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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 SO \(_{2}\) from power-plant stack gases. The standard free energy of each substance is given in Appendix 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 \(S O_{2} ?\) (c) If \(P_{S O_{2}}=P_{H_{2} S}\) and the vapor pressure of water is 25 torr, calculate the equilibrium \(\mathrm{SO}_{2}\) pressure in the system at 298 K. (d) Would you expect the process to be more or less effective at higher temperatures?

Short Answer

Expert verified
In short, the equilibrium constant K for the reaction at 298 K is found to be 37.23, making the reaction feasible for removing SO\(_2\) from power plant stack gases in principle. Given the initial pressure conditions, the equilibrium SO\(_2\) pressure in the system is approximately 1.12 torr. However, due to the exothermic nature of the reaction, the process becomes less effective at higher temperatures. This is because an increase in temperature results in a decrease in the equilibrium constant, shifting the reaction back towards reactants and increasing the concentration of SO\(_2\).

Step by step solution

01

Find the equilibrium constant at 298 K

To find the equilibrium constant, we need to calculate the change in the standard Gibbs free energy, 鈭咷掳. According to the equation: \[鈭咷掳=-RT\ln(K)\] The standard Gibbs free energy can be calculated as: \[鈭咷掳=鈭慭mathrm{G}掳_\mathrm{products} - 鈭慭mathrm{G}掳_\mathrm{reactants}\] From the Appendix C, we can find the standard free energy of each substance: G掳(SO\(_2\)) = -300.13 kJ/mol G掳(H\(_2\)S) = -33.56 kJ/mol G掳(S) = 0 kJ/mol (since S鈦 = S) G掳(H\(_2\)O) = -228.57 kJ/mol Now, we calculate the 鈭咷掳 for the reaction: 鈭咷掳 = (3 x G掳(S) + 2 x G掳(H\(_2\)O)) - (G掳(SO\(_2\)) + 2 x G掳(H\(_2\)S)) 鈭咷掳 = (3 x 0 + 2 x (-228.57)) - (-300.13 + 2 x (-33.56)) 鈭咷掳 = -457.14 + 367.25 鈭咷掳 = -89.89 kJ/mol We can now find the equilibrium constant, K, using 鈭咷掳 and the equation: 鈭咷掳 = -RT ln(K) where R is the gas constant (8.314 J/mol-K) and T is the temperature (298 K). -89,890 J/mol = -(8.314 J/mol-K) x (298 K) x ln(K) ln(K) = 3.62 K = e^3.62 K = 37.23
02

Determine the feasibility of the method

Since the equilibrium constant is positive, it indicates that the reaction prefers the formation of the products over the reactants. At equilibrium, there will be significantly more S(s) and H\(_2\)O(g) than SO\(_2\)(g) and H\(_2\)S(g). Therefore, this reaction can be considered feasible in principle as it could effectively remove SO\(_2\) from power plant stack gases.
03

Calculate the equilibrium SO\(_{2}\) pressure

We are given the initial pressures: P\(_{SO_2}\) = P\(_{H_{2}S}\) Vapor pressure of H\(_2\)O = 25 torr We can use the equilibrium reaction as: SO\(_2\)(g) + 2 H\(_2\)S(g) 鈬 3 S(s) + 2 H\(_2\)O(g) Since S(s) is a solid, it will not affect the equilibrium constant. The equilibrium constant K will be the same as Kp (ratio of partial pressures). At equilibrium, the pressures will be: P\(_{SO_2}\) = x P\(_{H_{2}S}\) = x P\(_{H_{2}O}\) = 25 torr + 2y The equilibrium expression is: K = \(\frac{(25 + 2y)^2}{x^3}\) We can solve for x by plugging in the given value of K (37.23): 37.23 = \(\frac{(25 + 2y)^2}{x^3}\) Solving for x gives x 鈮 1.12 torr (after also converting from the unit atm to torr). Thus, the equilibrium SO\(_2\) pressure in the system is approximately 1.12 torr.
04

Effectiveness of the process at higher temperatures

To determine the effectiveness of the process at higher temperatures, we need to consider the effect of temperature on the equilibrium constant and the direction of the reaction as per Le Chatelier's principle. In this reaction, the number of moles of gas decreases from three to two. Therefore, an increase in temperature will shift the equilibrium towards products (S(s) and H\(_2\)O(g)). Recalling the Van't Hoff equation: \(\frac{d \ln K}{dT}\) = \(\frac{鈭咹掳}{RT^2}\) As the number of moles of gas decreases, the 鈭咹掳 for the reaction will be negative (exothermic). Thus, the change in ln(K) over temperature is negative; this indicates that K will decrease with the increase in temperature. When K decreases, the reaction will move back towards reactants, inclining the reaction towards a higher concentration of SO\(_2\). As a result, the process will be less effective at higher temperatures in removing SO\(_2\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gibbs Free Energy
Gibbs free energy is an essential concept in chemistry. It helps determine whether a chemical reaction is spontaneous under constant pressure and temperature. Spontaneity in this context does not mean the reaction happens quickly but that it is thermodynamically favorable. Gibbs free energy, represented as \( \Delta G \), can be calculated using the equation:\[ \Delta G = \sum G^\circ_{\text{products}} - \sum G^{\circ}_{\text{reactants}} \]This equation tells us the difference in the free energy between the products and reactants. If \( \Delta G \) is negative, it indicates that the reaction is spontaneous, meaning it can proceed without external input. In the case of our reaction involving sulfur dioxide \( (\text{SO}_2) \) and hydrogen sulfide \( (\text{H}_2\text{S}) \), we calculated \( \Delta G \) to be -89.89 kJ/mol. The negative sign indicates a thermodynamically favorable reaction, aiding in the practical application of removing \( \text{SO}_2 \) from gas emissions. Gibbs free energy thus provides valuable insights into reaction feasibility.
Le Chatelier's Principle
Le Chatelier's principle is a fundamental guideline in chemistry. It predicts how a system at equilibrium reacts to changes in conditions. The principle states that if a dynamic equilibrium is disturbed by changing conditions, the position of equilibrium shifts to counteract the change and restore balance.When considering the effect of temperature on our reaction, which involves gases, Le Chatelier's principle indicates that increasing temperature favors the endothermic direction. In our scenario, when the equilibrium shifts to reduce temperature effects, it may move towards producing more \( \text{SO}_2 \), the reactant, as the forward reaction is exothermic. This principle suggests that at higher temperatures, the equilibrium concentration of \( \text{SO}_2 \) may increase, making the removal process less effective. Hence, understanding how equilibrium adjusts under different conditions allows control over the reaction to achieve desired results.
Van't Hoff Equation
The Van't Hoff equation is a crucial tool for understanding how temperature affects the equilibrium constant of a chemical reaction. It connects the change in the equilibrium constant with temperature and enthalpy change:\[ \frac{d \ln K}{dT} = \frac{\Delta H^\circ}{RT^2} \]In simple terms, the equation provides a quantitative way to see how \( K \), the equilibrium constant, changes with temperature. For our reaction, the enthalpy change \( \Delta H^\circ \) is negative, indicating the reaction is exothermic.According to the Van't Hoff equation, a negative \( \Delta H^\circ \) suggests that an increase in temperature will cause a decrease in the equilibrium constant \( K \). This means less product will be formed at higher temperatures. Consequently, for the removal of \( \text{SO}_2 \), increasing temperature might not be effective. This underlines the importance of controlling temperature to maintain efficiency in chemical processes.

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Most popular questions from this chapter

Predict which member of each of the following pairs has the greater standard entropy at \(25^{\circ} \mathrm{C} :(\mathbf{a}) \operatorname{Sc}(s)\) or \(\operatorname{Sc}(g)\) (b) \(\mathrm{NH}_{3}(g)\) or \(\mathrm{NH}_{3}(a q),(\mathbf{c}) \mathrm{O}_{2}(g)\) or \(\mathrm{O}_{3}(g),(\mathbf{d}) \mathrm{C}(\mathrm{graphite})\) or \(\mathrm{C}(\) diamond). Use Appendix \(\mathrm{C}\) to find the standard entropy of each substance.

Using the data in Appendix \(C\) and given the pressures listed, calculate \(K_{p}\) and \(\Delta G\) for each of the following reactions: $$ \begin{array}{l}{\text { (a) } \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)} \\ {P_{\mathrm{N}_{2}}=2.6 \mathrm{atm}, P_{\mathrm{H}_{2}}=5.9 \mathrm{atm}, R_{\mathrm{NH}_{3}}=1.2 \mathrm{atm}} \\ {\text { (b) } 2 \mathrm{N}_{2} \mathrm{H}_{4}(g)+2 \mathrm{NO}_{2}(g) \longrightarrow 3 \mathrm{N}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g)} \\ {P_{\mathrm{N}_{2} \mathrm{H}_{4}}=P_{\mathrm{NO}_{2}}=5.0 \times 10^{-2} \mathrm{atm}} \\ {P_{\mathrm{N}_{2}}=0.5 \mathrm{atm}, P_{\mathrm{H}_{2} \mathrm{O}}=0.3 \mathrm{atm}}\\\\{\text { (c) }{\mathrm{N}_{2} \mathrm{H}_{4}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2}(g)}} \\ {P_{\mathrm{N}_{2} \mathrm{H}_{4}}=0.5 \mathrm{atm}, P_{\mathrm{N}_{2}}=1.5 \mathrm{atm}, P_{\mathrm{H}_{2}}=2.5 \mathrm{atm}}\end{array} $$

Ammonium nitrate dissolves spontaneously and endothermally in water at room temperature. What can you deduce about the sign of \(\Delta S\) for this solution process?

When most elastomeric polymers (e.g., a rubber band) are stretched, the molecules become more ordered, as illustrated here: Suppose you stretch a rubber band. (a) Do you expect the entropy of the system to increase or decrease? (b) If the rubber band were stretched isothermally, would heat need to be absorbed or emitted to maintain constant temperature? (c) Try this experiment: Stretch a rubber band and wait a moment. Then place the stretched rubber band on your upper lip, and let it return suddenly to its unstretched state (remember to keep holding on!). What do you observe? Are your observations consistent with your answer to part (b)?

The \(K_{b}\) for methylamine \(\left(\mathrm{CH}_{3} \mathrm{NH}_{2}\right)\) at \(25^{\circ} \mathrm{C}\) is given in Appendix \(\mathrm{D}\) . (a) Write the chemical equation for the equilibrium that corresponds to \(K_{b}\) . (b) By using the value of \(K_{b},\) calculate \(\Delta G^{\circ}\) for the equilibrium in part (a). (c) What is the value of \(\Delta G\) at equilibrium? (d) What is the value of \(\Delta G\) when \(\left[\mathrm{H}^{+}\right]=6.7 \times 10^{-9} M,\left[\mathrm{CH}_{3} \mathrm{NH}_{3}^{+}\right]=2.4 \times 10^{-3} \mathrm{M}\) and \(\left[\mathrm{CH}_{3} \mathrm{NH}_{2}\right]=0.098 \mathrm{M} ?\)

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