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 \(\mathrm{CO}(g) ?\)

In the first pair, we have two gaseous compounds: ethyne (\(\mathrm{C}_{2} \mathrm{H}_{2}(g)\)) and ethane (\(\mathrm{C}_{2} \mathrm{H}_{6}(g)\)). To determine which one has a higher standard molar entropy, we consider the molecular complexity and the number of atoms in each molecule. Ethyne has 4 atoms in total (2 carbon atoms and 2 hydrogen atoms) and has a triple bond between the carbon atoms while ethane has 8 atoms in total (2 carbon atoms and 6 hydrogen atoms) and a single bond between the carbon atoms. Molecules with more atoms have more ways of storing energy within their bonds and a higher degree of freedom, leading to greater entropy. Therefore, ethane (\(\mathrm{C}_{2} \mathrm{H}_{6}(g)\)) has a higher standard molar entropy.

In the second pair, we have two gaseous compounds: carbon dioxide (\(\mathrm{CO}_{2}(g)\)) and carbon monoxide (\(\mathrm{CO}(g)\)). In order to determine which one has a higher standard molar entropy, we consider the molecular complexity and the number of atoms in each molecule. Carbon dioxide has 3 atoms (1 carbon atom and 2 oxygen atoms) while carbon monoxide has 2 atoms (1 carbon atom and 1 oxygen atom). Molecules with more atoms have more ways to store energy and a higher degree of freedom, contributing to higher entropy. Thus, carbon dioxide (\(\mathrm{CO}_{2}(g)\)) has a higher standard molar entropy than carbon monoxide (\(\mathrm{CO}(g)\)).