# 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 – find all choices before answering. 001 12.0 points What is the activation energy for a reaction if its rate constant is found to triple when the temperature is raised from 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 parenleftbigg k 2 k 1 parenrightbigg = E a R parenleftbigg 1 T 1 - 1 T 2 parenrightbigg E a = R ln parenleftbigg k 2 k 1 parenrightbigg 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 found to obey the rate law: rate = k [N 2 O 5 ]. Does the reaction mechanism consist of only a single bimolecular step? 1. Yes, it must 2. It might, but you cannot tell from the information given. 3. No, it cannot correct Explanation: Single-step reactions have rate laws with order equal to the stiochiometry of the reac- tant(s) in the balanced equation. 003 10.0 points Generally, reaction rates increase when the temperature of the system is raised because 1. the concentration of reactants increases as the temperature increases. 2. the bond energy of the reactants decreases as the temperature increases. 3. the bond energy of the products decreases as the temperature increases. 4. the kinetic energy (and thus the speed) of 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, therefore a faster rate of reaction. 004 12.0 points Calculate the potential for 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 first 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 parenleftbigg [Fe 2+ ] [Pb 2+ ] parenrightbigg = 0 . 314 V - 0 . 05916 2 log parenleftbigg 10 3 10 5 parenrightbigg = 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. oxidation may take place without reduc- tion. 3. a substance is oxidized and gains control over electrons.
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