The table shows the formula, state, and standard molar entropies of the first five straight-chain alkanes.

Alkane	${\mathrm{CH}}_4\left(\mathrm{g}\right)$	${\mathrm{C}}_2{\mathrm{H}}_6\left(\mathrm{g}\right)$	${\mathrm{C}}_3{\mathrm{H}}_8\left(\mathrm{g}\right)$	${\mathrm{C}}_4{\mathrm{H}}_{10}\left(\mathrm{g}\right)$	${\mathrm{C}}_5{\mathrm{H}}_{12}\left(\mathrm{l}\right)$
$\frac{∆{\mathrm{S}}^{⊝}}{\mathrm{J} {\mathrm{K}}^{-1} {\mathrm{mol}}^{-1}}$	$186.2$	$229.5$	$269.9$	$310.1$	$261.2$

Estimate the value of the standard molar entropy of the straight-chain alkane with the formula.

 

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 $-173.7 \mathrm{J} {\mathrm{K}}^{-1} {\mathrm{mol}}^{-1}.$.

State the relationship between standard free energy change of formation, standard entropy change of the system, and enthalpy change.

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.

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