Chapter 19: Problem 77

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

Expert verified

For each reaction: (a) As the partial pressure of \(O_{2}\) increases, the reaction will shift towards the products, making it more spontaneous and causing \(\Delta G\) to decrease. (b) With an increased partial pressure of \(O_{2}\), the reaction will shift towards the reactants, resulting in a decrease in spontaneity and an increase in \(\Delta G\). (c) Similar to reaction (b), increasing the partial pressure of \(O_{2}\) will shift the reaction towards the reactants, leading to a decrease in spontaneity and an increase in \(\Delta G\).

Step by step solution

ΔG is the Gibbs free energy change, which can be used to determine the spontaneity of a reaction. Mathematically, ΔG is related to the equilibrium constant (K) and the reaction quotient (Q) by the equation: \[ ΔG = ΔG^⦵ + RT\ln(Q) \] where ΔG^⦵ is the standard free energy change, R is the gas constant, and T is the temperature in Kelvin. As the partial pressure of O₂ changes, so will Q, causing ΔG to vary. We will apply Le Chatelier's principle, which states that when a system at equilibrium is subjected to a change in pressure, temperature, or concentration, it will shift to counteract the change and restore equilibrium. In this case, we will focus on the effect of increased partial pressure of O₂ on each reaction. #Step 2: Analyze reaction (a) with increased partial pressure of O₂#

The first reaction is given by: \[ 2CO(g) + O₂(g) \longrightarrow 2CO₂(g) \] In this reaction, as we increase the partial pressure of O₂ (a reactant), the reaction will try to counteract the change and restore equilibrium, according to Le Chatelier's principle. This leads to a shift in the reaction towards the products (CO₂). As a result, this reaction will become more spontaneous, meaning that ΔG will decrease. #Step 3: Analyze reaction (b) with increased partial pressure of O₂#

The second reaction is given by: \[ 2H₂O₂(l) \longrightarrow 2H₂O(l) + O₂(g) \] In this reaction, increasing the partial pressure of O₂ (a product) leads to a shift in the reaction in order to counteract the change. According to Le Chatelier's principle, the reaction will shift towards the reactants (H₂O₂) in order to restore equilibrium. Consequently, this reaction will become less spontaneous, meaning that ΔG will increase. #Step 4: Analyze reaction (c) with increased partial pressure of O₂#

The third reaction is given by: \[ 2KClO₃(s) \longrightarrow 2KCl(s) + 3O₂(g) \] In this reaction, increasing the partial pressure of O₂ (a product) also causes a shift back towards the reactants (KClO₃), as predicted by Le Chatelier's principle. Therefore, the reaction will become less spontaneous, implying an increase in ΔG.