In each of the following pairs, which compound would you expect to have the higher standard molar entropy: (a) \(\mathrm{C}_{2} \mathrm{H}_{2}(g)\) or \(\mathrm{C}_{2} \mathrm{H}_{6}(g),(\mathbf{b}) \mathrm{CO}_{2}(g)\) or \(\mathrm{CO}(g) ?\) Explain.

In this pair, we have two gaseous hydrocarbons – ethyne (C2H2) and ethane (C2H6). The first thing to consider is the molecular complexity and number of atoms in each molecule. Ethyne has 4 atoms and is a linear molecule, while ethane has 8 atoms and is non-linear. A molecule with more atoms generally has more vibrational modes, which can contribute to more randomness and, consequently, a higher entropy value. Moreover, ethane (C2H6) is non-linear, which adds even more vibrational modes compared to the linear ethyne (C2H2). Therefore, we would expect ethane (C2H6) to have higher standard molar entropy than ethyne (C2H2).

In this pair, we have two gaseous carbon oxides – carbon dioxide (CO2) and carbon monoxide (CO). Carbon dioxide has 3 atoms in its molecule and forms a linear molecule, while carbon monoxide only has 2 atoms forming a linear molecule. In this case, the difference in the number of atoms results in a difference in their molecular complexity and the number of vibrational modes. Carbon dioxide (CO2), having 3 atoms, will have more vibrational modes than carbon monoxide (CO), which only has 2 atoms. This difference in vibrational modes will lead to a higher standard molar entropy for carbon dioxide (CO2) compared to carbon monoxide (CO). In conclusion, for the given pairs: (a) Ethane (C2H6) has a higher standard molar entropy compared to ethyne (C2H2). (b) Carbon dioxide (CO2) has a higher standard molar entropy compared to carbon monoxide (CO).