# Exam04answers pg3 - Cheung Anthony – Exam 4 – Due Dec 6...

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Unformatted text preview: Cheung, Anthony – Exam 4 – Due: Dec 6 2006, 1:00 pm – Inst: McCord 008 (part 1 of 1) 12 points Trapped on a desert island with a desperate need to know the boiling point of chloroform (CHCl3 ), you come across some old thermodynamic tables that include the following data: For CHCl3 ( ): ∆Hf0 = −134.5 kJ/mol, S 0 = 202 J/mol · K For CHCl3 (g): ∆Hf0 = −103.1 kJ/mol, S 0 = 295.6 J/mol · K Using the data, what is the approximate boiling point of chloroform? 1. 273 K 2. 665 K 3. 349 K 4. 335 K correct Explanation: ∆Hf0,CHCl3 ( ) = −134.5 kJ/mol 0 SCH3 Cl3 ( ) = 202 J/mol · K ∆Hf0,CHCl3 (g) = −103.1 kJ/mol 0 SCHCl3 (g) = 295.6 J/mol · K ∆Hvap = n ∆Hf ,prod − n ∆Hf ,rct 009 (part 1 of 1) 12 points Consider the reaction CH4 (g) + X2 (g) → CH3 X(g) + HX(g) , 3 where X is a halogen-like atom (Group 7A). Calculate ∆H for this reaction if bond energy tables give the following values: C H H : 416 kJ/mol X : 322 kJ/mol X C X : 235 kJ/mol X : 258 kJ/mol Correct answer: 71 kJ/mol. Explanation: H H C H H H+X −− X −−→ H C H 0 ∆Hrxn = X+H X BErct − BEprod X) X) = 4 (C − 3 (C + (H H) + (X H) + (C X) = (−103.1 kJ/mol) − (−134.5 kJ/mol) = 31.4 kJ/mol ∆Svap = n ∆Sf ,prod − n ∆Sf ,rct = (C − (C H) + (X X) + (H X) X) = (295.6 J/mol · K) − (202 J/mol · K) = 93.6 J/mol · K = 0.0936 kJ/mol · K ∆G = ∆H − T ∆S At the boiling point ∆G = 0, so solve for T of the boiling point: T= ∆H 31.4 kJ/mol = ∆S 0.0936 kJ/mol · K = 335.47 K = (416 kJ/mol) + (235 kJ/mol) − (258 kJ/mol) + (322 kJ/mol) = 71 kJ/mol 010 (part 1 of 1) 10 points Which of the following would probably have a positive ∆S value? ...
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