Question

Hydrogen sulfide can be removed from natural gas by the reaction $$2 \mathrm{H}_{2} \mathrm{~S}(g)+\mathrm{SO}_{2}(g) \rightleftharpoons 3 \mathrm{~S}(s)+2 \mathrm{H}_{2} \mathrm{O}(g)$$ Calculate \(\Delta G^{\circ}\) and \(K\) (at \(298 \mathrm{~K}\) ) for this reaction. Would this reaction be favored at a high or low temperature?

Short answer

The standard Gibbs free energy change for the given reaction is \(\Delta G^{\circ} = -89.9 \,\text{kJ}\,\text{mol}^{-1}\), and the equilibrium constant at 298 K is \(K \approx 2.03 \times 10^6\). The reaction is favored at low temperatures.

Step-by-step solution

Determine the standard Gibbs free energy change of the reaction

To find \(\Delta G^{\circ}\) for the reaction, we use the following relationship for reactions involving gases: $$\Delta G^{\circ} = \sum_{products}(\Delta G^{\circ}_{prod} . \nu_{prod}) - \sum_{reactants}(\Delta G^{\circ}_{reac} . \nu_{reac})$$ where \(\Delta G^{\circ}_{prod}\) and \(\Delta G^{\circ}_{reac}\) are the standard Gibbs free energy of formation values for the products and reactants, respectively, and \(\nu_{prod}\) and \(\nu_{reac}\) are the stoichiometric coefficients for each species in the balanced chemical equation. We can look up the standard Gibbs free energy of formation values for each species from a standard thermodynamics table, which gives: \(\Delta G^{\circ}_{H_{2}S(g)} = -33.6 \,\text{kJ}\,\text{mol}^{-1}\) \(\Delta G^{\circ}_{SO_{2}(g)} = -300.1 \,\text{kJ}\,\text{mol}^{-1}\) \(\Delta G^{\circ}_{S(s)} = 0 \,\text{kJ}\,\text{mol}^{-1}\) (Since the element in its most stable state has \(\Delta G^{\circ} = 0\)) \(\Delta G^{\circ}_{H_{2}O(g)} = -228.6 \,\text{kJ}\,\text{mol}^{-1}\) Now, we can substitute these values into the equation for \(\Delta G^{\circ}\): $$\Delta G^{\circ} = (3\Delta G^{\circ}_{S(s)} + 2\Delta G^{\circ}_{H_{2}O(g)}) - (2\Delta G^{\circ}_{H_{2}S(g)} + \Delta G^{\circ}_{SO_{2}(g)})$$

Calculate the standard Gibbs free energy change of the reaction

Now, we can plug in the values for the standard Gibbs free energy of formation for each species and solve for \(\Delta G^{\circ}\): $$\Delta G^{\circ} = (3(0) + 2(-228.6)) - (2(-33.6) + (-300.1))$$ $$\Delta G^{\circ} = (-457.2) - (-367.3)$$ $$\Delta G^{\circ} = -89.9 \, \text{kJ}\, \text{mol}^{-1}$$

Find the equilibrium constant K using the standard Gibbs free energy change

We have the following relationship between the standard Gibbs free energy change, the gas constant \(R\), the temperature \(T\), and the equilibrium constant \(K\): $$\Delta G^{\circ} = -RT\ln K$$ We are asked to find \(K\) at \(T = 298 \,\text{K}\), and we know the gas constant \(R = 8.314 \,\text{J} \, \text{mol}^{-1} \, \text{K}^{-1}\). Rearranging the equation to solve for \(K\) and converting the units of \(\Delta G^{\circ}\) to J/mol to match the units of \(R\), we get: $$K = e^{\frac{-\Delta G^{\circ}}{RT}}$$ $$K = e^{\frac{-(-89.9 \, \times \, 1000)}{(8.314)(298)}}$$

Calculate the equilibrium constant K

Now, using the exponential function, we can calculate the equilibrium constant \(K\): $$K = e^{\frac{89.9 \, \times \, 1000}{8.314 \times 298}}$$ $$K \approx 2.03 \times 10^6$$

Determine if the reaction is favored at high or low temperatures

To determine whether the reaction is favored at high or low temperatures, we need to examine the van 't Hoff equation, which relates the equilibrium constant \(K\) to the temperature \(T\), the standard enthalpy change \(\Delta H^{\circ}\), and the gas constant \(R\): $$\frac{d \ln K}{dT} = \frac{\Delta H^{\circ}}{RT^2}$$ As the sign of \(\frac{d\ln K}{dT}\) depends only on the sign of \(\Delta H^{\circ}\), we can determine the temperature dependence of the reaction based on the standard enthalpy change alone. Using the relationship between \(\Delta G^{\circ}\), \(\Delta H^{\circ}\), and the standard entropy change \(\Delta S^{\circ}\): $$\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}$$ We see that the \(\Delta G^{\circ}\) is negative for the given reaction and entropy change \(\Delta S^{\circ}\) is also usually negative for exothermic reactions with fewer moles of gas in the product side. Therefore, the reaction is exothermic (\(\Delta H^{\circ} < 0\)). By examining the van 't Hoff equation, we can deduce that if \(\Delta H^{\circ} < 0\), the equilibrium constant \(K\) will decrease with increasing temperature. Hence, the reaction is favored at low temperatures. In conclusion, the standard Gibbs free energy change for the given reaction is \(\Delta G^{\circ} = -89.9 \,\text{kJ}\,\text{mol}^{-1}\), and the equilibrium constant at 298 K is \(K \approx 2.03 \times 10^6\). The reaction is favored at low temperatures.

Key concepts

Equilibrium Constant

In chemical reactions, the equilibrium constant, denoted as \(K\), is a fundamental concept that helps us understand how reactions are balanced between reactants and products at equilibrium. The equilibrium constant is derived from the concentrations of products and reactants at equilibrium according to the balanced chemical equation of a reaction. For a reaction such as \[ aA + bB \rightleftharpoons cC + dD \] the equilibrium constant \(K\) is expressed as \[ K = \frac{{[C]^c [D]^d}}{{[A]^a [B]^b}} \]where \([X]\) is the concentration of substance \(X\). In thermodynamics, \(K\) has a special relationship with the standard Gibbs free energy change \(\Delta G^{\circ}\). This relationship is given by \[ \Delta G^{\circ} = -RT\ln K \]where \(R\) is the universal gas constant and \(T\) is the temperature in Kelvin. This equation shows that a negative \(\Delta G^{\circ}\) implies a positive \(K\), indicating a reaction that favors the formation of products at equilibrium. Knowing \(K\) is crucial because it predicts the direction the reaction will proceed and how far it will go towards completion. A large \(K\) value (\(K \gg 1\)) implies that products are favored, while a small \(K\) value (\(K \ll 1\)) suggests that reactants are favored.

van 't Hoff Equation

The van 't Hoff equation provides insights into how the equilibrium constant \(K\) changes with temperature and allows us to infer whether the reaction is exothermic or endothermic. This equation is particularly useful for understanding temperature-dependent behavior of chemical reactions. The equation is written as: \[ \frac{d\ln K}{dT} = \frac{\Delta H^{\circ}}{RT^2} \]where \(\Delta H^{\circ}\) represents the standard enthalpy change, \(R\) is the gas constant, and \(T\) is the temperature in Kelvin. The sign of \(\Delta H^{\circ}\) dictates how \(K\) varies with temperature:

• If \(\Delta H^{\circ} > 0\) (endothermic reaction), \(K\) increases with increasing temperature.
• If \(\Delta H^{\circ} < 0\) (exothermic reaction), \(K\) decreases with increasing temperature.

This insight allows chemists to manipulate reaction conditions in industrial processes to optimize yield by simply adjusting the temperature. For instance, if a reaction is exothermic and the goal is to produce more products, lowering the temperature might move the equilibrium toward the products, as seen in our reference exercise.

Standard Enthalpy Change

The standard enthalpy change, denoted as \(\Delta H^{\circ}\), represents the heat absorbed or released during a reaction under standard conditions (1 atm pressure and 298 K temperature). This thermodynamic quantity is pivotal as it helps categorize reactions as either endothermic or exothermic. Endothermic reactions have a positive \(\Delta H^{\circ}\), indicating the absorption of heat, resulting in a cooling effect in the surroundings. Conversely, exothermic reactions have a negative \(\Delta H^{\circ}\), releasing heat and often warming the surroundings. In the context of the given reaction involving hydrogen sulfide and sulfur dioxide, \(\Delta H^{\circ}\) helps predict how changes in temperature affect the equilibrium position. According to the van 't Hoff equation explained earlier, an exothermic reaction (\(\Delta H^{\circ} < 0\)) often leads to a decrease in the equilibrium constant \(K\) with increased temperature, favoring the production of reactants. However, at lower temperatures, products are typically favored. Understanding \(\Delta H^{\circ}\) can also guide chemists in selecting appropriate reaction conditions, particularly in industrial applications where energy efficiency and safety are paramount.

Entropy Change

Entropy change, symbolized as \(\Delta S^{\circ}\), refers to the change in disorder or randomness within a system during a chemical reaction. Entropy is a critical factor in understanding reaction spontaneity when combined with enthalpy change \(\Delta H^{\circ}\) to determine the Gibbs free energy change \(\Delta G^{\circ}\). The relationship is embodied in the equation: \[ \Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ} \]where \(\Delta G^{\circ} < 0\) indicates a spontaneous reaction, suggesting a net favorability towards the formation of products. Free energy is minimized when a reaction proceeds, and both enthalpy and entropy contribute to this minimization. A positive \(\Delta S^{\circ}\) usually favors spontaneous reactions because the system becomes more disordered. Conversely, negative \(\Delta S^{\circ}\) might mean the system becomes more ordered and often involves fewer gaseous moles on the product side, which is significant in the reaction described in our exercise. Understanding \(\Delta S^{\circ}\) equips us with the capability to manage and predict how changes in reaction conditions, like pressure and temperature, affect both equilibrium and reaction rates.