Problem 48 - Chemistry The Central Science

Question

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.

Step-by-Step Solution

Answer

In conclusion, for the given pairs: (a) Ethane (C2H6) has a higher standard molar entropy compared to ethyne (C2H2) due to its higher molecular complexity and number of vibrational modes. (b) Carbon dioxide (CO2) has a higher standard molar entropy compared to carbon monoxide (CO) due to its greater molecular complexity and number of vibrational modes.

Step 1: (a) Comparing standard molar entropy of C2H2(g) and C2H6(g)

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).

Step 2: (b) Comparing standard molar entropy of CO2(g) and CO(g)

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).

Key Concepts

Molecular Complexity

When discussing molecular complexity, what we mean is the structure and number of atoms that compose a molecule. Complex molecules typically consist of a larger number of atoms. This complexity plays a crucial role in determining the standard molar entropy of a substance in a gaseous state.

Molecules like ethane (\( \mathrm{C}_{2}\mathrm{H}_{6} \)) have more atoms (in this case, 8) and a non-linear structure. The non-linear nature of such molecules results in more ways for the molecule to arrange its atoms and carry out rotational and vibrational motions. This ultimately increases the molecule’s overall entropy.

• More atoms increase the potential arrangements of the molecule.
• Non-linear structures enhance chaos by allowing more movement possibilities.

Contrarily, simpler molecules such as ethyne (\( \mathrm{C}_{2}\mathrm{H}_{2} \)) with only 4 atoms and a linear structure, offer fewer ways to distribute energy. Thus, they tend to have a comparatively lower molar entropy.

Vibrational Modes

Vibrational modes refer to the different ways the atoms in a molecule can vibrate relative to each other. The number and type of these modes greatly affect a gas molecule’s entropy.

A molecule with more atoms and complex structure naturally has more vibrational modes. These modes are significant as they enable higher energy storage within the molecule, contributing to an increase in entropy values. For instance, ethane (\( \mathrm{C}_{2}\mathrm{H}_{6} \)), which is non-linear with 8 atoms, has more vibrational modes compared to ethyne (\( \mathrm{C}_{2}\mathrm{H}_{2} \)).

• Linear molecules have fewer vibrational modes compared to non-linear molecules.
• Vibrational modes allow for higher energy dispersion within the molecule.

As with carbon dioxide (\( \mathrm{CO}_{2} \)) versus carbon monoxide (\( \mathrm{CO} \)), CO2, with its additional atom and added vibrational modes, stores energy more dynamically, leading to higher entropy.

Gaseous Compounds

Gaseous compounds, unlike their liquid or solid counterparts, are significantly affected by their molecular arrangement in terms of entropy. The entropy of a gas is largely determined by the flexibility and freedom of its molecules to move and occupy various energy states.

• In gaseous ethane (\( \mathrm{C}_{2}\mathrm{H}_{6} \)), the molecules are dispersed and free to rotate and vibrate, allowing higher entropy.
• Conversely, gaseous ethyne (\( \mathrm{C}_{2}\mathrm{H}_{2} \)) is more constrained in movement, with linear chains limiting vibrational freedom.

Carbon dioxide (\( \mathrm{CO}_{2} \)) experiences a similar effect, having higher entropy due to the additional atom and its capability to engage in more extensive vibrational motions compared to carbon monoxide (\( \mathrm{CO} \)). This intrinsic property demonstrates why gaseous compounds with more complex and varied structures tend to exhibit higher standard molar entropy.