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Chapter 19: Problem 7

For each of the following reactions, indicate whether \(\Delta S\) for the reaction should be positive or negative. If it is not possible to determine the sign of \(\Delta S\) from the information given, indicate why. (a) \(\mathrm{CaO}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{s})\) (b) \(2 \mathrm{HgO}(\mathrm{s}) \longrightarrow 2 \mathrm{Hg}(1)+\mathrm{O}_{2}(\mathrm{g})\) (c) \(2 \mathrm{NaCl}(1) \longrightarrow 2 \mathrm{Na}(1)+\mathrm{Cl}_{2}(\mathrm{g})\) (d) \(\mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{s})+3 \mathrm{CO}(\mathrm{g}) \longrightarrow 2 \mathrm{Fe}(\mathrm{s})+3 \mathrm{CO}_{2}(\mathrm{g})\) (e) \(\operatorname{Si}\left(\text { s) }+2 \mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow \mathrm{SiCl}_{4}(\mathrm{g})\right.\)

Short Answer

Expert verified
(a) Not determinable (approximately zero or slightly negative), (b) Positive, (c) Positive, (d) Not determinable (approximately zero or slightly negative or positive), (e) Positive.

Step by step solution

01

Analyze reaction (a)

For the reaction \(\mathrm{CaO}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{s})\), we can see that the number of molecules in reactant and product side is same (2) and the state of matter doesn't change. Hence, there is no increase in disorder, indicating that \(\Delta S\) for this reaction could be approximately zero or slightly negative.
02

Analyze reaction (b)

For the reaction \(2 \mathrm{HgO}(\mathrm{s}) \longrightarrow 2\mathrm{Hg}(1)+\mathrm{O}_{2}(\mathrm{g})\), the reactants are in solid state and the products include a gas which is more disorderly than a solid. There's an increase in disorder, indicating that \(\Delta S\) for this reaction should be positive.
03

Analyze reaction (c)

For the reaction \(2 \mathrm{NaCl}(1) \longrightarrow 2\mathrm{Na}(1)+\mathrm{Cl}_{2}(\mathrm{g})\), the reactants are in liquid state and the products include a gas. This equates to an increase in disorder, suggesting that the \(\Delta S\) for this reaction should be positive.
04

Analyze reaction (d)

For the reaction \(\mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{s})+3 \mathrm{CO}(\mathrm{g}) \longrightarrow 2 \mathrm{Fe}(\mathrm{s})+3 \mathrm{CO}_{2}(\mathrm{g})\), the overall number of molecules remain same but solid reactant turns to solid product while gas remains gas, the state of matter doesn't change much. Hence, \(\Delta S\) for this reaction could be approximately zero or slightly negative or positive, exact can't be predicted without more data.
05

Analyze reaction (e)

For the reaction \(\operatorname{Si}\left(\text { s) }+2 \mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow \mathrm{SiCl}_{4}(\mathrm{g})\right.\), the reactants include a solid and the product is a gas. There's an increase in disorder, indicating that the \(\Delta S\) for this reaction should be positive.

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Most popular questions from this chapter

For a process to occur spontaneously, (a) the entropy of the system must increase; (b) the entropy of the surroundings must increase; (c) both the entropy of the system and the entropy of the surroundings must increase; (d) the net change in entropy of the system and surroundings considered together must be a positive quantity; (e) the entropy of the universe must remain constant.

Which of the following substances would obey Trouton's rule most closely: HF, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{3}\) (toluene), or \(\mathrm{CH}_{3} \mathrm{OH}\) (methanol)? Explain your reasoning.

Use the following data to estimate the standard molar entropy of gaseous benzene at \(298.15 \mathrm{K} ;\) that is, \(S^{\circ}\left[\mathrm{C}_{6} \mathrm{H}_{6}(\mathrm{g}, 1 \mathrm{atm})\right] .\) For \(\mathrm{C}_{6} \mathrm{H}_{6}(\mathrm{s}, 1 \mathrm{atm})\) at its melting point of \(5.53^{\circ} \mathrm{C}, S^{\circ}\) is \(128.82 \mathrm{Jmol}^{-1} \mathrm{K}^{-1}\). The enthalpy of fusion is \(9.866 \mathrm{kJ} \mathrm{mol}^{-1} .\) From the melting point to 298.15 K, the average heat capacity of liquid benzene is \(134.0 \mathrm{JK}^{-1} \mathrm{mol}^{-1} .\) The enthalpy of vaporization of \(\mathrm{C}_{6} \mathrm{H}_{6}(1)\) at \(298.15 \mathrm{K}\) is \(33.85 \mathrm{kJ} \mathrm{mol}^{-1},\) and in the vapor- ization, \(\mathrm{C}_{6} \mathrm{H}_{6}(\mathrm{g})\) is produced at a pressure of 95.13 Torr. Imagine that this vapor could be compressed to 1 atm pressure without condensing and while behaving as an ideal gas. Calculate \(S^{\circ}\left[\mathrm{C}_{6} \mathrm{H}_{6}(\mathrm{g}, 1 \text { atm) }] .[ \text { Hint: Refer to }\right.\) Exercise \(88,\) and note the following: For infinitesimal quantities, \(d S=d q / d T ;\) for the compression of an ideal gas, \(d q=-d w ;\) and for pressure-volume work, \(d w=-P d V\).

For the following equilibrium reactions, calculate \(\Delta G^{\circ}\) at the indicated temperature. [Hint: How is each equilibrium constant related to a thermodynamic equilibrium constant, \(K ?]\) (a) \(\mathrm{H}_{2}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g}) \quad K_{\mathrm{c}}=50.2\) at \(445^{\circ} \mathrm{C}\) (b) \(\mathrm{N}_{2} \mathrm{O}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})\) \(K_{c}=1.7 \times 10^{-13} \mathrm{at} 25^{\circ} \mathrm{C}\) (c) \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{g})\) \(K_{c}=4.61 \times 10^{-3}\) at \(25^{\circ} \mathrm{C}\) (d) \(2 \mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{Hg}_{2}^{2+}(\mathrm{aq}) \rightleftharpoons\) \(2 \mathrm{Fe}^{2+}(\mathrm{aq})+2 \mathrm{Hg}^{2+}(\mathrm{aq})\) \(K_{\mathrm{c}}=9.14 \times 10^{-6} \mathrm{at} 25^{\circ} \mathrm{C}\)

If \(\Delta G^{\circ}=0\) for a reaction, it must also be true that (a) \(K=0 ;\) (b) \(K=1 ;\) (c) \(\Delta H^{\circ}=0 ;\) (d) \(\Delta S^{\circ}=0\) (e) the equilibrium activities of the reactants and products do not depend on the initial conditions.
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