Problem 114 - Chemistry The Central Science

Question

The reaction $$ \mathrm{SO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{~S}(g) \rightleftharpoons 3 \mathrm{~S}(s)+2 \mathrm{H}_{2} \mathrm{O}(g) $$ is the basis of a suggested method for removal of \(\mathrm{SO}_{2}\) from power-plant stack gases. The standard free energy of each substance is given in Appendix \(\mathrm{C}\). (a) What is the equilibrium constant for the reaction at \(298 \mathrm{~K}\) ? (b) In principle, is this reaction a feasible method of removing \(\mathrm{SO}_{2} ?\) (c) If \(P_{\mathrm{SO}_{2}}=P_{\mathrm{H}_{2} \mathrm{~S}}\) and the vapor pressure of water is 25 torr, calculate the equilibrium \(\mathrm{SO}_{2}\) pressure in the system at \(298 \mathrm{~K}\). (d) Would you expect the process to be more or less effective at higher temperatures?

Step-by-Step Solution

Answer

(a) Using provided standard free energies and the equation, the equilibrium constant (K) at 298 K can be calculated using \(K = e^{\frac{-\Delta G^{\circ}}{RT}}\). (b) The reaction is feasible for removing SO2 if K > 1, which would indicate that the forward reaction is favored at equilibrium. (c) The equilibrium SO2 pressure at 298 K can be found by substituting the given conditions into the K expression: \(K = \frac{(25 \,\text{torr})^{2}}{P_{\mathrm{SO}_{2}}(P_{\mathrm{H}_{2}\mathrm{S}})^{2}}\) and solving for \(P_{\mathrm{SO}_{2}}\). (d) Based on the calculated standard enthalpy change (∆H°), the temperature dependency of the process's effectiveness can be determined. If the reaction is endothermic (positive ∆H°), it will be more effective at higher temperatures; if exothermic (negative ∆H°), it will be less effective at higher temperatures.

Step 1: Determine the standard free energy change of reaction

Using the standard free energies of each substance provided in Appendix C, we can find the standard free energy change of the reaction (∆G°) using the equation: \(\Delta G^{\circ} = \sum{n_i \Delta G^{\circ}_i(F)} - \sum{n_i \Delta G^{\circ}_i(I)}\) where \(n_i\) refers to the stoichiometric coefficients, and F and I denote the final and the initial reactants and products. For the given reaction: \(\Delta G^{\circ} = [3 \Delta G^{\circ}(\mathrm{S}_{(s)}) + 2 \Delta G^{\circ}(\mathrm{H}_{2} \mathrm{O}_{(g)})] - [\Delta G^{\circ}(\mathrm{SO}_{2}_{(g)})+ 2 \Delta G^{\circ}(\mathrm{H}_{2} \mathrm{S}_{(g)})]\)

Step 2: Compute equilibrium constant at 298K

Using the standard free energy change of reaction (∆G°), we can calculate the equilibrium constant (K) at 298 K using the relationship: \(K = e^{\frac{-\Delta G^{\circ}}{RT}}\) where R is the ideal gas constant, 8.314 J/(mol K), and T = 298 K. #Part b: Analyze the feasibility of the reaction as a method to remove SO2#

Step 3: Check if the reaction can proceed forward at equilibrium

To check if the reaction can proceed forward at equilibrium, verify if the equilibrium constant (K) is greater than 1. If K > 1, the forward reaction will be favored at equilibrium, making the method feasible for removing SO2. #Part c: Calculate the equilibrium SO2 pressure at 298K#

Step 4: Write the expression for equilibrium constant

The expression for the equilibrium constant K can be expressed in terms of partial pressures of the reactants and products: \(K = \frac{\left[\mathrm{S}_{(s)}\right]^{3}\left[\mathrm{H}_{2} \mathrm{O}_{(g)}\right]^{2}}{\left[\mathrm{SO}_{2}_{(g)}\right]\left[\mathrm{H}_{2} \mathrm{S}_{(g)}\right]^{2}}\) Since the solid S does not contribute to the K expression, it can be simplified to: \(K = \frac{\left[\mathrm{H}_{2} \mathrm{O}_{(g)}\right]^{2}}{\left[\mathrm{SO}_{2}_{(g)}\right]\left[\mathrm{H}_{2} \mathrm{S}_{(g)}\right]^{2}}\)

Step 5: Substitute given conditions and solve for SO2 pressure

We are given that \(P_{\mathrm{SO}_{2}} = P_{\mathrm{H}_{2} \mathrm{S}}\) and the vapor pressure of water is 25 torr. With these conditions, we can substitute into the K expression and solve for the equilibrium SO2 pressure: \(K = \frac{(25 \,\text{torr})^{2}}{P_{\mathrm{SO}_{2}}(P_{\mathrm{H}_{2}\mathrm{S}})^{2}}\) #Part d: Effectiveness at higher temperatures#

Step 6: Determine the reaction's temperature dependency

To analyze the effectiveness of the process at higher temperatures, we first need to know if the reaction is endothermic or exothermic. This can be determined by calculating the standard enthalpy change of the reaction (∆H°) using the standard enthalpies of formation of the reactants and products provided in Appendix C. If the reaction is endothermic (positive ∆H°), the process will tend to be more effective at higher temperatures as more heat will provide more energy for the reaction to proceed forward. On the other hand, if the reaction is exothermic (negative ∆H°), the process will tend to be less effective at higher temperatures. From the calculated ∆H°, we can then determine whether the process will be more or less effective at higher temperatures.

Key Concepts

Free Energy Change

In chemical reactions, the change in free energy, denoted as \(\Delta G\), is a crucial parameter. It helps us understand the energy dynamics of the reaction. To compute the standard free energy change (\(\Delta G^\circ\)), we use the formula:

• \(\Delta G^{\circ} = \sum{n_i \Delta G^{\circ}_i(F)} - \sum{n_i \Delta G^{\circ}_i(I)}\).

Here, \(n_i\) represents the stoichiometric coefficients, while \(F\) and \(I\) stand for the final and initial reactants and products.
A negative \(\Delta G^\circ\) typically indicates that the reaction is spontaneous, meaning it can proceed without external energy. Meanwhile, a positive \(\Delta G^\circ\) suggests that the reaction is non-spontaneous and requires energy input.

Reaction Feasibility

To determine if a reaction can effectively proceed, we look at its equilibrium constant (\(K\)). The key relationship here is between \(\Delta G^\circ\) and \(K\), given by:

• \(K = e^{-\frac{\Delta G^{\circ}}{RT}}\)

where \(R\) is the ideal gas constant and \(T\) is the temperature in Kelvin.
If \(K\) is greater than 1, it implies that the reaction favors the formation of products, thus making it feasible. On the other hand, \(K < 1\) means the reaction doesn't proceed efficiently in the forward direction.
This helps in assessing whether the reaction can be used to remove \(\mathrm{SO}_2\) effectively.

Temperature Dependence

The effectiveness of a reaction can change with temperature due to the enthalpy change (\(\Delta H^\circ\)). By determining if the reaction is endothermic (\(\Delta H^\circ > 0\)) or exothermic (\(\Delta H^\circ < 0\)), we can predict how it will behave at different temperatures.
An endothermic reaction absorbs heat, making it more efficient at higher temperatures. Conversely, an exothermic reaction releases heat, and higher temperatures can drive the equilibrium left, reducing reaction effectiveness.
Understanding these dependencies helps predict the reaction's utility in varying thermal conditions.

Equilibrium Pressure

Calculating equilibrium pressure involves understanding how partial pressures of gases influence the reaction's direction. The relationship:

• \(K = \frac{(\text{Pressure of products})}{(\text{Pressure of reactants})}\)

allows us to determine the equilibrium concentrations.
For the given reaction, assuming equal pressures for \(\mathrm{SO}_2\) and \(\mathrm{H}_2\mathrm{S}\), and a known vapor pressure for water, we can solve for \(\mathrm{SO}_2\) pressure using:

• \(K = \frac{(25\, \text{torr})^2}{P_{\mathrm{SO}_2} (P_{\mathrm{H}_2\mathrm{S}})^2}\)

This equation helps in calculating how much \(\mathrm{SO}_2\) remains in the system, informing us on the reaction's progress.