Question

Sulfur exists as \(S_{2}\) vapor at temperatures between \(700^{\circ} \mathrm{C}\) and \(1500^{\circ} \mathrm{C}\). At \(1473 \mathrm{k}\) it combines with hydrogen according to the equation $$ \mathrm{H}_{2}(\mathrm{~g})+(1 / 2) \mathrm{S}_{2}(\mathrm{~g}) \rightarrow \mathrm{H}_{2} \mathrm{~S}(\mathrm{~g}) $$ At \(750^{\circ} \mathrm{C}\) the equilibrium constant is \(1.07 \times 10^{2}\) and at \(1200^{\circ} \mathrm{C}\) it is \(4.39\). Determine the heat of reaction in the temperature range \(750^{\circ} \mathrm{C}\) to \(1200^{\circ} \mathrm{C}\), and the change in free energy at each of these temperatures.

Short answer

The heat of reaction in the temperature range of \(750^{\circ} \mathrm{C}\) to \(1200^{\circ} \mathrm{C}\) is approximately -32684.26 J/mol. The change in free energy at \(750^{\circ} \mathrm{C}\) is approximately -92017.93 J/mol, and at \(1200^{\circ} \mathrm{C}\) it is approximately -91690.34 J/mol.

Step-by-step solution

Convert temperatures to Kelvin

To work with the given temperatures, we must first convert them to Kelvin. We can do this by adding 273.15 to each given temperature in Celsius: At \(750^{\circ} \mathrm{C}\): \(T_1 = 750 + 273.15 = 1023.15 \mathrm{K}\) At \(1200^{\circ} \mathrm{C}\): \(T_2 = 1200 + 273.15 = 1473.15 \mathrm{K}\)

Use Van't Hoff equation to determine the heat of reaction

We can use Van't Hoff equation to determine the heat of reaction (∆H) from the equilibrium constants given at two different temperatures: \[ \frac{-\Delta H}{R} = \frac{\ln(K_2) - \ln(K_1)}{\frac{1}{T_2} - \frac{1}{T_1}} \] where \(R\) is the ideal gas constant, \(K_1\) is the equilibrium constant at \(T_1\), \(K_2\) is the equilibrium constant at \(T_2\), and ∆H is the heat of reaction. Plugging in the given values and constants: \[ \frac{-\Delta H}{8.314} = \frac{\ln(4.39) - \ln(1.07 \times 10^{2})}{\frac{1}{1473.15} - \frac{1}{1023.15}} \] Now, we can solve for ∆H: \[ -\Delta H = 8.314(\frac{\ln(4.39) - \ln(1.07 \times 10^{2})}{\frac{1}{1473.15} - \frac{1}{1023.15}}) \] \[ \Delta H = -32684.26 \mathrm{J/mol} \] So, the heat of reaction in the temperature range of \(750^{\circ} \mathrm{C}\) to \(1200^{\circ} \mathrm{C}\) is approximately -32684.26 J/mol.

Calculate the change in free energy at each temperature

Now that we have the heat of reaction, we can use the Gibbs free energy equation to calculate the change in free energy (∆G) at each temperature: \[ \Delta G = \Delta H - T\Delta S \] The Gibbs free energy equation can also be expressed in terms of the equilibrium constant: \[ \Delta G = -RT\ln(K) \] Since we have the equilibrium constant for both the temperatures, we can calculate ∆G at each temperature: At \(T_1 = 1023.15 \mathrm{K}\): \[ \Delta G_1 = -8.314(1023.15)\ln(1.07 \times 10^{2}) \] \[ \Delta G_1 = -92017.93 \mathrm{J/mol} \] At \(T_2 = 1473.15 \mathrm{K}\): \[ \Delta G_2 = -8.314(1473.15)\ln(4.39) \] \[ \Delta G_2 = -91690.34 \mathrm{J/mol} \] So, the change in free energy at \(750^{\circ} \mathrm{C}\) is approximately -92017.93 J/mol, and at \(1200^{\circ} \mathrm{C}\) it is approximately -91690.34 J/mol.

Key concepts

Van't Hoff Equation

Understanding the Van't Hoff equation is key when exploring how equilibrium constants are affected by temperature. This equation demonstrates the relationship between the change in temperature and the change in the equilibrium constant of a chemical reaction. Specifically, it details how the heat of reaction, or enthalpy change \( \Delta H \), can be determined from the equilibrium constants at different temperatures.

The Van't Hoff equation is mathematically represented as:\[ \frac{-\Delta H}{R} = \frac{\ln(K_2) - \ln(K_1)}{\frac{1}{T_2} - \frac{1}{T_1}} \]where \( R \) is the gas constant, \( K_1 \) and \( K_2 \) are the equilibrium constants at temperatures \( T_1 \) and \( T_2 \) respectively. Using this equation requires both temperatures to be converted into Kelvin before doing the calculations. The equation suggests that a plot of \( \ln(K) \) against \( \frac{1}{T} \) will be linear, with a slope equal to \( -\frac{\Delta H}{R} \). This linearity helps chemists estimate the enthalpy change for a given reaction over a temperature range.

Gibbs Free Energy

Gibbs free energy, \( \Delta G \), is a thermodynamic quantity that indicates the amount of energy available to do work during a chemical reaction at constant temperature and pressure. It's a central concept in predicting the spontaneity of a reaction; if \( \Delta G \) is negative, the reaction is spontaneous. The equation to calculate Gibbs free energy is:\[ \Delta G = \Delta H - T\Delta S \]where \( \Delta H \) is the enthalpy change, \( T \) is the temperature in Kelvin, and \( \Delta S \) is the entropy change.

In context with equilibrium constants, \( \Delta G \) is related to \( K \) by the equation:\[ \Delta G = -RT\ln(K) \]This relation allows us to calculate the change in free energy from a known equilibrium constant at a given temperature. The Gibbs free energy is particularly useful because it integrates all aspects of thermodynamics into one value, representing the total measure of a reaction's tendency to proceed.

Heat of Reaction

The heat of reaction, or enthalpy change \( \Delta H \), is the heat exchanged during a chemical reaction at constant pressure. It is the difference between the enthalpy of the products and the reactants. A negative value for \( \Delta H \) indicates an exothermic reaction, which releases heat into the surroundings, while a positive value indicates an endothermic reaction, which absorbs heat.

Using the Van't Hoff equation mentioned previously, chemists can calculate the heat of reaction over a temperature range, which provides insight into the energetics of the reaction. It's important to recognize that \( \Delta H \) is a key factor in the Gibbs free energy equation as well. The heat of reaction ultimately affects the Gibbs free energy and, therefore, the spontaneity and equilibrium position of the reaction. Accurate determination of \( \Delta H \) is crucial for predicting how reactions will behave when subjected to various temperatures.