Workshop1_132_Ch9_key

Chem 132 Dr. Yang Chapter 9 Workshop #1 b. F 2 ; ionic forces in LiF are much stronger than the molecular forces in F 2 and HCl. HCl has dipole forces, whereas the nonpolar F 2 does not exhibit these. So F 2 has the weakest intermolecular forces and the lowest freezing point. c. CH 3 CH 2 OH; can form H-bonding interactions, unlike the other covalent compounds. d. H 2 O 2 ; the H-O-O-H structure has twice the number of H-bonding sites as compared to HF, so H 2 O 2 has the stronger H-bonding interactions and the greatest viscosity. e. H 2 CO; H 2 CO is polar, so it has dipole forces, unlike the other nonpolar covalent compounds, so H 2 CO will have the highest enthalpy of vaporization. f. I 2 ; I 2 has only LD forces, whereas CsBr and CaO have much stronger ionic forces. I 2 has the weakest intermolecular forces, so it has smallest ΔH fusion . 5. Ethanol has a heat of vaporization of 38.56 kJ/mol and a normal boiling point of 78.4  C. What is the vapor pressure of ethanol at 15  C? Clausius-Clapeyron equation relates vapor pressures at two different temperatures. ∆ H vap = 38.56 kJ mol = 38560 J mol T 1 = 78.4 ℃ = 351.55 K T 2 = 15 ℃ = 288.15 K P 1 = 760 torr ( atmospheric pressure ) ln P 2 P 1 = − ∆ H vap R ∗ ( 1 T 2 − 1 T 1 ) ln P 2 ( 760 torr ) = − ( 38560 J mol ) ( 8.314 J mol ∗ K ) ∗ ( 1 288.15 K − 1 351.55 K ) P 2 = 41.70 torr 6. The molar heat of fusion of benzene (C 6 H 6 ) is 9.92 kJ/mol. Its molar heat of vaporization is 30.7 kJ/mol. Calculate the heat required to melt 8.25 g benzene at its normal melting point. Calculate the heat required to vaporize 8.25 g benzene at its normal boiling point. Why is the heat of vaporization more than three times the heat of fusion? Melt: 8.25 g C 6 H 6 × 1 molC 6 H 6 78.11 g × 9.92 kJ 1 molC 6 H 6 = 1.05 kJ Vaporize: 8.25 g C 6 H 6 × 1 molC 6 H 6 78.11 g × 30.7 kJ 1 molC 6 H 6 = 3.24 kJ