Q16.4-67E - Chemistry (OpenStax)

Question

What happens to \({\bf{\Delta G}}_{{\bf{298}}}^{\bf{^\circ }}\) (becomes more negative or more positive) for the following chemical reactions when the partial pressure of oxygen is increased?

(a) \({\bf{S(s) + }}{{\bf{O}}_{\bf{2}}}{\bf{(g)}} \to {\bf{S}}{{\bf{O}}_{\bf{2}}}{\bf{(g)}}\)

(b) \({\bf{2S}}{{\bf{O}}_{\bf{2}}}{\bf{(g) + }}{{\bf{O}}_{\bf{2}}}{\bf{(g)}} \to {\bf{S}}{{\bf{O}}_{\bf{3}}}{\bf{(g)}}\)

(c) \({\bf{HgO(s)}} \to {\bf{Hg(l) + }}{{\bf{O}}_{\bf{2}}}{\bf{(g)}}\)

Step-by-Step Solution

Answer

1. As the partial pressure of oxygen is increased the ratio of pressures decreases which would in effect decrease the value of \(\Delta G\)   and make it more negative.
2. As the partial pressure of oxygen is increased the ratio of pressures decreases which would in effect decrease the value of \(\Delta G\) and make it more negative.
3. As the partial pressure of oxygen is increased the value of \(\ln \left( {{{\rm{P}}_{{\rm{O}}2}}} \right)\)  will also increase which would in effect increase the value of  \(\Delta G\) and make it more positive.

Step1: Define free energy change

The free energy change is related to the partial pressures of the reactants and products through the following equation:

\(\Delta {\rm{G}} = \Delta {{\rm{G}}^0} + {\rm{RTln}}\left[ {\frac{{{\rm{ Partial pressure of products }}}}{{{\rm{ partial pressure of reactants }}}}} \right] -  -   - (1)\)

\(\Delta {{\rm{G}}^0} = \) standard free energy change

\({\rm{R}} = \) gas constant and \({\rm{T}} = \) temperature

Step 2: a) Determine the decomposition reaction .

The given reaction is:

\({\rm{S}}({\rm{s}}) + {{\rm{O}}_2}(\;{\rm{g}}) \to {\rm{S}}{{\rm{O}}_2}(\;{\rm{g}})\)

Based on equation (1)

\(\Delta {\rm{G}} = \Delta {{\rm{G}}^0} + {\rm{RT}}\ln \left[ {\frac{{{{\rm{P}}_{{\rm{S}}{{\rm{O}}_2}}}}}{{{{\rm{P}}_{{{\rm{O}}_2}}}}}} \right]\)

As the partial pressure of oxygen is increased the ratio of pressures decreases which would in effect decrease the value of \(\Delta G\) and make it more negative.

Step 3: b) Determine the free energy of the decomposition reaction .

The given reaction is:

\(2{\rm{S}}{{\rm{O}}_2}(\;{\rm{g}}) + {{\rm{O}}_2}(\;{\rm{g}}) \to 2{\rm{S}}{{\rm{O}}_3}(\;{\rm{g}})\)

Based on equation (1)

\(\Delta {\rm{G}} = \Delta {{\rm{G}}^0} + {\rm{RTln}}\left[ {\frac{{{\rm{P}}_{{\rm{S}}{{\rm{O}}_3}}^2}}{{{\rm{P}}_{{\rm{S}}{{\rm{O}}_2}}^2{{\rm{P}}_{{{\rm{O}}_2}}}}}} \right]\)

As the partial pressure of oxygen is increased the ratio of pressures decreases which would in effect decrease the value of \(\Delta G\) and make it more negative.

Step 4: Determine the value of standard Gibbs free energy change

The given reaction is:

\({\rm{HgO}}({\rm{s}}) \to {\rm{Hg}}({\rm{l}}) + {{\rm{O}}_2}(\;{\rm{g}})\)

Based on equation (1)

\(\Delta {\rm{G}} = \Delta {{\rm{G}}^0} + {\rm{RTln}}\left[ {{{\rm{P}}_{{{\rm{O}}_2}}}} \right]\)

As the partial pressure of oxygen is increased the value of \(\ln \left( {{{\rm{P}}_{{\rm{O}}2}}} \right)\) will also increase which would in effect increase the value of \(\Delta G\) and make it more positive.