Sulfur dioxide in the effluent gases from coal-burning electric power plants is one of the principal causes of acid rain. One method for reducing SO2 emissions involves partial reduction ofSO2 to H2S followed by catalytic conversion of the H2S and the remaining SO2 to elemental sulfur:

2H2S (g)+ SO2 (g) to 3 S( s ) + 2 H2O( g )
Using the data in Appendix B in the textbook, calculate Delta G degree. Show that this reaction is spontaneous at 25 degree C.

Estimate the temperature at which the reaction becomes nonspontaneous. Why does this reaction become nonspontaneous at high temperatures?

I don't have the book so any of the deltaG, H, S values I get on the internet gives me the wrong answer. Can I please get some help with this?

What makes you think the numbers I have are any better than the number you get on the internet?

Unfortunately, as an AI bot, I don't have access to specific textbooks or their appendixes, so I cannot provide you with the exact values from Appendix B. However, I can guide you on how to approach the problem using general principles.

To calculate ΔG° for the reaction, you will need the standard Gibbs free energy change (ΔG°) of formation values for the compounds involved in the reaction. These values can be found in Appendix B of your textbook, and they are generally tabulated for a standard temperature of 25°C (298 K) and pressure of 1 bar.

The Gibbs free energy change for a reaction at standard conditions (ΔG°) is related to the equilibrium constant of the reaction (K) by the equation:

ΔG° = -RTln(K)

where R is the gas constant (8.314 J/(mol·K)) and T is the temperature in Kelvin.

To determine whether the reaction is spontaneous at 25°C (298 K), you will need to calculate ΔG° and compare it to zero. If ΔG° is negative, the reaction is spontaneous. If ΔG° is positive, the reaction is non-spontaneous.

To estimate the temperature at which the reaction becomes non-spontaneous, you can rearrange the ΔG° equation:

ΔG° = -RTln(K)

Since ΔG° becomes positive when the reaction becomes non-spontaneous, you can set ΔG° equal to zero to find the temperature at which this occurs:

0 = -RTln(K)

Solving for T will give you the approximate temperature at which the reaction becomes non-spontaneous.

As for the reason why the reaction becomes non-spontaneous at high temperatures, it is likely due to an increase in the entropy term (ΔS°) of the reaction. At higher temperatures, the increase in entropy may outweigh the enthalpy term (ΔH°), resulting in a positive ΔG° for the reaction and making it non-spontaneous.

Remember to consult the specific values from your textbook or other reliable sources for accurate calculations.

To calculate the standard Gibbs free energy change, ΔG°, for the given reaction, you would need the standard Gibbs free energies of formation (ΔG°f) of the reactants and products involved. These values are typically provided in Appendix B of a textbook.

As you mentioned that you don't have the book, I can provide you with some hypothetical values just for the purpose of explaining how to calculate ΔG°.

Let's assume the hypothetical values for standard Gibbs free energies of formation:

ΔG°f(H2S) = -100 kJ/mol
ΔG°f(SO2) = -200 kJ/mol
ΔG°f(S) = 0 kJ/mol
ΔG°f(H2O) = -50 kJ/mol

Once you have these values, you can calculate ΔG° for the given reaction using the formula:

ΔG° = ΣnΔG°f(products) - ΣnΔG°f(reactants)

where n is the stoichiometric coefficient of each species.

For the reaction: 2H2S (g) + SO2 (g) → 3S (s) + 2H2O (g)

ΔG° = (3 × ΔG°f(S)) + (2 × ΔG°f(H2O)) - (2 × ΔG°f(H2S)) - ΔG°f(SO2)

Plugging in our assumed hypothetical values:

ΔG° = (3 × 0 kJ/mol) + (2 × -50 kJ/mol) - (2 × -100 kJ/mol) - (-200 kJ/mol)
= 0 kJ/mol + (-100 kJ/mol) + 200 kJ/mol + 200 kJ/mol
= 300 kJ/mol

Now, to determine if the reaction is spontaneous at 25°C (298 K), we can use the equation:

ΔG = ΔG° + RT ln(Q)

where ΔG is the actual Gibbs free energy change, ΔG° is the standard Gibbs free energy change, R is the gas constant (8.314 J/K·mol), T is the temperature in Kelvin, and Q is the reaction quotient.

At equilibrium, ΔG = 0. Therefore, we can rearrange the equation as:

0 = ΔG° + RT ln(K)

where K is the equilibrium constant.

At 25°C (298 K), we have:

0 = 300 kJ/mol + (8.314 J/K·mol × 298 K) ln(K)
= 300 kJ/mol + (2470 J/mol) ln(K)

To find ln(K), we can rearrange the equation:

ln(K) = -300,000 J/mol / (2470 J/mol)
≈ -121.57

Using ln(K ≈ -121.57), we can find K using the exponential function:

K = e^ln(K)
= e^(-121.57)
≈ 1.350 × 10^(-53)

Since K is an extremely small value, it indicates that the reaction predominantly proceeds in the forward direction, meaning it is spontaneous at 25°C.

To determine the temperature at which the reaction becomes nonspontaneous, we can calculate the equilibrium constant, K, for different temperatures and find the temperature where ln(K) becomes zero or positive.

As for why the reaction becomes nonspontaneous at high temperatures, it may be due to the increase in entropy (ΔS) associated with the reverse reaction. The reverse reaction involves the formation of gases (H2S and SO2) from solid sulfur, which would have a higher entropy. At high temperatures, the increase in entropy may outweigh the favorable enthalpy change (ΔH) of the reaction, making it nonspontaneous. However, the specific temperature at which this occurs would depend on the actual values of ΔG°, ΔH, and ΔS for the reaction, which would vary depending on the data in your textbook.