In your own words, explain why (a) \(\Delta S^{\circ}\) is negative for a reaction in which the number of moles of gas decreases. (b) we take \(\Delta S^{\circ}\) to be independent of \(T\), even though entropy increases with \(T\). (c) a solid has lower entropy than its corresponding liquid.

Short Answer

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Question: Explain why the standard entropy change is negative for a reaction in which the number of moles of gas decreases. Answer: The standard entropy change (ΔS°) is negative for reactions with a decrease in moles of gas because a reduction in moles results in a decreased number of possible microstates, leading to a decrease in entropy as entropy is a measure of disorder or randomness in a system.

Step by step solution

01

(a) Explanation of negative ΔS° for reactions with decreasing moles of gas

Entropy (S) is a measure of the degree of disorder or randomness in a system. A higher entropy corresponds to a higher number of possible microstates (the arrangements of atoms or molecules in a system). When the number of moles of gas decreases in a reaction, the total number of molecules in the system also decreases, which subsequently reduces the number of possible microstates. A reduction in possible microstates leads to a decrease in entropy. Therefore, the standard entropy change (ΔS°) is negative for reactions with a decrease in moles of gas.
02

(b) Explanation for considering ΔS° independent of T

Although entropy increases with temperature, we often treat the standard entropy change (ΔS°) as independent of temperature when dealing with relatively small or narrow temperature ranges. This is because, in such cases, the variation in entropy with temperature is negligible for practical purposes. This simplification allows us to more easily analyze and predict the behavior of chemical reactions without having to account for temperature changes constantly. For larger temperature ranges or systems with significant temperature sensitivity, a more accurate analysis involving the temperature dependence of ΔS° might be required.
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(c) Explanation for lower entropy of a solid compared to its corresponding liquid

In a solid, the particles are arranged in a highly ordered, fixed structure, with a relatively low number of possible microstates. The particles are held together by strong forces, and their movement is restricted to vibrations around fixed positions. As a result, the degree of disorder or randomness is quite low in a solid. In contrast, in a liquid, the particles are more randomly arranged and have more freedom to move around. The forces binding the particles together are weaker than in a solid, allowing for a greater number of possible microstates. Therefore, the degree of disorder or randomness is higher in a liquid compared to a solid. Due to the higher number of microstates and greater disorder in a liquid, its entropy is higher than its corresponding solid.

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Most popular questions from this chapter

Consider the following reaction with its thermodynamic data: \(2 \mathrm{~A}(\mathrm{~g})+\mathrm{B}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{AB}(\mathrm{g}) \quad \Delta H^{\circ}<0 ; \Delta S^{\circ}<0 ; \Delta G^{\circ}\) at \(60^{\circ} \mathrm{C}=+10 \mathrm{~kJ}\) Which statements about the reaction are true? (a) When \(\Delta G=1\), the reaction is at equilibrium. (b) When \(Q=1, \Delta G=\Delta G^{\circ}\). (c) At \(75^{\circ} \mathrm{C}\), the reaction is definitely nonspontaneous. (d) At \(100^{\circ} \mathrm{C}\), the reaction has a positive entropy change. (e) If \(\mathrm{A}\) and \(\mathrm{B}_{2}\) are elements in their stable states, \(S^{\circ}\) for \(\mathrm{A}\) and \(\mathrm{B}_{2}\) at \(25^{\circ} \mathrm{C}\) is \(0 .\) (f) \(K\) for the reaction at \(60^{\circ} \mathrm{C}\) is less than 1 .

Consider the following reactions at \(25^{\circ} \mathrm{C}\) : $$ \begin{aligned} &\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q)+6 \mathrm{O}_{2}(g) \longrightarrow 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O} \quad \Delta G^{\circ}=-2870 \mathrm{~kJ} \\ &\mathrm{ADP}(a q)+\mathrm{HPO}_{4}{ }^{2-}(a q)+2 \mathrm{H}^{+}(a q) \longrightarrow \mathrm{ATP}(a q)+\mathrm{H}_{2} \mathrm{O} \\ &\Delta G^{\circ}=31 \mathrm{~kJ} \end{aligned} $$ Write an equation for a coupled reaction between glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\), and ADP in which \(\Delta G^{\circ}=-390 \mathrm{~kJ}\).

The overall reaction that occurs when sugar is metabolized is $$ \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(s)+12 \mathrm{O}_{2}(g) \longrightarrow 12 \mathrm{CO}_{2}(g)+11 \mathrm{H}_{2} \mathrm{O}(l) $$ For this reaction, \(\Delta H^{\circ}\) is \(-5650 \mathrm{~kJ}\) and \(\Delta G^{\circ}\) is \(-5790 \mathrm{~kJ}\) at \(25^{\circ} \mathrm{C}\). (a) If \(25 \%\) of the free energy change is actually converted to useful work, how many kilojoules of work are obtained when one gram of sugar is metabolized at body temperature, \(37^{\circ} \mathrm{C}\) ? (b) How many grams of sugar would a 120 -lb woman have to eat to get the energy to climb the Jungfrau in the Alps, which is \(4158 \mathrm{~m}\) high? \(\left(w=9.79 \times 10^{-3} \mathrm{mh}\right.\), where \(w=\) work in kilojoules, \(m\) is body mass in kilograms, and \(h\) is height in meters.)

Discuss the effect of temperature change on the spontaneity of the following reactions at 1 atm. (a) \(\mathrm{Al}_{2} \mathrm{O}_{3}(s)+2 \mathrm{Fe}(s) \longrightarrow 2 \mathrm{Al}(s)+\mathrm{Fe}_{2} \mathrm{O}_{3}(s)\) $$ \Delta H^{\circ}=+851.5 \mathrm{~kJ} ; \Delta S^{\circ}=+38.5 \mathrm{~J} / \mathrm{K} $$ (b) \(\mathrm{N}_{2} \mathrm{H}_{4}(l) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2}(g)\) $$ \Delta H^{\circ}=-50.6 \mathrm{~kJ} ; \Delta S^{\circ}=0.3315 \mathrm{~kJ} / \mathrm{K} $$ (c) \(\mathrm{SO}_{3}(g) \longrightarrow \mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g)\) $$ \Delta H^{\circ}=98.9 \mathrm{~kJ} ; \Delta S^{\circ}=+0.0939 \mathrm{~kJ} / \mathrm{K} $$

I Sodium carbonate, also called "washing soda," can be made by heating sodium hydrogen carbonate: $$ \begin{gathered} 2 \mathrm{NaHCO}_{3}(s) \longrightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \\ \Delta H^{\circ}=+135.6 \mathrm{~kJ} ; \Delta G^{\circ}=+34.6 \mathrm{~kJ} \text { at } 25^{\circ} \mathrm{C} \end{gathered} $$ (a) Calculate \(\Delta S^{\circ}\) for this reaction. Is the sign reasonable? (b) Calculate \(\Delta G^{\circ}\) at \(0 \mathrm{~K} ;\) at \(1000 \mathrm{~K}\).

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