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Chapter 19: Problem 20

What values of \(\Delta H, \Delta S,\) and \(\Delta G\) would you expect for the formation of an ideal solution of liquid components? (Is each value positive, negative, or zero?)

Short Answer

Expert verified
\(\Delta H = 0\) (neither absorbs nor releases heat), \(\Delta S > 0\) (entropy increases upon mixing), \(\Delta G < 0\) (process is spontaneous) for an ideal solution.

Step by step solution

01

Determine \(\Delta H\)

For an ideal solution, the change in enthalpy (\(\Delta H\)) should be zero. This is because, in an ideal solution, the interactions between the two types of particles (solvent and solute) are assumed to be equal before and after the mixture. If the intermolecular forces do not change upon mixing, no energy is absorbed or released, hence \(\Delta H = 0\).
02

Determine \(\Delta S\)

When two substances mix to form a solution, the entropy (\(\Delta S\)) typically increases because the mix represents a more disordered state compared to the separate pure components. Therefore, \(\Delta S\) for formation of an ideal solution should be positive.
03

Determine \(\Delta G\)

Free energy change (\(\Delta G\)) can be calculated by the formula \(\Delta G = \Delta H - T\Delta S\). From steps 1 and 2, we know that \(\Delta H = 0\) and \(\Delta S > 0\). For a spontaneous process, \(\Delta G\) must be negative or zero. Since the \(\Delta S\) value is positive, and multiplying it by any temperature will be positive, subtracting a positive number from zero will give a negative \(\Delta G\), indicating a spontaneous process. Therefore, the \(\Delta G\) for the formation of an ideal solution should be negative.

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Most popular questions from this chapter

The following equilibrium constants have been determined for the reaction \(\mathrm{H}_{2}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})\) \(K_{\mathrm{p}}=50.0\) at \(448^{\circ} \mathrm{C}\) and 66.9 at \(350^{\circ} \mathrm{C} .\) Use these data to estimate \(\Delta H^{\circ}\) for the reaction.

Why is \(\Delta G^{\circ}\) such an important property of a chemical reaction, even though the reaction is generally carried out under nonstandard conditions?

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) \(2 \mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{Hg}_{2}^{2+}(\mathrm{aq}) \rightleftharpoons\) \(2 \mathrm{Fe}^{2+}(\mathrm{aq})+2 \mathrm{Hg}^{2+}(\mathrm{aq})\) \(K_{\mathrm{c}}=9.14 \times 10^{-6} \mathrm{at} 25^{\circ} \mathrm{C}\)

Briefly describe each of the following ideas, methods, or phenomena: (a) absolute molar entropy; (b) coupled reactions; (c) Trouton's rule; (d) evaluation of an equilibrium constant from tabulated thermodynamic data.

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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