Question

Consider two reactions for the production of ethanol

$\begin{array}{l}{\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{4}\mathbf{\left(}\mathbf{g}\mathbf{\right)}}\mathbf{+}{\mathbf{H}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{\left(}\mathbf{g}\mathbf{\right)}}\mathbf{\to }\mathbf{C}{\mathbf{H}}_{\mathbf{3}}\mathbf{C}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}{\mathbf{H}}_{\mathbf{\left(}\mathbf{l}\mathbf{\right)}}\\ {\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{6}\mathbf{\left(}\mathbf{g}\mathbf{\right)}}\mathbf{+}{\mathbf{H}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{\left(}\mathbf{g}\mathbf{\right)}}\mathbf{\to }\mathbf{C}{\mathbf{H}}_{\mathbf{3}}\mathbf{C}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}{\mathbf{H}}_{\mathbf{\left(}\mathbf{l}\mathbf{\right)}}\mathbf{+}{\mathbf{H}}_{\mathbf{2}\mathbf{\left(}\mathbf{g}\mathbf{\right)}}\end{array}$

Which would be more thermodynamically feasible? Why? Assume standard conditions and assume that $\mathit{\Delta }{\mathit{H}}^{\mathbf{o}}$and $\mathit{\Delta }{\mathit{S}}^{\mathbf{o}}$ are temperature-independent.

Short answer

$C_2{H}_{4\left(g\right)}+H_2{O}_{\left(g\right)}\to C{H}_{3}C{H}_{2}O{H}_{\left(l\right)}$ would be more thermodynamically feasible.

Step-by-step solution

Thermodynamically feasible at C2H4(g)+H2O(g)→CH3CH2OH(l)

Between the standard entropy and Gibbs free energy

$\begin{array}{c}\Delta H_R=\Delta H_f\left(C{H}_{3}C{H}_{2}O{H}_{\left(l\right)}\right)-\left\{\Delta H_f(C_2{H}_{4\left(g\right)}+\Delta H_f(H_2{O}_{\left(g\right)})\right\}\\ =-278-(52-242)\\ =-88kJ\end{array}$

$\begin{array}{c}\Delta S_R=\Delta S^0\left(C{H}_{3}C{H}_{2}O{H}_{\left(l\right)}\right)-\left\{\Delta S^0(C_2{H}_{4\left(g\right)}+\Delta S^0(H_2{O}_{\left(g\right)})\right\}\\ =161-(219-189)\\ =-247kJ\end{array}$

Thermodynamically feasible at C2H4(g)+H2O(g)→CH3CH2OH(l)

According to Gibbs free energy

$\Delta G^0=0,\Delta H^0=T.\Delta S^0$

Then,

$\begin{array}{c}T=\frac{\Delta H^{0 }}{\Delta S^0}\\ =\frac{-88000J}{-247J/K}\\ =360K\end{array}$

So, the standard condition is $\Delta G=\Delta G^0$but here, $\Delta G^0<0$, Hence this reaction is spontaneous at $360K$of temperature.

Thermodynamically feasible at C2H6(g)+H2O(g)→CH3CH2OH(l)+H2(g)

Between the standard entropy and Gibbs free energy

$\begin{array}{c}\Delta H_R=\Delta H_f\left(C{H}_{3}C{H}_{2}O{H}_{\left(l\right)}\right)-\left\{\Delta H_f(C_2{H}_{6\left(g\right)}+\Delta H_f(H_2{O}_{\left(g\right)})\right\}\\ =-278-(-84.7-242)\\ =49kJ\end{array}$

$\begin{array}{c}\Delta S_R=\left\{\Delta S^0\left(C{H}_{3}C{H}_{2}O{H}_{\left(l\right)}\right)+\Delta S^0\left({H_2}_{\left(g\right)}\right)\right\}-\left\{\Delta S^0(C_2{H}_{6\left(g\right)}+\Delta S^0(H_2{O}_{\left(g\right)})\right\}\\ =161+131-(229.5-189)\\ =-127kJ\end{array}$

Thermodynamically feasible at C2H6(g)+H2O(g)→CH3CH2OH(l)+H2(g)

According to Gibbs free energy

$\Delta G^0=0,\Delta H^0=T.\Delta S^0$

So, The standard condition is$\Delta G=\Delta G^0$ but here, $\Delta G^0>0$, Hence this reaction is not spontaneous at $-127kJ$of temperature.

Conclusion

Therefore, $C_2{H}_{4\left(g\right)}+H_2{O}_{\left(g\right)}\to C{H}_{3}C{H}_{2}O{H}_{\left(l\right)}$would be more thermodynamically feasible.