{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Exam 4 Key

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

This preview shows pages 1–4. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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
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.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern