Exam04answers pg3 - Cheung, Anthony – Exam 4 – Due: Dec...

Info iconThis preview shows page 1. Sign up to view the full content.

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

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? ...
View Full Document

Ask a homework question - tutors are online