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.

Drag the appropriate items to their respective bins.
THE ITEMS ARE: 1) Pressure increases by decreasing the container size, 2) Temperature increases while maintaining the container size, 3) Pressure decreases by increasing the container size, and 4) Temperature decreases while maintaining the container size.

THE BINS ARE: MORE EFFECTIVE, LESS EFFECTIVE, EQUALLY EFFECTIVE.

The question is incomplete, the complete question is:

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/mol

H2O(g) | -228.6 | -241.8

H2O(l) | -237.1 | -285.8

SO2(g) | -300.4 | -296.9

SO3(g) | -370.4 | -395.2

H2S(g) | -33.01 | -20.17

S(s) | 0 | 0

Categorize into: "More Effective" ~ "Less Effective" ~ "Equally Effective"

- Temp. decreases while maintaining container size

- Temp. increases while maintaining container size

- Pressure decreases while maintaining container size

- Pressure increases while maintaining container size

Explanation:

ΔH∘rxn=ΔH∘f(products)−ΔH∘f(reactants)

ΔH∘rxn= [0+2(-241.8)] - [(-296.9) + 2(-20.7)]= -145.6KJ/mol

ΔG∘rxn=[(0+ 2(-228.6) -[(-300.4) + 2(-33.01)]= -90.78KJ/mol

The reaction is spontaneous since ΔG∘rxn is negative

The reaction is exothermic since ΔH∘rxn= negative

1) increase in pressure is less effective because it favours the reverse reaction.

2) increase in temperature is less effective since it favours the reverse reaction given that the reaction is exothermic in nature

3) decrease in pressure is more effective, it favours the forward reaction

4)Decrease in temperature is more effective because it favours the forward reaction.