Using relative enthalpy and entropy values, determine how the process is affected after each of the following temperature or pressure changes. Consider that a more effective reaction produces more product or more product in a shorter amount of time.

Reaction: SO₂ (g) + 2H₂S (g) ↔ 3S(s) + 2H₂O (g)Substance | ΔG kJ/mol | ΔH kJ/molH₂O(g) | -228.6 | -241.8H₂O(l) | -237.1 | -285.8SO₂(g) | -300.4 | -296.9SO₃(g) | -370.4 | -395.2H₂S(g) | -33.01 | -20.17S(s) | 0 | 0Categorize into: "More Effective" ~ "Less Effective" ~ "Equally Effective"a. Temp. decreases while maintaining container sizeb. Temp. increases while maintaining container sizec. Pressure decreases while maintaining container sized. Pressure increases while maintaining container size

It is possible to obtain ΔG and ΔH of the reaction from the standard ΔG and ΔH of products minus standard ΔG and ΔH of reactants. Thus, ΔG of reaction is:

ΔGr = 3ΔG S(s) + 2ΔG H₂O (g) - ( ΔG SO₂ (g) + 2ΔG H₂S (g))

ΔGr = 3×0 kJ/mol + 2×-228,6 kJ/mol - (-300,4 kJ/mol + 2×-33,01 kJ/mol)

ΔGr = -90,78 kJ/mol

In the same way, ΔHr is:

ΔHr = ΔH S(s) + 2ΔH H₂O (g) - ( ΔH SO₂ (g) + 2ΔH H₂S (g))

ΔHr = 3×0 kJ/mol + 2×-241,8 kJ/mol - (-296,9 kJ/mol + 2×-20,17 kJ/mol)

ΔHr = -146,36 kJ/mol

A ΔH < 0 means that the reaction is producing heat. By Le Chatelier's principle (Changes in the temperature, pressure, volume, or concentration of a system will result in predictable and opposing changes in the system), the increasing of temperature will produce less product, in the same way, the decreasing of temperature will produce more product.

Thus,

a. More effective.

b. Less efective.

As 2 moles of gas produce 5 moles of gas if the pressure decreases the system will try to produce more gas opposing the change, and again if the pressure increases the system will try to produce less gas.

Thus:

c. More effective.

d. Less effective.

I hope it helps!