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For each of the following pairs of substances, which substance has the greater value of \(S^{\circ} ?\) a. \(\mathrm{C}_{\text { graphite }}(s)\) or \(\mathrm{C}_{\text { diamond }}(s)\) b. \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)\) or \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(g)\) c. \(\mathrm{CO}_{2}(s)\) or \(\mathrm{CO}_{2}(g)\)

Short Answer

Expert verified
In summary, the substances with greater standard molar entropy (S°) are: a. C (graphite) b. \(C_2H_5OH\) (gas) c. \(CO_2\) (gas)

Step by step solution

01

Identify the substances and the realm they belong to

The three pairs of substances given are: a. C (graphite, solid) and C (diamond, solid) b. C2H5OH (ethanol, liquid) and C2H5OH (ethanol, gas) c. CO2 (carbon dioxide, solid) and CO2 (carbon dioxide, gas) These substances differ in their molecular structure or states of matter.
02

Compare the entropies of the substances in each pair

For each pair, we need to determine which substance has a higher entropy by considering the general rules of entropy: a. C (graphite, solid) or C (diamond, solid): Both substances are solids, so we need to consider their molecular structure. Graphite has a more disordered structure than diamond, so its entropy (S°) is greater. Answer: C (graphite) b. C2H5OH (ethanol, liquid) or C2H5OH (ethanol, gas): The only difference between these two substances is their physical state. Gases have higher entropy than liquids. Answer: C2H5OH (gas) c. CO2 (carbon dioxide, solid) or CO2 (carbon dioxide, gas): Again, the only difference between these two substances is their physical state. Gases have higher entropy than solids. Answer: CO2 (gas)
03

Summarize the results

In conclusion, the substances with greater values of the standard molar entropy (S°) for each pair are: a. C (graphite) b. C2H5OH (gas) c. CO2 (gas)

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Key Concepts

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

Graphite vs Diamond
Graphite and diamond are two different solid forms of carbon, known as allotropes. While they both consist entirely of carbon atoms, their difference lies in the arrangement of these atoms. In graphite, the carbon atoms form layers that can slide over each other, making it more disordered compared to diamond. Each carbon atom in graphite bonds with three others, creating sheets that are weakly held together. This structure is why pencil "lead," which is actually graphite, leaves a mark.
  • Graphite's structure is layered and allows for more atomic movements.
  • Diamond has a rigid three-dimensional lattice where each carbon atom is bonded to four other carbon atoms, creating a repeating pattern that extends throughout the solid. This strong bonding and rigid structure is why diamond is one of the hardest known substances.
  • Because graphite's atoms can move more freely, it has a higher standard molar entropy (S°) compared to diamond.
In summary, graphite has a more disordered structure compared to diamond, which translates to a greater entropy.
States of Matter
States of matter refer to the distinct forms that different phases of matter take on, primarily solid, liquid, and gas. These states are differentiated by the behavior and arrangement of particles.
For a given substance, moving from solid to gas increases the entropy, or the disorder within a system. Particles in a solid are tightly packed in a regular pattern, liquids are less orderly, and gases are highly disordered with particles moving freely.
  • Solids: Particles are fixed in place, have the least amount of space between them, and exhibit minimal movement.
  • Liquids: Particles are closer together but can slide past each other, allowing for more motion than solids.
  • Gases: Particles move independently of each other, filling available space completely, which results in the greatest degree of randomness.
Thus, when comparing different states of the same substance, gases typically exhibit higher entropy than both liquids and solids due to increased freedom of movement and positions available to the particles.
Standard Molar Entropy
Standard molar entropy \(S^{\circ}\) refers to the entropy content of one mole of a substance at a standard state. It is an absolute measure that considers the disorder and distribution of energy within a substance. \(S^{\circ}\) values are typically expressed in joules per mole per kelvin (J/mol·K).
Various factors affect the standard molar entropy of a substance, including:
  • Molecular complexity: Bigger or more complex molecules usually have higher entropy due to more ways energy can be distributed among atomic motions.
  • Physical state: As noted, gases have higher \(S^{\circ}\) compared to liquids and solids.
  • Atomic structure: Solids with more atomic freedom or less ordered structures, like graphite compared to diamond, have higher entropy.
Understanding \(S^{\circ}\) helps in predicting how a substance may behave under different conditions and is an invaluable tool in the study of thermodynamics and chemistry.

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Most popular questions from this chapter

Carbon monoxide is toxic because it bonds much more strongly to the iron in hemoglobin (Hgb) than does \(\mathrm{O}_{2} .\) Consider the following reactions and approximate standard free energy changes: $$\mathrm{Hgb}+\mathrm{O}_{2} \longrightarrow \mathrm{HgbO}_{2} \quad \Delta G^{\circ}=-70 \mathrm{kJ}$$ $$\mathrm{Hgb}+\mathrm{CO} \longrightarrow \mathrm{HgbCO} \quad \Delta G^{\circ}=-80 \mathrm{kJ} $$ Using these data, estimate the equilibrium constant value at \(25^{\circ} \mathrm{C}\) for the following reaction: $$\mathrm{HgbO}_{2}+\mathrm{CO} \rightleftharpoons \mathrm{HgbCO}+\mathrm{O}_{2}$$

Using the free energy profile for a simple one-step reaction, show that at equilibrium \(K=k_{f} / k_{\mathrm{r}},\) where \(k_{\mathrm{f}}\) and \(k_{\mathrm{r}}\) are the rate constants for the forward and reverse reactions. Hint: Use the relationship \(\Delta G^{\circ}=-R T \ln (K)\) and represent \(k_{\mathrm{f}}\) and \(k_{\mathrm{r}}\) using the Arrhenius equation \(\left(k=A e^{-E_{2} / R T}\right)\) b. Why is the following statement false? "A catalyst can increase the rate of a forward reaction but not the rate of the reverse reaction."

Calculate the entropy change for the vaporization of liquid methane and liquid hexane using the following data. $$\begin{array}{lll} {\text { }} & {\text { Boiling Point (1 atm)}} & { \Delta H_{\mathrm{vap}} } \\ \hline {\text { Methane }} & \quad\quad\quad {112 \mathrm{K}} & {8.20 \mathrm{kJ} / \mathrm{mol}} \\ {\text { Hexane }} & \quad\quad\quad {342 \mathrm{K}} & {28.9 \mathrm{kJ} / \mathrm{mol}}\end{array}$$ Compare the molar volume of gaseous methane at 112 \(\mathrm{K}\) with that of gaseous hexane at 342 \(\mathrm{K}\) . How do the differences in molar volume affect the values of \(\Delta S_{\mathrm{vap}}\) for these liquids?

Predict the sign of \(\Delta S^{\circ}\) and then calculate \(\Delta S^{\circ}\) for each of the following reactions. a. \(\mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)\) b. \(2 \mathrm{CH}_{3} \mathrm{OH}(g)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g)\) c. \(\mathrm{HCl}(g) \longrightarrow \mathrm{H}^{+}(a q)+\mathrm{Cl}^{-}(a q)\)

Some nonelectrolyte solute (molar mass \(=142 \mathrm{g} / \mathrm{mol} )\) was dissolved in \(150 . \mathrm{mL}\) of a solvent (density \(=0.879 \mathrm{g} / \mathrm{cm}^{3} )\) The elevated boiling point of the solution was 355.4 \(\mathrm{K}\) . What mass of solute was dissolved in the solvent? For the solvent, the enthalpy of vaporization is 33.90 \(\mathrm{kJ} / \mathrm{mol}\) , the entropy of vaporization is 95.95 \(\mathrm{J} / \mathrm{K} \cdot\) mol, and the boiling- point elevation constant is 2.5 \(\mathrm{K} \cdot \mathrm{kg} / \mathrm{mol}\) .

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