exam 4 - Version 115 – Exam 4 – Fakhreddine –(52395...

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Version 115 – Exam 4 – Fakhreddine – (52395) 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 5.0 points The reaction between nitrogen dioxide and carbon monoxide is thought to occur by the following mechanism: 2 NO 2 (g) NO 3 (g) + NO(g) k 1 , slow NO 3 (g) + CO(g) NO 2 (g) + CO 2 (g) k 2 , fast What is the rate law for this mechanism? 1. rate = k 2 [NO 3 ] [CO] 2. rate = k 1 k 2 [NO 2 ] 2 [CO] 3. rate = k 1 [NO 3 ] [NO] 4. rate = k 1 [NO 2 ] 2 correct 5. rate = k 1 k 2 [NO 2 ] 2 [CO] Explanation: 002 5.0 points For the cell diagram Pt | H 2 (g) , H + (aq) || Cu 2+ (aq) | Cu(s) , which reaction occurs at the anode? 1. 2 H + (aq) + 2 e H 2 (g) 2. 2 H + (aq) + Cu(s) H 2 (g) + Cu 2+ (aq) 3. H 2 (g) 2 H + (aq) + 2 e correct 4. Cu 2+ (aq) + 2 e Cu(s) 5. Cu(s) Cu 2+ (aq) + 2 e Explanation: 003 10.0 points The following data were collected for the re- action 2 A + B 2 + C D Initial Initial Initial Initial [A] [B 2 ] [C] rate (M) (M) (M) (M/s) 1 0 . 01 0 . 01 0 . 01 1 . 250 × 10 3 2 0 . 02 0 . 01 0 . 01 5 . 000 × 10 3 3 0 . 03 0 . 01 0 . 05 1 . 125 × 10 4 4 0 . 04 0 . 02 0 . 01 8 . 000 × 10 4 Find the rate law. 1. Rate = (1 . 25 × 10 7 ) [A] [B 2 ] [C] 2 2. Rate = (1 . 25 × 10 9 ) [A] 2 [B 2 ] 3. Rate = (1 . 25 × 10 11 ) [A] 2 [B 2 ] 2 correct 4. Rate = (1 . 25 × 10 9 ) [A] [B 2 ] 2 5. Rate = (1 . 25 × 10 7 ) [A] 2 [B 2 ] 2 Explanation: Rate = k [A] x [B 2 ] y [C] z Rate 2 Rate 1 = k [A] x 2 [B 2 ] y 2 [C] z 2 k [A] x 1 [B 2 ] y 1 [C] z 1 5 . 000 × 10 3 1 . 250 × 10 3 = parenleftbigg 0 . 02 0 . 01 parenrightbigg x parenleftbigg 0 . 01 0 . 01 parenrightbigg y parenleftbigg 0 . 01 0 . 01 parenrightbigg z 4 = 2 2 = 2 x x = 2 Rate 4 Rate 1 = k [A] 2 4 [B 2 ] y 4 [C] z 4 k [A] 2 1 [B 2 ] y 1 [C] z 1 8 . 000 × 10 4 1 . 250 × 10 3 = parenleftbigg 0 . 04 0 . 01 parenrightbigg 2 parenleftbigg 0 . 02 0 . 01 parenrightbigg y parenleftbigg 0 . 01 0 . 01 parenrightbigg z 64 = 4 2 · 2 y 2 y = 64 16 = 4 = 2 2 y = 2 Rate 3 Rate 1 = k [A] 2 3 [B 2 ] y 3 [C] z 3 k [A] 2 1 [B 2 ] y 1 [C] z 1 1 . 125 × 10 4 1 . 250 × 10 3 = parenleftbigg 0 . 03 0 . 01 parenrightbigg 2 parenleftbigg 0 . 01 0 . 01 parenrightbigg y parenleftbigg 0 . 05 0 . 01 parenrightbigg z 9 = 9(5) z 5 z = 5 0 = 1 z = 0
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Version 115 – Exam 4 – Fakhreddine – (52395) 2 k = Rate [A] 2 [B] 2 [C] 0 = 1 . 250 × 10 3 M / s (0 . 01) 4 M 4 = 1 . 250 × 10 11 M 3 · s 1 Rate = 1.250 × 10 11 [A] 2 [B] 2 004 10.0 points Consider the voltaic cell Pt | Sn 4+ (0.0010 M), Sn 2+ (0.10 M) || Ag + (0.010 M) | Ag Sn 4+ + 2 e Sn 2+ E 0 = 0 . 15 V Ag + + 1 e Ag(s) E 0 = 0 . 8 V The experimental cell potential for the cell is approximately 1. 0.65 V. 2. 0.62 V. 3. 0.68 V. 4. 0.59 V. correct 5. 0.72 V. Explanation: The species in contact with the electrode surfaces are Sn 2+ and Ag + . Note under standard conditions: Sn 2+ Sn 4+ + 2 e E 0 anode = - 0 . 15 V Ag + + 1 e Ag(s) E 0 cathode = 0 . 8 V E 0 cell = 0 . 65 V But these aren’t standard conditions, as the concentrations are not at 1 M. Use the Nernst equation to calculate E ’s at these concentra- tions: E = E 0 - 0 . 0592 V n log parenleftbigg [Red] y [Ox] x parenrightbigg For Sn 4+ + 2 e Sn 2+ , E = 0 . 15 V - 0 . 0592 V 2 log parenleftbigg 0 . 10 M 0 . 0010 M parenrightbigg = 0 . 0908 V For Ag + + 1 e Ag(s), E = 0 . 8 V - 0 . 0592 V 1 log parenleftbigg 1 0 . 010 parenrightbigg = 0 . 6816 V So, using the appropriate values of E , Sn 2+ Sn 4+ + 2 e E 0 anode = - 0 . 0908 V Ag + + 1 e Ag(s) E 0 cathode = + 0 . 6816 V E cell = 0 . 5908 V 005 5.0 points Which of the following is the strongest oxidiz- ing agent?
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