2010 Exam 4 Key - Name Section Number Chemistry 1314 Honors...

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Name Section Number Chemistry 1314 Honors Midterm Exam #4 December 3, 2010 This exam is closed book, notes, papers, etc. You may use a calculator. If you need a physical constant or unit conversion factor that is not provided, please ask. This exam must represent your own work. Scores: 1. a. / 4 b. / 6 c. / 4 d. / 11 2. a. / 6 b. / 12 c. / 5 d. / 2 3. a. / 4 b. / 8 c. / 5 d. / 8 4. a. / 6 b. / 4 c. / 4 d. / 11 Total: / 100
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1. The bond enthalpy Δ H B is the change in enthalpy when a bond is broken. Δ H B is always positive for any stable bond, revealing that the bond breaking process is always endothermic. a. Using Molecular Orbital Theory arguments, explain why breaking a bond always requires adding energy to the atoms in the bond. The electrons in the bonding region between the nuclei are at a lower energy than in separated atoms due to attraction to two nuclei. Pulling the nuclei apart elevates the energy of the electrons. Electrons in a bond are at a lower energy than those on the atom: 2 points Pulling the atoms apart raises the energy of the electrons: 2 points b. Δ H B = 269 kJ/mol for H 2 + . Again using Molecular Orbital Theory arguments, explain why a rough estimate for Δ H B for H 2 should be two times Δ H B (H 2 + ). (Hint: What is the electron configuration of each system?) Electron configuration comparison: 4 points 2 electrons in same orbital OR bond order comparison: 2 points The electron configuration for H 2 + is 1s σ 1 . The electron configuration for H 2 is 1s σ 2 . Thus lowering of electron energy for the 2 electrons in H 2 should be about double the lowering for 1 electron in H 2 + . An alternate explanation comes from comparing bond orders (1/2 vs. 1). c. In reality, Δ H B (H 2 ) = 485 kJ/mol, which is less than 2 x Δ H B (H 2 + ). Give a reason why this comparison makes sense (i.e., why the bond energy is less than the comparison). Although the attraction of the second electron to the two nuclei roughly equals the attraction of the first electron, electron/electron repulsion partly offsets this, raising the total two-electron energy. Electron-electron repulsion raises the energy: 4 points d. Using the given table of bond enthalpies, calculate Δ H for the following reaction: ( ) ( ) ( ) ( ) ( ) ( g HCl g HF g Cl CF g F g Cl g CH 2 2 2 2 2 2 2 2 4 + + ⎯→ + + ) Give a one sentence statement, based on this calculation, as to why this reaction is exothermic. Δ H = 4 Δ H B (C-H) + 2 Δ H B (Cl-Cl) + 2 Δ H B (F-F) – 2 Δ H B (C-F) – 2 Δ H B (C-Cl) – 2 Δ H B (H-F) – 2 Δ H B (H-Cl) = (finish calculation) An equal number of bonds are broken as are formed, but the bonds formed are stronger than the bonds broken.
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