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Chapter 17: Problem 14

Predict the sign of \(\Delta S^{\circ}\) for each of the following reactions. (a) \(\mathrm{H}_{2}(g)+\mathrm{Ni}^{2+}(a q) \longrightarrow 2 \mathrm{H}^{+}(a q)+\mathrm{Ni}(s)\) (b) \(\mathrm{Cu}(s)+2 \mathrm{H}^{+}(a q) \longrightarrow \mathrm{H}_{2}(g)+\mathrm{Cu}^{2+}(a q)\) (c) \(\mathrm{N}_{2} \mathrm{O}_{4}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)\)

Short Answer

Expert verified
Question: Predict the signs of the standard entropy change (\(\Delta S^{\circ}\)) for the following reactions: (a) H\(_{2}\)(g) + Ni\(^{2+}\)(aq) → 2 H\(^{+}\)(aq) + Ni(s) (b) Cu(s) + 2 H\(^{+}\)(aq) → H\(_{2}\)(g) + Cu\(^{2+}\)(aq) (c) N\(_{2}\)O\(_{4}\)(g) → 2 NO\(_{2}\)(g) Answer: (a) For reaction (a), the standard entropy change (\(\Delta S^{\circ}\)) is predicted to be negative: \(\Delta S^{\circ} < 0\). (b) For reaction (b), the standard entropy change (\(\Delta S^{\circ}\)) is predicted to be positive: \(\Delta S^{\circ} > 0\). (c) For reaction (c), the standard entropy change (\(\Delta S^{\circ}\)) is predicted to be positive: \(\Delta S^{\circ} > 0\).

Step by step solution

01

Reaction (a) Analysis

In this reaction, hydrogen gas reacts with Ni\(^{2+}\) ions in aqueous solution and forms 2 H\(^{+}\) ions in aqueous solution and solid Ni. The gaseous reactant changes to the solid and aqueous products, which decreases the disorder. Moreover, two moles of products are formed from one mole of hydrogen gas.
02

Predicting the Sign of \(\Delta S^{\circ}\) for Reaction (a)

Since the reaction involves a gas changing to solid and aqueous states, and the number of moles does not significantly increase, it is reasonable to predict that the standard entropy change for reaction (a) will be negative, \(\Delta S^{\circ} < 0\).
03

Reaction (b) Analysis

In this reaction, solid Cu reacts with two moles of H\(^{+}\) ions in aqueous solution and forms hydrogen gas and Cu\(^{2+}\) ion in aqueous solution. The solid reactant changes to gaseous and aqueous products, which increases the disorder. Regarding the number of moles, two moles of reactants change to two moles of products.
04

Predicting the Sign of \(\Delta S^{\circ}\) for Reaction (b)

Since the reaction involves a solid changing to gaseous and aqueous states, and the number of moles remains the same, it is reasonable to predict that the standard entropy change for reaction (b) will be positive, \(\Delta S^{\circ} > 0\).
05

Reaction (c) Analysis

In this reaction, one mole of N\(_2\)O\(_4\) gas decomposes into 2 moles of NO\(_2\) gas. There's no phase change in this reaction; both reactants and products are in gaseous states. However, the number of moles of reactants increases from one to two.
06

Predicting the Sign of \(\Delta S^{\circ}\) for Reaction (c)

Since there is no phase change, but an increase in the number of moles, it is reasonable to predict that the standard entropy change for reaction (c) will be positive, \(\Delta S^{\circ} > 0\).

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Most popular questions from this chapter

A student is asked to prepare a \(0.030 \mathrm{M}\) aqueous solution of \(\mathrm{PbCl}_{2}\) (a) Is this possible at \(25^{\circ} \mathrm{C} ?\) (Hint: Is dissolving \(0.030 \mathrm{~mol}\) of \(\mathrm{PbCl}_{2}\) at \(25^{\circ} \mathrm{C}\) possible? \()\) (b) If the student used water at \(100^{\circ} \mathrm{C}\), would this be possible?

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 \(1000 \mathrm{~K}\).

Predict the sign of \(\Delta S^{\circ}\) for each of the following reactions. (a) \(2 \mathrm{Na}(s)+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{NaCl}(s)\) (b) \(2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)\) (c) \(\mathrm{C}_{2} \mathrm{H}_{4}(g)+3 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{CO}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(l)\) (d) \(\mathrm{NH}_{4} \mathrm{NO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{NH}_{3}(g)+\mathrm{O}_{2}(g)\)

Which of the following quantities can be taken to be independent of temperature? Independent of pressure? (a) \(\Delta H\) for a reaction (b) \(\Delta S\) for a reaction (c) \(\Delta G\) for a reaction (d) \(S\) for a substance

Pencil "lead" is almost pure graphite. Graphite is the stable elemental form of carbon at \(25^{\circ} \mathrm{C}\) and 1 atm. Diamond is an allotrope of graphite. Given $$ \text { diamond: } \Delta H_{\mathrm{f}}^{\circ}=1.9 \mathrm{~kJ} / \mathrm{mol} ; S^{\circ}=2.4 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K} $$ at what temperature are the two forms in equilibrium at 1 atm? \(\mathrm{C}\) (graphite) \(\rightleftharpoons \mathrm{C}\) (diamond)
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