Ch10_GilbertChem2E_StudentSolutions

Ch10_GilbertChem2E_StudentSolutions - 542 CHAPTER 10 |...

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Unformatted text preview: 542 CHAPTER 10 | Forces between Ions and Molecules and Colligative Properties 10.1. Collect and Organize In Figure P10.1, two similarly sized spheres of opposite charge represent KF. In KI, the iodide ion is shown negatively charged and larger than the K + ion. We are to determine which substance would have a higher melting point by virtue of having stronger ion–ion forces. Analyze Coulomb’s law describes the strength of the ion–ion interactions in these substances as E ~ Q 1 Q 2 d In the case of KF and KI, Q 1 and Q 2 are the same ( Q K + = +1, Q F – and Q I – = –1). What does change is the distance between the ion centers. Solve Coulomb’s law states that as distance between the ions increases, the energy of the interaction decreases. Because I – is larger than F – , the KI interaction is weaker. Therefore, the stronger ion–ion interaction results in a higher melting point for KF than KI. Think About It The strength of the ion–ion interaction is even more influenced by the charge on the ions. The melting point of CaF 2 (where Q 1 = +2 and Q 2 = –1) would be predicted to be even higher than that of KF. 10.3. Collect and Organize Given the boiling points of two trigonal pyramidal molecules, we are to determine which substance is NH 3 and which is PH 3 . Analyze Both molecules might be polar due to their pyramidal geometry. The more polar molecule has the stronger intermolecular forces. The dipole strength may be estimated using electronegativity values for N (3.0), P (2.1), and H (2.1). Solve Because the electronegativity of P is equal to that of H, the phosphine, PH 3 , is predicted to be less polar. Ammonia, NH 3 , on the other hand, is more polar; morever, because H is bonded to the very electronegative N atom, it forms strong hydrogen bonds with other NH 3 molecules. Ammonia therefore has the higher boiling point (–33 ˚C YH 3 in Figure P10.3). Phosphine, with only weaker forces between its atoms, has a low (–88 ˚C) boiling point and is represented as XH 3 in Figure P10.3. Think About It The phosphine molecule is not strictly nonpolar. The lone pair will “pull” differently than the N atoms do on the electrons on the phosphorus atom, so PH 3 is slightly polar. 10.5. Collect and Organize From the boiling points for X and Y that are read from the plot in Figure P10.5, we are to predict which substance has stronger intermolecular forces between its molecules. Analyze The normal boiling point is defined as the temperature at which the vapor pressure of a substance is equal to 1.00 atm. Forces between Ions and Molecules and Colligative Properties | 543 Solve From Figure P10.5, we see that the boiling point for X is about 5 ˚C and that the boiling point for Y is about 20 ˚C at 1.00 atm. The substance with the stronger intermolecular forces has the higher boiling point, so Y has the stronger intermolecular forces....
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This note was uploaded on 06/12/2011 for the course CHE 132 taught by Professor Hanson during the Spring '08 term at SUNY Stony Brook.

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Ch10_GilbertChem2E_StudentSolutions - 542 CHAPTER 10 |...

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