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Exam 4 Key - Chem 10181 Fall 2007(1 EXAM 4 KEY 1 Some...

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Chem 10181, Fall 2007 EXAM 4 KEY November 29, 2007 1 (1) Some thermodynamic properties of aluminum trichloride are given in the table below: Species f (kJ/mol) Sº (J/mol•K) f (kJ/mol) AlCl 3 (s) –704.2 110.67 –628.8 AlCl 3 (l) -669.2 185.8 -616.2 (a) (5 pts.) The heat of fusion of AlCl 3 is +35.0 kJ/mol: AlCl 3 (s) AlCl 3 (l) Hº = +35.0 kJ/mol Calculate f of AlCl 3 (l) (at 298 K). H o rxn = H o f [AlCl 3 (l)] - H o f [AlCl 3 (s)] 35.0 kJ/mol = H o f [AlCl 3 (l)] - (–704.2 kJ/mol) H o f [AlCl 3 (l)] = -669.2 kJ/mol (b) (7 pts.) The melting point of AlCl 3 is 192.4 ºC. Using this information and the information in part (a), calculate Sº of AlCl 3 (l). At 192.4 o C, G o rxn = 0. 0 = H o - T S o = 35000 J/mol - (465.6 K)( S o ) S o rxn = (35000 J/mol)/(465.6 K) S o rxn = + 75.2 J/mol•K S o rxn = S o [AlCl 3 (l)] - S o [AlCl 3 (s)] 75.2 J/mol•K = S o [AlCl 3 (l)] - 110.67 J/mol•K S o [AlCl 3 (l)] = 185.8 J/mol•K
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Chem 10181, Fall 2007 EXAM 4 KEY November 29, 2007 2 (c) (5 pts.) Calculate f of AlCl 3 (l) (at 298 K). [If you were UNABLE to do parts (a) or (b), you may assume that the H º f of AlCl 3 (l) is –700 kJ/mol and its Sº is 125 J/mol•K.] For the melting reaction, G o rxn = H o rxn - T S o rxn = G o f [AlCl 3 (l)] - G o f [AlCl 3 (s)] G o rxn = 35.0 kJ/mol - (298 K)(.0752 kJ/mol•K) G o rxn = 12.6 kJ/mol 12.6 kJ/mol = G o f [AlCl 3 (l)] - G o f [AlCl 3 (s)] 12.6 kJ/mol = G o f [AlCl 3 (l)] - (-628.8 kJ/mol) G o f [AlCl 3 (l)] = -616.2 kJ/mol (d) (7 pts.) The standard enthalpy change for the gas-phase reaction of Al atoms with elemental chlorine to form monomeric AlCl 3 (g) is given below: Al (g) + 1.5 Cl 2 (g) AlCl 3 (g) Hº = –765.1 kJ/mol Calculate the BDE of an Al–Cl bond in AlCl 3 (g). The BDE of the Cl–Cl bond is 243.4 kJ/mol. H o rxn = Σ (BDE of bonds broken) - Σ (BDE of bonds formed) H o rxn = 1.5(BDE Cl–Cl) - 3(BDE Al–Cl) -765.1 kJ/mol = 1.5(243.4 kJ/mol) - 3(BDE Al–Cl) 3(BDE Al–Cl) = 1130.2 kJ/mol BDE Al–Cl = 376.7 kJ/mol
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Chem 10181, Fall 2007 EXAM 4 KEY November 29, 2007 3 (2) Both elemental aluminum (Al (s)) and aluminum phosphide (AlP) crystallize in cubic lattices, as illustrated below (one complete unit cell is shown for each structure). The edge lengths a of the unit cells are 4.050 Å and 5.451 Å, respectively. Al (s) (dark circles Al, open circles P) AlP (a) (12 pts.) For each of the two structures, indicate the following characteristics: Al (s) AlP Number of Al atoms per unit 4 4 cell: Number of nearest neighbors of each 12 4 Al atom: Type of unit cell (primitive cubic, f.c.c. f.c.c.
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