Question

What happens to ΔG° (becomes more negative or more positive) for the following chemical reaction when the partial pressure of oxygen is increased?

S(s) + O₂ (g) ==== SO₂ (g)


a) Increasing P of O₂ will shift the equilibrium toward the reactants, which decreases the value of K. therefore ΔG° becomes more positive.

b) Increasing P of O₂ will shift the equilibrium toward the reactants, which increases the value of K. therefore ΔG° becomes more negative.

C) Increasing P of O₂ will shift the equilibrium toward the products, which increases the value of K. therefore ΔG° becomes more negative.

d) None of the above

Answer 1

Increasing the Po2 shifts the equilibrium toward the products, increasing K and resulting in a more negative ΔG° for the reaction of sulfur and oxygen forming sulfur dioxide.

The correct answer is: (c) Increasing P of O₂ will shift the equilibrium toward the products, which increases the value of K. therefore ΔG° becomes more negative.

Step-by-step explanation:

What happens to ΔG° (Delta G naught) for the chemical reaction S(s) + O2(g) ⟶ SO2(g) when the partial pressure of oxygen is increased?

Increasing the partial pressure of oxygen (Po2) will shift the equilibrium toward the products, which increases the value of the equilibrium constant (K). According to the Gibbs free energy equation, ΔG° = -RT × ln(K), where R is the universal gas constant and T is the temperature in Kelvin, an increase in the value of K correlates with a more negative value of ΔG°.

Hence, the correct answer is: (c) Increasing Po2 will shift the equilibrium toward the products, which increases the value of K. Therefore ΔG° becomes more negative.