Partial pressure is the pressure that a single gas in a mixture of gases would exert if it alone occupied the entire volume. In the context of chemical reactions involving gases, the equilibrium constant Kp is expressed in terms of the partial pressures of reactants and products. The equilibrium expression incorporates these pressures raised to the power of their stoichiometric coefficients.

In our example, we needed to solve for the equilibrium partial pressure of sulfur dioxide (SO2) using the established Kp and the given partial pressures of oxygen (O2) and sulfur trioxide (SO3). The final step of the calculation involves setting up an equilibrium expression and rearranging it to solve for the unknown partial pressure of SO2. Knowledge of partial pressures is crucial to understanding and predicting the behaviour of gaseous systems at equilibrium.

Le Chatelier's Principle

Le Chatelier's principle is a qualitative tool that predicts the response of a system at equilibrium when it is subjected to an external change. According to this principle, if a system at equilibrium is disturbed, it will shift in a direction that counteracts the disturbance. Common disturbances include changes in concentration, pressure, and temperature.

In the example provided, analyzing the effect of temperature on our reaction is pertinent. Since the formation of sulfur trioxide from sulfur dioxide and oxygen is exothermic (releases heat), an increase in temperature would shift the equilibrium to favour the reactants, as indicated by a negative ΔH° value. Consequently, a lower temperature would promote the formation of products. Thus, this reaction is most favored at a low temperature, aligning with Le Chatelier's principle that suggests the system adjusts to minimize temperature changes by absorbing additional heat.