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Chapter 17: Problem 47

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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Most popular questions from this chapter

Which of the following reactions (or processes) are expected to have a negative value for \(\Delta S^{\circ} ?\) a. $\mathrm{SiF}_{6}(a q)+\mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{HF}(g)+\mathrm{SiF}_{4}(g)$ b. $4 \mathrm{Al}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{Al}_{2} \mathrm{O}_{3}(s)$ c. \(\mathrm{CO}(g)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{COCl}_{2}(g)\) d. $\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)$ e. \(\mathrm{H}_{2} \mathrm{O}(s) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)\)

Given the values of \(\Delta H\) and \(\Delta S,\) which of the following changes will be spontaneous at constant \(T\) and \(P ?\) a. $\Delta H=+25 \mathrm{kJ}, \Delta S=+5.0 \mathrm{J} / \mathrm{K}, T=300 . \mathrm{K}$ b. $\Delta H=+25 \mathrm{kJ}, \Delta S=+100 . \mathrm{J} / \mathrm{K}, T=300 . \mathrm{K}$ c. $\Delta H=-10 . \mathrm{kJ}, \Delta S=+5.0 \mathrm{J} / \mathrm{K}, T=298 \mathrm{K}$ d. $\Delta H=-10 . \mathrm{kJ}, \Delta S=-40 . \mathrm{J} / \mathrm{K}, T=200 . \mathrm{K}$

You have a 1.00 -L sample of hot water \(\left(90.0^{\circ} \mathrm{C}\right)\) sitting open in a \(25.0^{\circ} \mathrm{C}\) room. Eventually the water cools to \(25.0^{\circ} \mathrm{C}\) while the temperature of the room remains unchanged. Calculate \(\Delta S_{\mathrm{sur}}\) for this process. Assume the density of water is 1.00 $\mathrm{g} / \mathrm{cm}^{3}$ over this temperature range, and the heat capacity of water is constant over this temperature range and equal to 75.4 $\mathrm{J} / \mathrm{K} \cdot$ mol.

When most biological enzymes are heated, they lose their catalytic activity. This process is called denaturing. The change original enzyme \(\rightarrow\) new form that occurs on heating is endothermic and spontaneous. Is the structure of the original enzyme or its new form more ordered (has the smaller positional probability)? Explain.

In the text, the equation $$\Delta G=\Delta G^{\circ}+R T \ln (Q)$$ was derived for gaseous reactions where the quantities in \(Q\) were expressed in units of pressure. We also can use units of mol/L for the quantities in \(Q\)specifically for aqueous reactions. With this in mind, consider the reaction $$\mathrm{HF}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{F}^{-}(a q)$$ for which \(K_{\mathrm{a}}=7.2 \times 10^{-4}\) at \(25^{\circ} \mathrm{C}\) . Calculate \(\Delta G\) for the reaction under the following conditions at \(25^{\circ} \mathrm{C} .\) a. $[\mathrm{HF}]=\left[\mathrm{H}^{+}\right]=\left[\mathrm{F}^{-}\right]=1.0 \mathrm{M}$ b. $[\mathrm{HF}]=0.98 M,\left[\mathrm{H}^{+}\right]=\left[\mathrm{F}^{-}\right]=2.7 \times 10^{-2} M$ c. $[\mathrm{HF}]=\left[\mathrm{H}^{+}\right]=\left[\mathrm{F}^{-}\right]=1.0 \times 10^{-5} \mathrm{M}$ d. $[\mathrm{HF}]=\left[\mathrm{F}^{-}\right]=0.27 M,\left[\mathrm{H}^{+}\right]=7.2 \times 10^{-4} M$ e. $[\mathrm{HF}]=0.52 M,\left[\mathrm{F}^{-}\right]=0.67 M,\left[\mathrm{H}^{+}\right]=1.0 \times 10^{-3} \mathrm{M}$ Based on the calculated DG values, in what direction will the reaction shift to reach equilibrium for each of the five sets of conditions?
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