Question

Using data from Appendix 4, Calculate $\mathbf{∆}{\mathbf{G}}^{\mathbf{°}}$for the following reaction that produce acetic acid:

Which reaction would you choose as a commercial method for producing acetic acid $\mathbf{\left(}\mathit{C}{\mathit{H}}_{\mathbf{3}}\mathit{C}{\mathit{O}}_{\mathbf{2}}\mathit{H}\mathbf{\right)}$? What temperature conditions would you choose for the reaction? Assume standard conditions and assume that $\mathit{\Delta }{\mathit{H}}^{\mathbf{o}}\mathit{}\mathit{a}\mathit{n}\mathit{d}\mathit{}\mathit{\Delta }{\mathit{G}}^{\mathit{o}}$are temperature-independent.

Short answer

Both can be chosen for producing acetic acid commercially. But the second reaction can be preferable as the temperature below 622K will be spontaneous.

Step-by-step solution

 

Standard expression for

Standard free energy at room temperature

$$\begin{gathered} \Delta G^o=\Delta {G^o}_{298} \\ =\sum n{G^o}_{298}\left(Products\right)-\sum p{G^o}_{298}\left(Reactants\right) \end{gathered}$$

Standard free energy foris $-389kJ/\text{mol}$

Standard free energy for $C{H}_{4\left(g\right)}$ is $-51kJ/\text{mol}$

Standard free energy for$C{O}_{2\left(g\right)}$is $-394kJ/\text{mol}$

$$\begin{gathered} \Delta {G^o}_{298}=(1\times {G^o}_{298}(C{H}_{3}COO{K}_{\left(l\right)})-(1\times {G^o}_{298}\left(C{H_4}_{\left(g\right)}\right)+1\times {G^o}_{298}\left(C{O_2}_{\left(g\right)}\right) \\ =\left[(-389)-(-51-394)\right] \\ \Delta {G^o}_{298} =56kJ/mol \end{gathered}$$

Step 2: 

Standard expression for

Standard free energy at room temperature:

$$\begin{gathered} \Delta S^o=\Delta {S^o}_{298} \\ =\sum n{S^o}_{298}\left(Products\right)-\sum p{S^o}_{298}\left(Reactants\right) \end{gathered}$$

Standard free energy for

Standard free energy for $C{H}_{4\left(g\right)}$ is $-186kJ/\text{mol}$

Standard free energy for $C{O}_{2\left(g\right)}$is$-214kJ/\text{mol}$

$$\begin{gathered} \Delta {S^o}_{298}=(1\times {S^o}_{298}(C{H}_{3}COO{H}_{\left(l\right)})-(1\times {S^o}_{298}\left(C{H_4}_{\left(g\right)}\right)+1\times {S^o}_{298}\left(C{O_2}_{\left(g\right)}\right) \\ =\left[\left(160\right)-(186+214)\right] \\ \Delta {S^o}_{298} =240kJ/mol \end{gathered}$$

Step 3:

Standard expression for

Standard free energy at room temperature

$$\begin{gathered} \Delta H^o=\Delta {H^o}_{298} \\ =\sum n{H^o}_{298}\left(Products\right)-\sum p{H^o}_{298}\left(Reactants\right) \end{gathered}$$

Standard free energy for is $-484kJ/\text{mol}$

Standard free energy for $C{H}_{4\left(g\right)}$is $-75kJ/\text{mol}$

Standard free energy for $C{O}_{2\left(g\right)}$is $-393kJ/\text{mol}$

$$\begin{gathered} \Delta {H^o}_{298}=(1\times {H^o}_{298}(C{H}_{3}COO{H}_{\left(l\right)})-(1\times {H^o}_{298}\left(C{H_4}_{\left(g\right)}\right)+1\times {H^o}_{298}\left(C{O_2}_{\left(g\right)}\right) \\ =\left[(-484)-(-75+-393.5)\right] \\ \Delta {H^o}_{298} =-16kJ/mol \end{gathered}$$

Step 4:

Standard expression for

Standard free energy at room temperature

$$\begin{gathered} \Delta G^o=\Delta {G^o}_{298} \\ =\sum n{G^o}_{298}\left(Products\right)-\sum p{G^o}_{298}\left(Reactants\right) \end{gathered}$$

Standard free energy for is localid="1648702762329" $-389kJ/\text{mol}$

Standard free energy for $C{H}_{4\left(g\right)}$ is localid="1648702783154" $-163kJ/\text{mol}$

Standard free energy for $C{O}_{2\left(g\right)}$is localid="1648702801410" $-137kJ/\text{mol}$

$$\begin{gathered} \Delta {G^o}_{298}=(1\times {G^o}_{298}(C{H}_{3}COO{H}_{\left(l\right)})-(1\times {G^o}_{298}\left(C{H}_{3}O{H}_{\left(g\right)}\right)+1\times {G^o}_{298}\left(C{O_{}}_{\left(g\right)}\right) \\ =\left[(-389)-(-163+-137)\right] \\ \Delta {G^o}_{298} =-89kJ/mol \end{gathered}$$

Step 5:

Standard expression for

Standard free energy at room temperature

$$\begin{gathered} \Delta H^o=\Delta {H^o}_{298} \\ =\sum n{H^o}_{298}\left(Products\right)-\sum p{H^o}_{298}\left(Reactants\right) \end{gathered}$$

Standard free energy for is localid="1648702853879" $160kJ/\text{mol}$

Standard free energy for $C{H}_{4\left(g\right)}$ is localid="1648702863224" $240kJ/\text{mol}$

Standard free energy for $C{O}_{2\left(g\right)}$is localid="1648702873279" $198kJ/\text{mol}$

$$\begin{gathered} \Delta {S^o}_{298}=(1\times {S^o}_{298}(C{H}_{3}COO{H}_{\left(l\right)})-(1\times {S^o}_{298}\left(C{H}_{3}O{H}_{\left(g\right)}\right)+1\times {G^o}_{298}\left(C{O}_{\left(g\right)}\right) \\ =\left[(-389)-(-163-137)\right] \\ \Delta {S^o}_{298} =-278kJ/mol \end{gathered}$$

Step 6:

Standard expression for

Standard free energy at room temperature

$$\begin{gathered} \Delta S^o=\Delta {S^o}_{298} \\ =\sum n{S^o}_{298}\left(Products\right)-\sum p{S^o}_{298}\left(Reactants\right) \end{gathered}$$

Standard free energy foris $-484kJ/\text{mol}$

Standard free energy for $C{H}_{4\left(g\right)}$is localid="1648702959291" $-201kJ/\text{mol}$

Standard free energy for $C{O}_{2\left(g\right)}$is localid="1648702980328" $-110.5kJ/\text{mol}$

$$\begin{gathered} \Delta {H^o}_{298}=(1\times {H^o}_{298}(C{H}_{3}COO{H}_{\left(l\right)})-(1\times {H^o}_{298}\left(C{H}_{3}O{H_4}_{\left(g\right)}\right)+1\times {H^o}_{298}\left(C{O_2}_{\left(g\right)}\right) \\ =\left[(-484)-(-201+(-110.5\left)\right)\right] \\ \Delta {H^o}_{298} =-173kJ/mol \end{gathered}$$

Conclusion

Therefore, $\Delta {G°}_{298} =56kJ/\text{mol}$, data-custom-editor="chemistry" $\Delta {S°}_{298} =240kJ/\text{mol}$, anddata-custom-editor="chemistry" $\Delta {H°}_{298} =-16kJ/\text{mol}$

Here both can be chosen for producing acetic acid commercially. But the second reaction can be preferable as the temperature below 622K will be spontaneous.