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) \(2 \mathrm{KClO}_{3}(s) \longrightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g)\)

Short answer

For each reaction: (a) Increasing the partial pressure of O₂ shifts the reaction towards the right side, making ΔG more negative. (b) Increasing the partial pressure of O₂ shifts the reaction towards the left side, making ΔG less negative or more positive. (c) Increasing the partial pressure of O₂ shifts the reaction towards the left side, making ΔG less negative or more positive.

Step-by-step solution

(a) Reaction 1: \(2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)\)

(For reaction 1, increasing the partial pressure of O₂ would cause the reaction to shift towards the right side, i.e., towards the formation of CO₂ gas. This is because, an increase in the partial pressure of O₂, a reactant, would result in an increase in the rate of the forward reaction. This would help maintain equilibrium and hence, ΔG would become more negative as more reactants are converted into products (CO₂).)

(b) Reaction 2: \(2 \mathrm{H}_{2} \mathrm{O}_{2}(l) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g)\)

(For reaction 2, increasing the partial pressure of O₂ would cause the reaction to shift towards the left side, i.e., towards the formation of hydrogen peroxide (H₂O₂). This is because an increase in the partial pressure of O₂, a product, would result in an increase in the rate of the reverse reaction. This would help maintain equilibrium and hence, ΔG would become less negative or more positive as more products are converted back to reactants (H₂O₂).)

(c) Reaction 3: \(2 \mathrm{KClO}_{3}(s) \longrightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g)\)

(For reaction 3, increasing the partial pressure of O₂ would cause the reaction to shift towards the left side, i.e., towards the formation of potassium chlorate (KClO₃). This is because an increase in the partial pressure of O₂, a product, would result in an increase in the rate of the reverse reaction. This would help maintain equilibrium and hence, ΔG would become less negative or more positive as more products are converted back to reactants (KClO₃).)