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Chapter 19: Problem 45

In each of the following pairs, which compound would you expect to have the higher standard molar entropy: (a) \(\mathrm{C}_{3} \mathrm{H}_{\mathrm{s}}(g)\) or $\mathrm{C}_{4} \mathrm{H}_{10}(g)$, (b) \(\mathrm{C}_{4} \mathrm{H}_{10}(l)\) or \(\mathrm{C}_{4} \mathrm{H}_{10}(g)\)

Short Answer

Expert verified
In conclusion: - For pair (a): Butane, \(\mathrm{C}_{4} \mathrm{H}_{10}(g)\), has a higher standard molar entropy due to its higher molecular complexity compared to propane, \(\mathrm{C}_{3} \mathrm{H}_{8}(g)\). - For pair (b): The gaseous phase of butane, \(\mathrm{C}_{4} \mathrm{H}_{10}(g)\), has a higher standard molar entropy compared to the liquid phase, \(\mathrm{C}_{4} \mathrm{H}_{10}(l)\), due to its increased freedom of movement and spacing between particles.

Step by step solution

01

Identify the compounds

We have two compounds: \(\mathrm{C}_{3} \mathrm{H}_{8}(g)\), which is propane, and \(\mathrm{C}_{4} \mathrm{H}_{10}(g)\), which is butane.
02

Analyze the phase and molecular complexity

Both are in the gaseous phase, so there is no difference in entropy due to phase. Molecular complexity is higher in butane since it has one extra carbon atom and two extra hydrogen atoms compared to propane.
03

Determine the higher entropy compound

Since both compounds are in the same phase, the only factor determining their entropy is the molecular complexity. In this case, butane (\(\mathrm{C}_{4} \mathrm{H}_{10}(g)\)) has a higher molecular complexity, resulting in a higher standard molar entropy. For pair (b):
04

Identify the compounds and phases

We have the same compound (\(\mathrm{C}_{4} \mathrm{H}_{10}\)) in two different phases: liquid and gas.
05

Analyze the influence of phase on entropy

A gaseous phase compound has more freedom of movement and spacing between particles, which leads to a higher entropy compared to a liquid phase compound.
06

Determine the higher entropy compound

In this case, since the substance is the same in both cases, the difference in entropy is determined by the phase. The gaseous phase of butane (\(\mathrm{C}_{4} \mathrm{H}_{10}(g)\)) has a higher standard molar entropy compared to the liquid phase (\(\mathrm{C}_{4} \mathrm{H}_{10}(l)\)). In conclusion: - For pair (a): \(\mathrm{C}_{4} \mathrm{H}_{10}(g)\) has a higher standard molar entropy; - For pair (b): \(\mathrm{C}_{4} \mathrm{H}_{10}(g)\) has a higher standard molar entropy.

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Most popular questions from this chapter

Carbon disulfide \(\left(C S_{2}\right)\) is a toxic, highly flammable substance. The following thermodynamic data are available for \(\mathrm{CS}_{2}(I)\) and \(\mathrm{CS}_{2}(g)\) at \(298 \mathrm{~K}\) \begin{tabular}{lcc} \hline & \(\Delta H_{i}(\mathrm{k} / \mathrm{mol})\) & $\Delta G_{i}^{\prime}(\mathrm{kJ} / \mathrm{mol})$ \\ \hline\(C S_{2}(l)\) & 89.7 & 65.3 \\ \(C S_{2}(g)\) & 117.4 & 67.2 \\ \hline \end{tabular} (a) Draw the Lewis structure of the molecule. What do you predict for the bond order of the \(\mathrm{C}-\mathrm{S}\) bonds? \((\mathbf{b})\) Use the VSEPR method to predict the structure of the \(\mathrm{CS}_{2}\) molecule. (c) Liquid \(\mathrm{CS}_{2}\) burns in \(\mathrm{O}_{2}\) with a blue flame, forming \(\mathrm{CO}_{2}(g)\) and \(\mathrm{SO}_{2}(g)\). Write a balanced equation for this reaction. (d) Using the data in the preceding table and in Appendix \(C,\) calculate \(\Delta H^{\circ}\) and \(\Delta G^{\circ}\) for the reaction in part \((c) .\) Is the reaction exothermic? Is it spontaneous at \(298 \mathrm{~K} ?\) (e) Use the data in the table to calculate \(\Delta S^{\circ}\) at $298 \mathrm{~K}\( for the vaporization of \)\mathrm{CS}_{2}(I) .$ Is the sign of \(\Delta S^{\circ}\) as you would expect for a vaporization? (f) Using data in the table and your answer to part (e), estimate the boiling point of \(\mathrm{CS}_{2}(l)\). Do you predict that the substance will be a liquid or a gas at \(298 \mathrm{~K}\) and \(101.3 \mathrm{kPa}\) ?

The potassium-ion concentration in blood plasma is about $5.0 \times 10^{-3} \mathrm{M}$, whereas the concentration in muscle-cell fluid is much greater \((0.15 \mathrm{M})\). The plasma and intracellular fluid are separated by the cell membrane, which we assume is permeable only to \(\mathrm{K}^{+}\). (a) What is \(\Delta G\) for the transfer of \(1 \mathrm{~mol}\) of \(\mathrm{K}^{+}\) from blood plasma to the cellular fluid at body temperature \(37^{\circ} \mathrm{C} ?\) (b) What is the minimum amount of work that must be used to transfer this \(\mathrm{K}^{+} ?\)

Most liquids follow Trouton's rule (see Exercise 19.93 ), which states that the molar entropy of vaporization is approximately $88 \pm 5 \mathrm{~J} / \mathrm{mol}-\mathrm{K}$. The normal boiling points and enthalpies of vaporization of several organic liquids are as follows: \begin{tabular}{lcc} \hline & Normal Boiling & \\ Substance & Point \(\left({ }^{\circ} \mathrm{C}\right)\) & $\Delta H_{\text {vap }}(\mathrm{k} / / \mathrm{mol})$ \\ \hline Acetone, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CO}\) & 56.1 & 29.1 \\\ Dimethyl ether, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{O}\) & -24.8 & 21.5 \\\ Ethanol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) & 78.4 & 38.6 \\ Octane, \(\mathrm{C}_{\mathrm{s}} \mathrm{H}_{18}\) & 125.6 & 34.4 \\ Pyridine, \(\mathrm{C}_{5} \mathrm{H}_{\mathrm{S}} \mathrm{N}\) & 115.3 & 35.1 \\\ \hline \end{tabular} (a) Calculate \(\Delta S_{\text {vap }}\) for each of the liquids. Do all the liquids obey Trouton's rule? (b) With reference to intermolecular forces (Section 11.2), can you explain any exceptions to the rule? (c) Would you expect water to obey Trouton's rule? By using data in Appendix \(\mathrm{B}\), check the accuracy of your conclusion. (d) Chlorobenzene \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Cl}\right)\) boils at \(131.8^{\circ} \mathrm{C}\). Use Trouton's rule to estimate $\Delta H_{\text {vap }}$ for this substance.

The value of \(K_{a}\) for nitrous acid \(\left(\mathrm{HNO}_{2}\right)\) at \(25^{\circ} \mathrm{C}\) is given in Appendix D. (a) Write the chemical equation for the equilibrium that corresponds to \(K_{a}\). (b) By using the value of \(K_{a}\) calculate \(\Delta G^{\circ}\) for the dissociation of nitrous acid in aqueous solution. (c) What is the value of \(\Delta G\) at equilibrium? (d) What is the value of \(\Delta G\) when $\left[\mathrm{H}^{+}\right]=5.0 \times 10^{-2} \mathrm{M}\(, \)\left[\mathrm{NO}_{2}^{-}\right]=6.0 \times 10^{-4} \mathrm{M},\( and \)\left[\mathrm{HNO}_{2}\right]=0.20 \mathrm{M} ?$

Sulfur dioxide reacts with strontium oxide as follows: $$ \mathrm{SO}_{2}(g)+\mathrm{SrO}(g) \longrightarrow \mathrm{SrSO}_{3}(s) $$ (a) Without using thermochemical data, predict whether \(\Delta G^{\circ}\) for this reaction is more negative or less negative than \(\Delta H^{\circ} .\) (b) If you had only standard enthalpy data for this reaction, how would you estimate the value of \(\Delta G^{\circ}\) at \(298 \mathrm{~K},\) using data from Appendix \(\mathrm{C}\) on other substances.
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