The enthalpy of formation steadily changes from \(–17.89\, \ k cal/mol\) to \(–49.82\, \ k cal/mol\) as we go from \(CH_4,\, \ C_2H_6\) to \(C_8H_{18}\). The value of \(\Delta G\), however, shows opposite trend from \(–12.12 kcal/mol\) for \(CH_4\) to \(4.14\, \ kcal/mol\) for \(C_8H_{18}\). Why?

In the formation of \(C_n H_{2n + 2}\) from n carbon atoms and \((n + 1)\) hydrogen molecules there is large decrease in entropy.

The correct answer is option 3. In the formation of \(C_n H_{2n + 2}\) from n carbon atoms and \((n + 1)\) hydrogen molecules there is large decrease in entropy.

Let us examine the given data and the potential reasons for the observed trends in the enthalpy of formation (\(\Delta H_f\)) and Gibbs free energy (\(\Delta G\)) for alkanes (\(CH_4\) to \(C_8H_{18}\)):

Enthalpy of Formation (\(\Delta H_f\)):

It steadily decreases (more negative) from \(-17.89\, \text{kcal/mol}\) for \(CH_4\) to \(-49.82\, \text{kcal/mol}\) for \(C_8H_{18}\).

Gibbs Free Energy (\(\Delta G\)):

It shows an opposite trend, becoming less negative and eventually positive from \(-12.12\, \text{kcal/mol}\) for \(CH_4\) to \(4.14\, \text{kcal/mol}\) for \(C_8H_{18}\).

Analysis of Potential Reasons

Increase in Isomers:

This reason suggests that as the number of carbon atoms increases, the number of possible isomers increases, affecting \(\Delta G\). However, the trend in \(\Delta G\) is due to intrinsic thermodynamic properties of the most stable isomer, not the number of isomers. Hence, this is not the primary reason for the observed trend.

Change in \(C–C\) and \(C–H\) Bonds:

As the chain length increases, the ratio of \(C–C\) bonds to \(C–H\) bonds increases. \(C–C\) bonds have different bond energies compared to \(C–H\) bonds, and this can influence both \(\Delta H\) and \(\Delta G\). However, this change alone doesn't fully explain the observed trend in \(\Delta G\).

Entropy Changes:

The formation of \(C_n H_{2n+2}\) from carbon atoms and hydrogen molecules involves a large decrease in entropy (\(\Delta S\)). For larger alkanes, the decrease in entropy becomes more significant. Since \(\Delta G = \Delta H - T\Delta S\), a larger negative \(\Delta S\) (large decrease in entropy) means \(T\Delta S\) becomes more negative, leading to a less negative or even positive \(\Delta G\). This is consistent with the observed trend: \(\Delta G\) becomes less negative and eventually positive as the number of carbon atoms increases.

No Simple Reason:

While this suggests complexity, it is important to identify the dominant factors. Entropy considerations provide a clear explanation for the observed trend.

Conclusion

The most significant reason for the trend in \(\Delta G\) values is the entropy change associated with the formation of larger alkanes. As the number of carbon atoms increases, the decrease in entropy (\(\Delta S\)) becomes more substantial. This larger decrease in entropy makes \(T\Delta S\) more negative, which in turn causes \(\Delta G\) to become less negative and eventually positive.

Therefore, the correct answer is: (3) In the formation of \(C_nH_{2n+2}\) from \(n\) carbon atoms and \((n+1)\) hydrogen molecules, there is a large decrease in entropy.