The standard free energy change of formation of ethane is $-32.9 \mathrm{kJ} {\mathrm{mol}}^{-1}.$ The standard entropy change of the system for this reaction is $-173.7 \mathrm{J} {\mathrm{K}}^{-1} {\mathrm{mol}}^{-1}.$

Explain why the reaction can be feasible, even though the value of the standard entropy change of the system is negative.

Ethane undergoes combustion to form carbon dioxide and water.

${\mathrm{C}}_2{\mathrm{H}}_6\left(\mathrm{g}\right)+\frac{7}{2}{\mathrm{O}}_2\left(\mathrm{g}\right)\to 2{\mathrm{CO}}_2\left(\mathrm{g}\right)+3{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

Calculate the free energy change of combustion for this reaction.

$$\begin{gathered} \mathrm{Values} \mathrm{of} ∆{\mathrm{G}}_{\mathrm{f}}^{⊝} \mathrm{in} \mathrm{kJ} {\mathrm{mol}}^{-1}: \\ {\mathrm{C}}_2{\mathrm{H}}_6\left(\mathrm{g}\right)=-32.9, {\mathrm{CO}}_2\left(\mathrm{g}\right)=-394.4, {\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)=-237.2. \end{gathered}$$

Calculate the standard entropy change of the system in the given below reaction using the standard molar entropy values given here. (Values for ${\mathrm{S}}^{⊝}\mathrm{in} \mathrm{J} {\mathrm{K}}^{-1} {\mathrm{mol}}^{-1})$.

Given:

$$\begin{gathered} {\mathrm{H}}_2{\mathrm{O}}_2\left(\mathrm{l}\right)=109.6 \\ {\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)=69.90 \\ {\mathrm{O}}_2\left(\mathrm{g}\right)=205.0 \end{gathered}$$

$2{\mathrm{H}}_2{\mathrm{O}}_2\left(\mathrm{l}\right)\to 2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)$