Problem 73 - Chemistry: The Central Science

Question

Explain qualitatively how \(\Delta G\) changes for each of the following reactions as the partial pressure of \(\mathrm{O}_{2}\) is increased: (a) \(2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)\) (b) \(2 \mathrm{H}_{2} \mathrm{O}_{2}(l) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g)\) (c) \(2 \mathrm{KClO}_{3}(s) \longrightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g)\)

Step-by-Step Solution

Answer

As the partial pressure of \(\mathrm{O}_{2}\) increases: (a) \(\Delta G\) for reaction \(2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)\) becomes more negative, making the forward reaction more spontaneous. (b) \(\Delta G\) for reaction \(2 \mathrm{H}_{2} \mathrm{O}_{2}(l) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g)\) becomes less negative or more positive, making the forward reaction less spontaneous and the reverse reaction more spontaneous. (c) \(\Delta G\) for reaction \(2 \mathrm{KClO}_{3}(s) \longrightarrow 2 \mathrm{KCl}(s)+3\mathrm{O}_{2}(g)\) becomes less negative or more positive, making the forward reaction less spontaneous and the reverse reaction more spontaneous.

Step 1: Identify the change in partial pressure of \(\mathrm{O}_{2}\)

For this reaction, an increase in the partial pressure of \(\mathrm{O}_{2}\) will cause the concentration of one of the reactants to increase.

Step 2: Determine the effect on reaction quotient \(Q\)

As the partial pressure of \(\mathrm{O}_{2}\) increases, the reaction quotient \(Q\) increases, because \(Q = \frac{[\mathrm{CO}_2]^2}{[\mathrm{CO}]^2[\mathrm{O}_2]}\). The increase in partial pressure of \(\mathrm{O}_{2}\) will cause \(Q\) to decrease since the denominator will become larger.

Step 3: Determine the effect on the change in free energy \(\Delta G\)

According to the relationship \(\Delta G=\Delta G^{o}+RT \ln{Q}\), when \(Q\) decreases, \(\Delta G\) becomes more negative. So, as the partial pressure of \(\mathrm{O}_{2}\) is increased, \(\Delta G\) for the reaction will become more negative, indicating that the forward reaction will be increasingly spontaneous. (b) \(2 \mathrm{H}_{2} \mathrm{O}_{2}(l) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g)\)

Step 1: Identify the change in partial pressure of \(\mathrm{O}_{2}\)

For this reaction, an increase in the partial pressure of \(\mathrm{O}_{2}\) will cause the concentration of one of the products to increase.

Step 2: Determine the effect on reaction quotient \(Q\)

As the partial pressure of \(\mathrm{O}_{2}\) increases, the reaction quotient \(Q = \frac{[\mathrm{O}_2]}{[\mathrm{H}_2\mathrm{O}_2]^2}\) will increase, since the numerator will become larger.

Step 3: Determine the effect on the change in free energy \(\Delta G\)

According to the relationship \(\Delta G=\Delta G^{o}+RT \ln{Q}\), when \(Q\) increases, \(\Delta G\) becomes less negative or more positive. So, as the partial pressure of \(\mathrm{O}_{2}\) is increased, \(\Delta G\) for the reaction will become less negative or more positive, making the forward reaction less spontaneous and the reverse reaction more spontaneous. (c) \(2 \mathrm{KClO}_{3}(s) \longrightarrow 2 \mathrm{KCl}(s)+3\mathrm{O}_{2}(g)\)

Step 1: Identify the change in partial pressure of \(\mathrm{O}_{2}\)

For this reaction, an increase in the partial pressure of \(\mathrm{O}_{2}\) will cause the concentration of one of the products to increase.

Step 2: Determine the effect on reaction quotient \(Q\)

As the partial pressure of \(\mathrm{O}_{2}\) increases, the reaction quotient \(Q = \frac{[\mathrm{O}_2]^3}{[\mathrm{KClO}_3]^2}\) will increase since the numerator will become larger.

Step 3: Determine the effect on the change in free energy \(\Delta G\)

According to the relationship \(\Delta G=\Delta G^{o}+RT \ln{Q}\), when \(Q\) increases, \(\Delta G\) becomes less negative or more positive. So, as the partial pressure of \(\mathrm{O}_{2}\) is increased, \(\Delta G\) for the reaction will become less negative or more positive, making the forward reaction less spontaneous and the reverse reaction more spontaneous.