Exam 04 - Version 080 EXAM 4 Fakhreddine (53605) This...

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Version 080 – EXAM 4 – Fakhreddine – (53605) 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – fnd all choices be±ore answering. 001 12.0 points What is the activation energy ±or a reaction i± its rate constant is ±ound to triple when the temperature is raised ±rom 300 K to 310 K? 1. 418,400 J/mol 2. 195,600 J/mol 3. 84,900 J/mol correct 4. 20,300 J/mol 5. No other choice is within 3 percent. Explanation: T 1 = 300 K k 2 = 3 k 1 T 2 = 310 K ln p k 2 k 1 P = E a R p 1 T 1 - 1 T 2 P E a = R ln p k 2 k 1 P 1 T 1 + 1 T 2 The rate constant tripled, so k 2 = 3 k 1 : E a = (8 . 314 J / mol · K) ln 3 1 300 K - 1 310 K = 84944 J / mol 002 10.0 points The reaction 2 N 2 O 5 (g) 4 NO 2 (g) + O 2 (g) is ±ound to obey the rate law: rate = k [N 2 O 5 ]. Does the reaction mechanism consist o± only a single bimolecular step? 1. Yes, it must 2. It might, but you cannot tell ±rom the in±ormation given. 3. No, it cannot correct Explanation: Single-step reactions have rate laws with order equal to the stiochiometry o± the reac- tant(s) in the balanced equation. 003 10.0 points Generally, reaction rates increase when the temperature o± the system is raised because 1. the concentration o± reactants increases as the temperature increases. 2. the bond energy o± the reactants decreases as the temperature increases. 3. the bond energy o± the products decreases as the temperature increases. 4. the kinetic energy (and thus the speed) o± the reactants increases as the temperature increases. correct 5. the activation barrier decreases as the temperature increases. Explanation: Higher KE means more collisions and more properly oriented collisions, there±ore a ±aster rate o± reaction. 004 12.0 points Calculate the potential ±or the cell indicated: Fe | Fe 2+ (10 3 M) || Pb 2+ (10 5 M) | Pb Pb 2+ + 2 e Pb E 0 = - 0 . 126 V Fe 2+ + 2 e Fe E 0 = - 0 . 440 V 1. 0.373 V 2. 0.196 V 3. 0.432 V 4. 0.284 V 5. 0.255 V correct Explanation:
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Version 080 – EXAM 4 – Fakhreddine – (53605) 2 The overall reaction is Fe + Pb 2+ Fe 2+ + Pb Please notice that since the concentrations are not 1 M, the Nernst equation must be used. In this cell notation, the anode is located on the left of the salt bridge || and the cathode on the right. So ±rst calculate E 0 cell = E cathode - E 0 anode = - 0 . 126 V - ( - 0 . 440) V = 0 . 314 V Using the Nernst Equation E cell = E 0 cell - 0 . 05916 n log Q = 0 . 314 V - 0 . 05916 2 log p [Fe 2+ ] [Pb 2+ ] P = 0 . 314 V - 0 . 05916 2 log p 10 3 10 5 P = 0 . 25484 V 005 10.0 points During the chemical reaction in an electro- chemical cell, 1. electrons travel from the cathode to the anode. 2.
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This note was uploaded on 04/15/2009 for the course CH 53625 taught by Professor Fakhreddine during the Spring '09 term at University of Texas.

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Exam 04 - Version 080 EXAM 4 Fakhreddine (53605) This...

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