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Question: Consider the following reaction at \({\text{298}}\) K. \({\text{2S}}{{\text{O}}_{\text{2}}}{\text...

Consider the following reaction at 298{\text{298}} K.
2SO2(g)  +  O2(g) ⇌ 2SO3(g){\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  O2(g)\,{{\text{O}}_{\text{2}}}{\text{(g)}} and SO3(g){\text{S}}{{\text{O}}_{\text{3}}}{\text{(g)}} at partial pressure of 0.500.50 atm and 2.02.0 atm, respectively. Using data from appendix 4.4., determine the equilibrium partial pressure of SO2(g) {\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

Explanation

Solution

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 SO2(g) {\text{S}}{{\text{O}}_{\text{2}}}{\text{(g)}}\,.

Complete solution:
The formula to determine the standard Gibbs free change for a reaction as follows:
ΔGo= ∑ΔfG (products) − ∑ΔfG (reactants)\Delta {{\text{G}}^{\text{o}}} = \,\sum {{\Delta _{\text{f}}}{\text{G}}\,({\text{products)}}} \, - \,\sum {{\Delta _{\text{f}}}{\text{G}}\,({\text{reactants)}}}
Where,
ΔGo\Delta {{\text{G}}^{\text{o}}} is the change in standard Gibbs free energy
∑ΔfG (products) \sum {{\Delta _{\text{f}}}{\text{G}}\,({\text{products)}}} \,is the summation of enthalpy of products
∑ΔfG (reactants)\sum {{\Delta _{\text{f}}}{\text{G}}\,({\text{reactants)}}} is the summation of enthalpy of products
The given reaction is as follows:
2SO2(g)  +  O2(g) ⇌ 2SO3(g){\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 reactionΔGo{{\Delta }}{{\text{G}}^{\text{o}}}formula can be written as follows:
ΔGo= [2 ×ΔfH (SO3)]−[1  ×ΔfH (O2)]+[2 ×ΔfH (SO2)]\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 - 310for ΔfH (SO2){\Delta _{\text{f}}}{\text{H}}\,({\text{S}}{{\text{O}}_2}{\text{)}}, 0 forΔfH (O2){\Delta _{\text{f}}}{\text{H}}\,({{\text{O}}_{\text{2}}}{\text{)}}, and −371 - 371 for ΔfH (SO3){\Delta _{\text{f}}}{\text{H}}\,({\text{S}}{{\text{O}}_{\text{3}}}{\text{)}},
ΔGo= [2 ×(−371)]−[1  ×(0)]+[2 ×(−310)]\Delta {{\text{G}}^{\text{o}}} = \,\left[ {2\, \times ( - 371{\text{)}}} \right] - \left[ {1\,\, \times (0{\text{)}}} \right] + \left[ {2\, \times ( - 310{\text{)}}} \right]
ΔGo= −742 + 620\Delta {{\text{G}}^{\text{o}}} = \, - 742\, + \,620
ΔGo= −142\Delta {{\text{G}}^{\text{o}}} = \, - 142 kJ.
We will convert the Gibbs free energy change from kilojoule to joule as follows:
1 kJ  =  1000 J{\text{1}}\,{\text{kJ}}\,{\text{ = }}\,{\text{1000}}\,{\text{J}}
−142 kJ   =  −1.42× 105 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:
ΔG  =  ΔGo+ RT ln Kp\Delta {\text{G}}\,{\text{ = }}\,\Delta {{\text{G}}^{\text{o}}} + \,{\text{RT}}\,{\text{ln}}\,{{\text{K}}_{\text{p}}}
Where,
ΔG\Delta {\text{G}} is the change in Gibbs free energy
R is the gas constant
T is the temperature
Kp{{\text{K}}_{\text{p}}} is the equilibrium constant
At equilibrium ΔG\Delta {\text{G}}becomes zero so,
ΔGo= − RT ln Kp\Delta {{\text{G}}^{\text{o}}} = \, - \,{\text{RT}}\,{\text{ln}}\,{{\text{K}}_{\text{p}}}
On rearranging the above equation for Kp{{\text{K}}_{\text{p}}} is,
 ln Kp = −ΔGoRT \,{\text{ln}}\,{{\text{K}}_{\text{p}}}\, = \, - \dfrac{{\Delta {{\text{G}}^{\text{o}}}}}{{{\text{RT}}}}\,
On substituting −1.42× 105 J - {\text{1}}{\text{.42}} \times \,{\text{1}}{{\text{0}}^5}\,{\text{J}} for ΔGo\Delta {{\text{G}}^{\text{o}}}, 8.314 JK−18.314\,{\text{J}}{{\text{K}}^{ - 1}} for R, and 298{\text{298}} K for T,
 ln Kp = −−1.42× 105 J8.314 JK−1×298 K \,{\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}}}}\,
 ln Kp = −−1.42× 105 2277.5 \,{\text{ln}}\,{{\text{K}}_{\text{p}}}\, = \, - \dfrac{{ - {\text{1}}{\text{.42}} \times \,{\text{1}}{{\text{0}}^5}\,}}{{2277.5}}\,
 ln Kp = 57.31 \,{\text{ln}}\,{{\text{K}}_{\text{p}}}\, = \,57.31\,
Kp = e57.31 {{\text{K}}_{\text{p}}}\, = \,{{\text{e}}^{57.31}}\,
Kp = 8×1024 {{\text{K}}_{\text{p}}}\, = \,8 \times {10^{24}}\,
Now we will write the equilibrium constant expression for the reaction as follows:
2SO2(g)  +  O2(g) ⇌ 2SO3(g){\text{2S}}{{\text{O}}_{\text{2}}}{\text{(g)}}\,{\text{ + }}\,{{\text{O}}_{\text{2}}}{\text{(g)}}\, \rightleftharpoons \,{\text{2S}}{{\text{O}}_{\text{3}}}{\text{(g)}}
Kp =  [pSO3]2[pSO2]2[pO2]{{\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 0.500.50 atm for pO2{{\text{p}}_{{{\text{O}}_{\text{2}}}}}, 2.02.0 atm for pSO3{{\text{p}}_{{\text{S}}{{\text{O}}_{\text{3}}}}} , and 8×1024 8 \times {10^{24}}\,for Kp{{\text{K}}_{\text{p}}},
8×1024  =  [2.0]2[pSO2]2[0.5]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]}}
[pSO2]2 =  4.08×1024  ×0.5{\left[ {{{\text{p}}_{{\text{S}}{{\text{O}}_{\text{2}}}}}} \right]^2}\, = \,\,\dfrac{{4.0}}{{8 \times {{10}^{24}}\,\, \times 0.5}}
[pSO2]2 =  4.04.0×1024{\left[ {{{\text{p}}_{{\text{S}}{{\text{O}}_{\text{2}}}}}} \right]^2}\, = \,\,\dfrac{{4.0}}{{4.0 \times {{10}^{24}}}}
[pSO2]2 =  1.0×10−24{\left[ {{{\text{p}}_{{\text{S}}{{\text{O}}_{\text{2}}}}}} \right]^2}\, = \,\,1.0 \times {10^{ - 24}}
pSO2=  1.0×10−24{{\text{p}}_{{\text{S}}{{\text{O}}_{\text{2}}}}} = \,\,\sqrt {1.0 \times {{10}^{ - 24}}}
pSO2=  1.0×10−12{{\text{p}}_{{\text{S}}{{\text{O}}_{\text{2}}}}} = \,\,1.0 \times {10^{ - 12}}
So, the partial pressure of SO2{\text{S}}{{\text{O}}_{\text{2}}} is 1.0×10−121.0 \times {10^{ - 12}} atm.
Therefore, the equilibrium partial pressure of SO2(g) {\text{S}}{{\text{O}}_{\text{2}}}{\text{(g)}}\, in the mixture is 1.0×10−121.0 \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.