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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 \(\mathrm{C}_{2} \mathrm{H}_{6}(\mathrm{~g})\), (b) \(\mathrm{CO}_{2}(g)\) or \(\mathrm{CO}(g)\) ?

Short Answer

Expert verified
In each pair, the compound with a higher molar mass and more complex molecular structure is expected to have the higher standard molar entropy. Thus, for (a) C2H6(g) and for (b) CO2(g) would have higher standard molar entropies in comparison to their respective counterparts.

Step by step solution

01

Comparing C2H2(g) and C2H6(g)

First, let's analyze the two compounds: - C2H2(g): this is an ethylene molecule. It has a molar mass of approximately 26.04 g/mol. - C2H6(g): this is an ethane molecule. It has a molar mass of approximately 30.07 g/mol. Since C2H6(g) has a greater molar mass and more atoms in its structure, we would generally expect it to have a higher standard molar entropy than C2H2(g).
02

Comparing CO2(g) and CO(g)

Next, let's examine the comparison between CO2(g) and CO(g): - CO2(g): this is a carbon dioxide molecule. It has a molar mass of approximately 44.01 g/mol. - CO(g): this is a carbon monoxide molecule. It has a molar mass of approximately 28.01 g/mol. In this case, CO2(g) not only has a greater molar mass, but it also has a more complex linear molecular structure. Thus, it is expected that CO2(g) would have a higher standard molar entropy than CO(g). To summarize our findings: - Between C2H2(g) and C2H6(g), C2H6(g) is expected to have the higher standard molar entropy. - Between CO2(g) and CO(g), CO2(g) is expected to have the higher standard molar entropy.

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Key Concepts

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

Molar Mass
Molar mass is crucial in determining the standard molar entropy of substances. It represents the mass of one mole of a compound and is typically expressed in grams per mole (g/mol).

In our analysis, we compare two pairs of compounds based on molar mass. For instance, ethane (\(\mathrm{C}_2\mathrm{H}_6\)) has a higher molar mass (approximately 30.07 g/mol) compared to ethylene (\(\mathrm{C}_2\mathrm{H}_2\)) which is around 26.04 g/mol. This higher molar mass generally indicates a greater number of possible microstates that contribute to higher standard molar entropy.

Similarly, carbon dioxide (\(\mathrm{CO}_2\)) with a molar mass of about 44.01 g/mol surpasses carbon monoxide (\(\mathrm{CO}\)) at 28.01 g/mol. Higher molar mass in these instances tends to correlate with higher standard molar entropy due to increased molecular complexity.
Structure Complexity
The complexity of a molecule's structure significantly influences its standard molar entropy. More complex structures often have more positional and energetic arrangements available.

Take ethane \(\mathrm{C}_2\mathrm{H}_6\) as an example. It is a larger molecule than ethylene \(\mathrm{C}_2\mathrm{H}_2\). This complexity allows ethane to have more vibrational and rotational movements, contributing to a higher standard molar entropy.

In the comparison of carbon dioxide \(\mathrm{CO}_2\) and carbon monoxide \(\mathrm{CO}\), \(\mathrm{CO}_2\) is also more complex. Its linear structure enables additional flexibility in atomic motions, enhancing its ability to occupy a greater number of microstates. Thus, \(\mathrm{CO}_2\) is expected to exhibit higher standard molar entropy.
Comparative Entropy Analysis
When analyzing the standard molar entropy between pairs of molecules, a few key factors must be considered:
  • Molar Mass
  • Structural Complexity
In our examples, for both pairs – \(\mathrm{C}_2\mathrm{H}_2\) & \(\mathrm{C}_2\mathrm{H}_6\) and \(\mathrm{CO}_2\) & \(\mathrm{CO}\) – the molecule with the higher molar mass and more complex structure has a greater standard molar entropy.

This is fundamental in thermodynamic studies as it highlights that molecules with greater mass and intricate structures can disperse energy more efficiently.
  • \(\mathrm{C}_2\mathrm{H}_6\) has more complex structure than \(\mathrm{C}_2\mathrm{H}_2\), yielding higher entropy.
  • \(\mathrm{CO}_2\) is more complex than \(\mathrm{CO}\) and thus has a higher entropy.
Understanding these patterns helps in predicting material behavior in chemical processes, ensuring that entropy changes align with theoretical expectations.

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Most popular questions from this chapter

(a) Does the entropy of the surroundings increase for spontancous processes? (b) In a particular spontaneous process the entropy of the system decreases. What can you conclude about the sign and magnitude of \(\Delta S_{\text {gar }}\) ? (c) During a certain reversible process, the surroundings undergo an entropy change, \(\Delta S_{\text {sury }}=-78 \mathrm{~J} / \mathrm{K}\). What is the entropy change of the system for this process?

Consider a system consisting of an ice cube. (a) Under what conditions can the ice cube melt reversibly? (b) If the ice cube melts reversibly, is \(\Delta E\) zero for the process?

Would each of the following changes increase, decrease, or have no effect on the number of microstates available to a system: (a) increase in temperature, (b) decrease in volume, (c) change of state from liquid to gas?

Using data from Appendix \(C\), write the equilibrium-constant expression and calculate the value of the equilibrium constant and the free-energy change for these reactions at \(298 \mathrm{~K}\) : (a) \(\mathrm{NaHCO}_{3}(s) \rightleftharpoons \mathrm{NaOH}(s)+\mathrm{CO}_{2}(g)\) (b) \(2 \mathrm{HBr}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{HCl}(g)+\mathrm{Br}_{2}(g)\) (c) \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)\)

Using the data in Appendix \(\mathrm{C}\) and given the pressures listed, calculate \(K_{\rho}\) and \(\Delta G\) for each of the following reactions: (a) \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\) \(R_{\mathrm{N}_{2}}=2.6 \mathrm{~atm}, P_{\mathrm{H}_{2}}=5.9 \mathrm{~atm}, P_{\mathrm{NH}_{5}}=1.2 \mathrm{~atm}\) (b) \(2 \mathrm{~N}_{2} \mathrm{H}_{4}(g)+2 \mathrm{NO}_{2}(g) \longrightarrow 3 \mathrm{~N}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g)\) \(P_{\mathrm{N}_{2} \mathrm{H}_{4}}=\mathrm{PNo}_{1}=5.0 \times 10^{-2} \mathrm{~atm}\). \(P_{\mathrm{N}_{2}}=0.5 \mathrm{~atm}, P_{\mathrm{H}_{2} O}=0.3 \mathrm{~atm}\) (c) \(\mathrm{N}_{2} \mathrm{H}_{4}(\mathrm{~g}) \longrightarrow \mathrm{N}_{2}(\mathrm{~g})+2 \mathrm{H}_{2}(\mathrm{~g})\) \(P_{\mathrm{N}_{2} \mathrm{H}_{4}}=0.5 \mathrm{~atm}, P_{\mathrm{N}_{2}}=1.5 \mathrm{~atm}, P_{\mathrm{H}_{1}}=2.5 \mathrm{~atm}\)

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