Question

Consider the following reaction at ${\text{298}}$ K.
${\text{2S}}{{\text{O}}_{\text{2}}}{\text{(g)}}\,{\text{ + }}\,{{\text{O}}_{\text{2}}}{\text{(g)}}\, \rightleftharpoons \,{\text{2S}}{{\text{O}}_{\text{3}}}{\text{(g)}}$
At equilibrium mixture contains $\,{{\text{O}}_{\text{2}}}{\text{(g)}}$ and ${\text{S}}{{\text{O}}_{\text{3}}}{\text{(g)}}$ at partial pressure of $0.50$ atm and $2.0$ atm, respectively. Using data from appendix $4.$, determine the equilibrium partial pressure of ${\text{S}}{{\text{O}}_{\text{2}}}{\text{(g)}}\,$ in the mixture.
In the appendix I have the following
${\text{S}}{{\text{O}}_{\text{2}}}{\text{(g)}}\,$\[{{\Delta G(f)}}\,{\text{ = }}\, - 300\] kJ/mol
${{\text{O}}_{\text{2}}}{\text{(g)}}\,$\[{{\Delta G(f)}}\,{\text{ = }}\,{\text{0}}\] kJ/mol
${\text{S}}{{\text{O}}_{\text{3}}}{\text{(g)}}$\[{{\Delta G(f)}}\,{\text{ = }}\, - 371\] kJ/mol

Answer 1

Hint:First we will determine the Gibbs free energy change. Standard Gibbs free change of a reaction is determined by subtracting the standard Gibbs free change for the formation of reactants from the standard Gibbs free change for the formation of products. Then by using the Gibbs free energy and equilibrium constant relation we will determine the equilibrium constant. Then we will write the equilibrium constant expression and by substituting the given partial pressure we will calculate the partial pressure of ${\text{S}}{{\text{O}}_{\text{2}}}{\text{(g)}}\,$.

Complete solution:
The formula to determine the standard Gibbs free change for a reaction as follows:
$\Delta {{\text{G}}^{\text{o}}} = \,\sum {{\Delta _{\text{f}}}{\text{G}}\,({\text{products)}}} \, - \,\sum {{\Delta _{\text{f}}}{\text{G}}\,({\text{reactants)}}} $
Where,
$\Delta {{\text{G}}^{\text{o}}}$ is the change in standard Gibbs free energy
$\sum {{\Delta _{\text{f}}}{\text{G}}\,({\text{products)}}} \,$is the summation of enthalpy of products
$\sum {{\Delta _{\text{f}}}{\text{G}}\,({\text{reactants)}}} $is the summation of enthalpy of products
The given reaction is as follows:
${\text{2S}}{{\text{O}}_{\text{2}}}{\text{(g)}}\,{\text{ + }}\,{{\text{O}}_{\text{2}}}{\text{(g)}}\, \rightleftharpoons \,{\text{2S}}{{\text{O}}_{\text{3}}}{\text{(g)}}$
For the given reaction${{\Delta }}{{\text{G}}^{\text{o}}}$formula can be written as follows:
$\Delta {{\text{G}}^{\text{o}}} = \,\left[ {2\, \times {\Delta _{\text{f}}}{\text{H}}\,({\text{S}}{{\text{O}}_{\text{3}}}{\text{)}}} \right] - \left[ {1\,\, \times {\Delta _{\text{f}}}{\text{H}}\,({{\text{O}}_{\text{2}}}{\text{)}}} \right] + \left[ {2\, \times {\Delta _{\text{f}}}{\text{H}}\,({\text{S}}{{\text{O}}_2}{\text{)}}} \right]$
On substituting $ - 310$for ${\Delta _{\text{f}}}{\text{H}}\,({\text{S}}{{\text{O}}_2}{\text{)}}$, 0 for${\Delta _{\text{f}}}{\text{H}}\,({{\text{O}}_{\text{2}}}{\text{)}}$, and \[ - 371\] for ${\Delta _{\text{f}}}{\text{H}}\,({\text{S}}{{\text{O}}_{\text{3}}}{\text{)}}$,
$\Delta {{\text{G}}^{\text{o}}} = \,\left[ {2\, \times ( - 371{\text{)}}} \right] - \left[ {1\,\, \times (0{\text{)}}} \right] + \left[ {2\, \times ( - 310{\text{)}}} \right]$
$\Delta {{\text{G}}^{\text{o}}} = \, - 742\, + \,620$
$\Delta {{\text{G}}^{\text{o}}} = \, - 142$ kJ.
We will convert the Gibbs free energy change from kilojoule to joule as follows:
${\text{1}}\,{\text{kJ}}\,{\text{ = }}\,{\text{1000}}\,{\text{J}}$
$ - {\text{142}}\,{\text{kJ}}\,\,{\text{ = }}\, - {\text{1}}{\text{.42}} \times \,{\text{1}}{{\text{0}}^5}\,{\text{J}}$
The relation between Gibbs free energy change and equilibrium constant is as follows:
$\Delta {\text{G}}\,{\text{ = }}\,\Delta {{\text{G}}^{\text{o}}} + \,{\text{RT}}\,{\text{ln}}\,{{\text{K}}_{\text{p}}}$
Where,
$\Delta {\text{G}}$ is the change in Gibbs free energy
R is the gas constant
T is the temperature
${{\text{K}}_{\text{p}}}$ is the equilibrium constant
At equilibrium $\Delta {\text{G}}$becomes zero so,
$\Delta {{\text{G}}^{\text{o}}} = \, - \,{\text{RT}}\,{\text{ln}}\,{{\text{K}}_{\text{p}}}$
On rearranging the above equation for ${{\text{K}}_{\text{p}}}$ is,
$\,{\text{ln}}\,{{\text{K}}_{\text{p}}}\, = \, - \dfrac{{\Delta {{\text{G}}^{\text{o}}}}}{{{\text{RT}}}}\,$
On substituting $ - {\text{1}}{\text{.42}} \times \,{\text{1}}{{\text{0}}^5}\,{\text{J}}$ for $\Delta {{\text{G}}^{\text{o}}}$, $8.314\,{\text{J}}{{\text{K}}^{ - 1}}$ for R, and ${\text{298}}$ K for T,
$\,{\text{ln}}\,{{\text{K}}_{\text{p}}}\, = \, - \dfrac{{ - {\text{1}}{\text{.42}} \times \,{\text{1}}{{\text{0}}^5}\,{\text{J}}}}{{8.314\,{\text{J}}{{\text{K}}^{ - 1}} \times {\text{298}}\,{\text{K}}}}\,$
$\,{\text{ln}}\,{{\text{K}}_{\text{p}}}\, = \, - \dfrac{{ - {\text{1}}{\text{.42}} \times \,{\text{1}}{{\text{0}}^5}\,}}{{2277.5}}\,$
$\,{\text{ln}}\,{{\text{K}}_{\text{p}}}\, = \,57.31\,$
${{\text{K}}_{\text{p}}}\, = \,{{\text{e}}^{57.31}}\,$
${{\text{K}}_{\text{p}}}\, = \,8 \times {10^{24}}\,$
Now we will write the equilibrium constant expression for the reaction as follows:
${\text{2S}}{{\text{O}}_{\text{2}}}{\text{(g)}}\,{\text{ + }}\,{{\text{O}}_{\text{2}}}{\text{(g)}}\, \rightleftharpoons \,{\text{2S}}{{\text{O}}_{\text{3}}}{\text{(g)}}$
\[{{\text{K}}_{\text{p}}}\, = \,\,\dfrac{{{{\left[ {{{\text{p}}_{{\text{S}}{{\text{O}}_{\text{3}}}}}} \right]}^2}}}{{{{\left[ {{{\text{p}}_{{\text{S}}{{\text{O}}_{\text{2}}}}}} \right]}^2}\left[ {{{\text{p}}_{{{\text{O}}_{\text{2}}}}}} \right]}}\]
On substituting contains $0.50$ atm for \[{{\text{p}}_{{{\text{O}}_{\text{2}}}}}\], $2.0$ atm for \[{{\text{p}}_{{\text{S}}{{\text{O}}_{\text{3}}}}}\] , and $8 \times {10^{24}}\,$for ${{\text{K}}_{\text{p}}}$,
\[8 \times {10^{24}}\,\, = \,\,\dfrac{{{{\left[ {{\text{2}}{\text{.0}}} \right]}^2}}}{{{{\left[ {{{\text{p}}_{{\text{S}}{{\text{O}}_{\text{2}}}}}} \right]}^2}\left[ {{\text{0}}{\text{.5}}} \right]}}\]
\[{\left[ {{{\text{p}}_{{\text{S}}{{\text{O}}_{\text{2}}}}}} \right]^2}\, = \,\,\dfrac{{4.0}}{{8 \times {{10}^{24}}\,\, \times 0.5}}\]
\[{\left[ {{{\text{p}}_{{\text{S}}{{\text{O}}_{\text{2}}}}}} \right]^2}\, = \,\,\dfrac{{4.0}}{{4.0 \times {{10}^{24}}}}\]
\[{\left[ {{{\text{p}}_{{\text{S}}{{\text{O}}_{\text{2}}}}}} \right]^2}\, = \,\,1.0 \times {10^{ - 24}}\]
\[{{\text{p}}_{{\text{S}}{{\text{O}}_{\text{2}}}}} = \,\,\sqrt {1.0 \times {{10}^{ - 24}}} \]
\[{{\text{p}}_{{\text{S}}{{\text{O}}_{\text{2}}}}} = \,\,1.0 \times {10^{ - 12}}\]
So, the partial pressure of \[{\text{S}}{{\text{O}}_{\text{2}}}\] is \[1.0 \times {10^{ - 12}}\] atm.
Therefore, the equilibrium partial pressure of ${\text{S}}{{\text{O}}_{\text{2}}}{\text{(g)}}\,$ in the mixture is \[1.0 \times {10^{ - 12}}\] atm.

Note: At equilibrium, the rate of forward direction and backward direction reaction becomes equal, so the concentration on both sides of the reaction becomes equal so the free energy change also tends to zero. To determine the change in Gibbs free energy of reaction a balanced chemical equation is necessary. Standard Gibbs free energy of formation is measured in kJ/mol. Oxygen is found in a gaseous state. So, there is no change in enthalpy during the formation of gaseous oxygen.