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

Short Answer

Expert verified
In both pairs, the compounds with more atoms have a higher standard molar entropy due to increased molecular complexity and degrees of freedom. Thus, ethane (\(\mathrm{C}_{2} \mathrm{H}_{6}(g)\)) has a higher standard molar entropy than ethyne (\(\mathrm{C}_{2} \mathrm{H}_{2}(g)\)), and carbon dioxide (\(\mathrm{CO}_{2}(g)\)) has a higher standard molar entropy than carbon monoxide (\(\mathrm{CO}(g)\)).

Step by step solution

01

(a) \(\mathrm{C}_{2} \mathrm{H}_{2}(g)\) or \(\mathrm{C}_{2} \mathrm{H}_{6}(g)\)

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.
02

(b) \(\mathrm{CO}_{2}(g)\) or \(\mathrm{CO}(g)\)

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

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molecular Complexity
Molecular complexity plays a crucial role in determining the standard molar entropy of a compound. The complexity of a molecule refers to the number of different atoms bonded together and the types of bonds that exist between these atoms. The greater the number of different atoms in a molecule, the more complex it is considered to be.
In our example with ethyne (\(\text{C}_2\text{H}_2\)(g)) and ethane (\(\text{C}_2\text{H}_6\)(g)), ethane is more molecularly complex because it has more atoms - two carbons and six hydrogens compared to two carbons and two hydrogens in ethyne. Additionally, ethane has only single bonds allowing for more variability in internal motion compared to the triple bond present in ethyne. This higher complexity results in a greater number of potential molecular arrangements and therefore, a higher standard molar entropy for ethane.
Understanding molecular complexity helps explain why molecules with a greater array of elements and multi-type bonds can have higher entropies.
Degree of Freedom
The degree of freedom in a molecule refers to the number of different types of motions or ways in which a molecule can move. These include translations (movement from one place to another), rotations (spinning around an axis), and vibrations (oscillations of atoms within the molecule). More complex molecules with more atoms tend to have more degrees of freedom, which can contribute to higher entropy.
In the example of carbon dioxide (\(\text{CO}_2\)(g)) and carbon monoxide (\(\text{CO}\)(g)), carbon dioxide has three atoms, offering more vibrational modes compared to the two atoms in carbon monoxide. Each additional atom in a molecule adds potential ways for the molecule to vibrate and rotate, effectively increasing its degree of freedom.
Thus, a molecule with more degrees of freedom has more possible microstates, leading to a higher standard molar entropy. By increasing the number of atoms or by having different types of bonds, molecules expand their degrees of freedom, maximizing the potential energy distribution throughout the molecule.
Entropy Comparison
When comparing the standard molar entropy of different compounds, we focus on two main factors: molecular complexity and the degree of freedom. Entropy is essentially a measure of disorder or the number of possible ways energy can be distributed within a system. Hence, the comparison between molecules often hinges on how complex these molecules are and the variety of motion they allow.
With ethyne and ethane, for example, the presence of more atoms in ethane gives rise to higher entropy. Similarly, comparing carbon dioxide and carbon monoxide shows that CO₂, with more atoms, has a higher entropy due to increased molecular motion possibilities.
To effectively compare entropies, remember that:
  • Molecules with more atoms generally have higher entropies.
  • The nature of bonds (single, double, triple) plays a role in the molecule's complexity and entropy.
  • More complex molecules with varied atomic arrangements increase entropy due to greater freedom of motion.
Understanding these principles helps in predicting which molecules in a set will possess higher standard molar entropies.

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Most popular questions from this chapter

Consider what happens when a sample of the explosive TNT is detonated under atmospheric pressure. (a) Is the detonation a reversible process? (b) What is the sign of \(q\) for this process? (c) Is \(w\) positive, negative, or zero for the process?

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As shown here, one type of computer keyboard cleaner contains liquefied \(1,1\) -difluoroethane \(\left(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{F}_{2}\right),\) which is a gas at atmospheric pressure. When the nozzle is squeezed, the \(1,1\) -difluoroethane vaporizes out of the nozzle at high pressure, blowing dust out of objects. (a) Based on your experience, is the vaporization a spontaneous process at room temperature? (b) Defining the \(1,1\) -difluoroethane as the system, do you expect \(q_{s y s}\) for the process to be positive or negative? (c) Predict whether \DeltaS is positive or negative for this process. (d) Given your answers to (a), (b), and (c), do you think the operation of this product depends more on enthalpy or entropy? [Sections 19.1 and 19.2\(]\)

Indicate whether \(\Delta G\) increases, decreases, or does not change when the partial pressure of \(\mathrm{H}_{2}\) is increased in each of the following reactions: $$ \begin{array}{l}{\text { (a) } \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)} \\ {\text { (b) } 2 \mathrm{HBr}(g) \longrightarrow \mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g)} \\ {\text { (c) } 2 \mathrm{H}_{2}(g)+\mathrm{C}_{2} \mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)}\end{array} $$

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