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Foreach of the following pairs, indicate which substance possesses the larger standard entropy: (a) 1 mol of \(\mathrm{P}_{4}(g)\) at \(300{ }^{\circ} \mathrm{C}, 0.01 \mathrm{~atm}\), or \(1 \mathrm{~mol}\) of \(\mathrm{As}_{4}(g)\) at \(300^{\circ} \mathrm{C}, 0.01 \mathrm{~atm} ;\) (b) \(1 \mathrm{~mol}\) of \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) at \(100^{\circ} \mathrm{C}, 1 \mathrm{~atm}\), or \(1 \mathrm{~mol}\) of \(\mathrm{H}_{2} \mathrm{O}(l)\) at \(100^{\circ} \mathrm{C}, 1 \mathrm{~atm} ;\) (c) \(0.5 \mathrm{~mol}\) of \(\mathrm{N}_{2}(\mathrm{~g})\) at \(298 \mathrm{~K}, 20\) - \(\mathrm{L}\) volume, or \(0.5 \mathrm{~mol} \mathrm{CH}_{4}(g)\) at \(298 \mathrm{~K}, 20\) -L volume; (d) \(100 \mathrm{~g}\), \(\mathrm{Na}_{2} \mathrm{SO}_{4}(s)\) at \(30^{\circ} \mathrm{C}\) or \(100 \mathrm{~g} \mathrm{Na}_{2} \mathrm{SO}_{4}(a q)\) at \(30^{\circ} \mathrm{C}\).

Short Answer

Expert verified
For each pair, the substance with a larger standard entropy is: a) As4(g) b) H2O(g) c) CH4(g) d) Na2SO4(aq)

Step by step solution

01

(Pair A: P4(g) vs As4(g))

For this pair, we have 1 mol of P4(g) and 1 mol of As4(g) at the same temperature and pressure. Since both substances are in the gaseous state and have the same number of moles and conditions, we can compare their entropies by considering their atomic mass. As has a larger atomic mass than P, so As4(g) has a larger entropy.
02

(Pair B: H2O(g) vs H2O(l))

For this comparison, we have 1 mol of H2O in both the gaseous and liquid states, but at the same temperature and pressure. In general, gases have more entropy than liquids because the particles have more freedom to move in a gas than in a liquid. Therefore, H2O(g) has a larger standard entropy than H2O(l).
03

(Pair C: N2(g) vs CH4(g))

In this case, we need to compare entropy for 0.5 mol of N2(g) and 0.5 mol of CH4(g) at the same temperature and volume. N2(g) is a simple diatomic molecule and CH4(g) is a more complex tetrahedral molecule. More complex molecules usually have larger entropies because they have more possible arrangements and motions. Therefore, CH4(g) has a larger standard entropy than N2(g).
04

(Pair D: Na2SO4(s) vs Na2SO4(aq))

For this pair, we are comparing between 100 g of Na2SO4(s) at 30°C and 100 g of Na2SO4(aq) at 30°C. The difference between the two is their state: solid versus aqueous. An aqueous solution will have more entropy than a solid because solvation of ions in the aqueous solution increases the number of particles and their freedom to move. So, Na2SO4(aq) has a larger standard entropy than Na2SO4(s). In conclusion, for each pair, we have: a) As4(g) has a larger standard entropy than P4(g). b) H2O(g) has a larger standard entropy than H2O(l). c) CH4(g) has a larger standard entropy than N2(g). d) Na2SO4(aq) has a larger standard entropy than Na2SO4(s).

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Key Concepts

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

Entropy Comparison
Entropy, a measure of the disorder or randomness in a system, is a fundamental concept in thermodynamics. Entropy comparison involves evaluating which of two or more systems or states has higher randomness or potential for spontaneous change.

In the exercise, we compared the standard entropy of various substances under identical conditions to determine which has the greater entropy. Substance state, molecular weight, complexity, and phase change all play vital roles in determining entropy, just as seen in the provided solutions where gaseous and more complex molecules generally exhibited higher entropies than their counterparts.
Gases vs Liquids
The state of matter has profound effects on entropy. Gases have significantly higher standard entropy than liquids because of their greater freedom of movement. Molecules in a gas are much farther apart than those in a liquid and can move more independently, resulting in a higher number of microstates, hence a higher entropy.

This concept is reinforced in the textbook exercise where H2O(g) exhibits a higher entropy than H2O(l) when both are at the same temperature and pressure because the gas molecules are more dispersed and disorderly.
Molecular Complexity and Entropy
Molecular complexity refers to the number of atoms and the variety of ways they can be arranged within a molecule. A more complex molecule, like methane (CH4), has more possible arrangements (microstates) than a simpler diatomic molecule like nitrogen (N2).

Increased complexity leads to a higher number of vibrational, rotational, and translational motions, all of which contribute to a higher standard entropy. The example with CH4(g) having higher entropy than N2(g), despite both being gases at identical conditions, perfectly illustrates the impact of molecular complexity on entropy.
Enthalpy of Aqueous Solutions
Enthalpy changes are associated with the making and breaking of bonds during the dissolution process, and they can influence entropy. The process of solvation in forming an aqueous solution can increase the degree of disorder, as ions or molecules become surrounded by solvent molecules, thereby increasing the system’s entropy.

In our textbook problem, Na2SO4(aq) has a larger standard entropy than Na2SO4(s), as the solvated ions in the aqueous solution have more freedom to move relative to their positions in the solid crystal lattice, emphasizing how solvation affects the entropy of substances.

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Most popular questions from this chapter

A particular reaction is spontaneous at \(450 \mathrm{~K}\). The enthalpy change for the reaction is \(+34.5 \mathrm{~kJ}\). What can you conclude about the sign and magnitude of \(\Delta S\) for the reaction?

(a) State the third law of thermodynamics. (b) Distinguish between translational motion, vibrational motion, and rotational motion of a molecule. (c) Illustrate these three kinds of motion with sketches for the HCl molecule.

For a particular reaction, \(\Delta H=-32 \mathrm{~kJ}\) and \(\Delta S=\) \(-98 \mathrm{~J} / \mathrm{K}\). Assume that \(\Delta H\) and \(\Delta S\) do not vary with temperature. (a) At what temperature will the reaction have \(\Delta G=0 ?\) (b) If \(T\) is increased from that in part (a), will the reaction be spontaneous or nonspontaneous?

Predict the sign of the entropy change of the system for each of the following reactions: (a) \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g)\) (b) \(\mathrm{Ba}(\mathrm{OH})_{2}(s) \stackrel{\mathrm{L}}{\longrightarrow} \mathrm{BaO}(s)+\mathrm{H}_{2} \mathrm{O}(g)\) (c) \(\mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(l)\) (d) \(\mathrm{FeCl}_{2}(s)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{Fe}(s)+2 \mathrm{HCl}(g)\)

Consider the polymerization of ethylene to polyethylene. cos (Section 12.6) (a) What would you predict for the sign of the entropy change during polymerization ( \(\Delta S_{\text {poly }}\) )? Explain your reasoning, (b) The polymerization of ethylene is a spontaneous process at room temperature. What can you conclude about the enthalpy change during polymerization \(\left(\Delta H_{\text {poly }}\right) ?(\mathrm{c})\) Use average bond enthalpies (Table 8.4) to estimate the value of \(\Delta H_{\text {poly }}\) per ethylene monomer added. (d) Polyethylene is an addition polymer. By comparison, Nylon 66 is a condensation polymer. How would you expect \(\Delta S_{\text {poly }}\) for a condensation polymer to compare to that for an addition polymer? Explain.

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