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Chapter 13: Problem 20

The enthalpy of solution of \(\mathrm{KBr}\) in water is about \(+198 \mathrm{~kJ} / \mathrm{mol}\). Nevertheless, the solubility of \(\mathrm{KBr}\) in water is relatively high. Why does the solution process occur even though it is endothermic?

Short Answer

Expert verified
The solubility of KBr in water is relatively high despite its endothermic enthalpy of solution because the increase in entropy during the dissolution process compensates for the positive enthalpy change. This leads to a negative Gibbs Free Energy change, indicating a spontaneous process. The increased entropy comes from the disorder created when KBr molecules are dispersed in water and the favorable ion-dipole interactions between KBr and water molecules.

Step by step solution

01

Understanding Entropy

Entropy (\(\Delta S\)) is a measure of the randomness or disorder of a system. When a solute dissolves in a solvent, the particles become more dispersed, increasing the entropy of the system. The Second Law of Thermodynamics states that spontaneous processes will always result in an increase in the total entropy of the universe, which includes both the system and the surroundings.
02

Define Gibbs Free Energy

Gibbs Free Energy (\(\Delta G\)) is a thermodynamic concept that helps determine the spontaneity of a process. The change in Gibbs Free Energy is given by the equation: \[\Delta G = \Delta H - T\Delta S\] Where \(\Delta G\) is the change in Gibbs Free Energy, \(\Delta H\) is the change in enthalpy, \(T\) is the temperature in Kelvin, and \(\Delta S\) is the change in entropy.
03

Determine Spontaneity

For a process to be spontaneous, it should have a negative value for \(\Delta G\). In the case of KBr dissolving in water, the enthalpy of the reaction (\(\Delta H\)) is given as +198 kJ/mol, which indicates that the process is endothermic. However, as mentioned earlier, the entropy of the system will increase due to dissolution. If the entropy change (\(\Delta S\)) is high enough, it is possible for the process to be spontaneous even if the enthalpy change is positive.
04

Explanation for High Solubility

The solubility of KBr in water is relatively high because the entropy increase due to the dissolution process compensates for the positive enthalpy change, making the overall Gibbs Free Energy change negative and thus spontaneous. The increase in entropy arises from the disorder created when KBr molecules are dispersed in water, and the interactions between KBr and water molecules (ion-dipole forces) also contribute to the overall favorable energy change during the dissolution process.

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Most popular questions from this chapter

A saturated solution of sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) is made by dissolving excess table sugar in a flask of water. There are \(50 \mathrm{~g}\) of undissolved sucrose crystals at the bottom of the flask in contact with the saturated solution. The flask is stoppered and set aside. A year later a single large crystal of mass \(50 \mathrm{~g}\) is at the bottom of the flask. Explain how this experiment provides evidence for a dynamic equilibrium between the saturated solution and the undissolved solute.

List the following aqueous solutions in order of decreasing freezing point: \(0.040 \mathrm{~m}\) glycerin \(\left(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}_{3}\right), 0.020 \mathrm{~m}\) \(\mathrm{KBr}, 0.030 \mathrm{~m}\) phenol \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}\right)\).

Carbon disulfide \(\left(\mathrm{CS}_{2}\right.\) ) boils at \(46.30{ }^{\circ} \mathrm{C}\) and has a density of \(1.261 \mathrm{~g} / \mathrm{mL}\). (a) When \(0.250 \mathrm{~mol}\) of a nondissociating solute is dissolved in \(400.0 \mathrm{~mL}\) of \(\mathrm{CS}_{2}\), the solution boils at \(47.46^{\circ} \mathrm{C}\). What is the molal boiling-point-elevation constant for \(\mathrm{CS}_{2}\) ? (b) When \(5.39 \mathrm{~g}\) of a nondissociating unknown is dissolved in \(50.0 \mathrm{~mL}\) of \(\mathrm{CS}_{2}\), the solution boils at \(47.08^{\circ} \mathrm{C}\). What is the molecular weight of the unknown?

Calculate the number of moles of solute present in each of the following aqueous solutions: (a) \(600 \mathrm{~mL}\) of \(0.250 \mathrm{M} \mathrm{SrBr}_{2}\), (b) \(86.4 \mathrm{~g}\) of \(0.180 \mathrm{~m} \mathrm{KCl}\), (c) \(124.0 \mathrm{~g}\) of a solution that is \(6.45 \%\) glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) by mass.

Calculate the molality of each of the following solutions: (a) \(8.66 \mathrm{~g}\) benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) dissolved in \(23.6 \mathrm{~g}\) carbon tetrachloride \(\left(\mathrm{CCl}_{4}\right)\), (b) \(4.80 \mathrm{~g} \mathrm{NaCl}\) dissolved in \(0.350 \mathrm{~L}\) of water.
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