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Chapter 17: Problem 10

Predict the sign of \(\Delta S\) for the following. (a) a lake freezing (b) precipitating lead chloride (c) a candle burning (d) weeding a garden

Short Answer

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Question: Predict the sign of the change in entropy (ΔS) for each of the following processes: (a) a lake freezing, (b) precipitating lead chloride, (c) a candle burning, and (d) weeding a garden. Answer: (a) ΔS is negative for a lake freezing, as the water molecules form a more ordered structure. (b) ΔS is negative for precipitating lead chloride, due to the formation of a more ordered solid precipitate. (c) ΔS is positive for a candle burning, as the system experiences an increase in disorder with the formation of gaseous products. (d) ΔS is negative for weeding a garden, as organizing the remaining plants reduces the disorder within the garden.

Step by step solution

01

(a) Predicting the sign of ΔS for a lake freezing.

As a lake freezes, the water molecules slow down and arrange themselves into a more ordered crystal lattice structure. This decrease in disorder means that ΔS will be negative for this process.
02

(b) Predicting the sign of ΔS for precipitating lead chloride.

In the precipitation of lead chloride, ions in an aqueous solution combine to form a solid precipitate with a more ordered structure. This leads to a decrease in disorder in the system, so ΔS will be negative for this process.
03

(c) Predicting the sign of ΔS for a candle burning.

Burning a candle involves the combustion of the wax, which is a hydrocarbon. During combustion, the hydrocarbon molecules react with oxygen to form carbon dioxide and water, which are released as gases. This transformation from solid wax and gaseous oxygen to gaseous carbon dioxide and water represents an increase in disorder in the system. Thus, ΔS will be positive for this process.
04

(d) Predicting the sign of ΔS for weeding a garden.

Weeding a garden involves removing unwanted plants and organizing the remaining plants. Although this process does not strictly involve a thermodynamic change, we can consider the change in disorder from an information perspective. By removing the weeds and organizing the remaining plants, one is reducing the disorder in the garden. Thus, ΔS will be negative for this process.

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Most popular questions from this chapter

Given the following standard free energies at \(25^{\circ} \mathrm{C}\), $$ \begin{array}{ll} \mathrm{SO}_{2}(g)+3 \mathrm{CO}(g) \longrightarrow \operatorname{COS}(g)+2 \mathrm{CO}_{2}(g) & \Delta G^{\circ}=-246.5 \mathrm{~kJ} \\ \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g) & \Delta G^{\circ}=-28.5 \mathrm{~kJ} \end{array} $$ find \(\Delta G^{\circ}\) at \(25^{\circ} \mathrm{C}\) for the following reaction. $$ \mathrm{SO}_{2}(g)+\mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{COS}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) $$

For the reaction $$ 2 \mathrm{Cl}^{-}(a q)+\mathrm{Br}_{2}(l) \longrightarrow \mathrm{Cl}_{2}(g)+2 \mathrm{Br}^{-}(a q) $$ calculate the temperature at which \(\Delta G^{\circ}=0 .\)

Calculate \(\Delta G^{\circ}\) at \(45^{\circ} \mathrm{C}\) for reactions for which (a) \(\Delta H^{\circ}=293 \mathrm{~kJ} ; \Delta S^{\circ}=-695 \mathrm{~J} / \mathrm{K}\) (b) \(\Delta H^{\circ}=-1137 \mathrm{~kJ} ; \Delta S^{\circ}=0.496 \mathrm{~kJ} / \mathrm{K}\) (c) \(\Delta H^{\circ}=-86.6 \mathrm{~kJ} ; \Delta S^{\circ}=-382 \mathrm{~J} / \mathrm{K}\)

On the basis of your experience, predict which of the following reactions are spontaneous. (a) \(\mathrm{Zn}(s)+2 \mathrm{H}^{+}(a q) \longrightarrow \mathrm{Zn}^{2+}(a q)+\mathrm{H}_{2}(g)\) (b) \(\mathrm{CaCO}_{3}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(s)+\mathrm{H}_{2} \mathrm{CO}_{3}(a q)\) (c) \(\mathrm{CH}_{4}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)\) (d) \(\mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q) \longrightarrow \mathrm{AgCl}(s)\)

Determine whether each of the following statements is true or false. If false, modify it to make it true. (a) An exothermic reaction is spontaneous. (b) When \(\Delta G^{\circ}\) is positive, the reaction cannot occur under any conditions. (c) \(\Delta S^{\circ}\) is positive for a reaction in which there is an increase in the number of moles. (d) If \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are both negative, \(\Delta G^{\circ}\) will be negative.
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