solutionfinal_pdf - Version 074 – Final – Mccord...

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Version 074 – Final – Mccord – (52450) 1 This print-out should have 52 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. McCord CH302 1pm This exam is only for McCord’s MWF 1pm CH302 class. R = 8 . 314 J/mol · K R = 0 . 08206 L atm/mol · K R = 62 . 36 L · torr/mol · K 1 L · atm = 101.325 J Δ G = Δ H - T Δ S Δ T f = k f · m Δ T b = k b · m P A = x A · P A , pure Π = cRT ln parenleftbigg P 2 P 1 parenrightbigg = Δ H vap R parenleftbigg 1 T 1 - 1 T 2 parenrightbigg ln parenleftbigg K 2 K 1 parenrightbigg = Δ H rxn R parenleftbigg 1 T 1 - 1 T 2 parenrightbigg ln parenleftbigg k 2 k 1 parenrightbigg = E a R parenleftbigg 1 T 1 - 1 T 2 parenrightbigg pH = -log[H + ] K w =[H + ][OH ] K w = 1 . 0 × 10 14 at 25 C pH = p K a + log [ A ] [ HA ] Δ G = - nFE Δ G = - RT ln K anode | solution || solution | cathode E = E - RT nF ln Q E = E - 0 . 0257 n ln Q E = E - 0 . 05916 n log Q I · t n · F = moles ln parenleftbigg [A] 0 [A] parenrightbigg = kt t 1 / 2 = ln 2 ak 1 [A] - 1 [A] 0 = kt t 1 / 2 = 1 ak [A] 0 [A] 0 - [A] = kt t 1 / 2 = [A] 0 2 ak 001 10.0 points 2 . 37 moles of a weak electrolyte is dissolved into 588 grams of water. The freezing point of the solution is - 9 . 52111 C. What is the percent ionization for this substance in this solution? K f = 1 . 86 C /m for water. Assume that the electrolyte is a simple 1:1 ratio. 1. 50.0 2. 34.0 3. 25.0 4. 29.0 5. 42.0 6. 24.0 7. 43.0 8. 41.0 9. 33.0 10. 27.0 Correct answer: 27%. Explanation: T f = - 9 . 52111 C T 0 f = 0 C m H 2 O = 588 g n = 2 . 37 mol K f = 1 . 86 C /m Δ T f = T 0 f - T f = 0 . 0 C - ( - 9 . 52111 C) = 9 . 52111 C Δ T f = K f · m eff m eff = Δ T f K f = 9 . 52111 C 1 . 86 C /m = 5 . 11888 m
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Version 074 – Final – Mccord – (52450) 2 initial molality of solution = 2 . 37 mol MX 0 . 588 kg water = 4 . 03061 m MX M + + X ini, m 4 . 03061 0 0 Δ, m - x x x fin, m 4 . 03061 - x x x Total molality = 4 . 03061 m - x + x + x = 4 . 03061 m + x m eff = 4 . 03061 m + x = 5 . 11888 m x = 1 . 08827 m % ionization = x m ini × 100% = 1 . 08827 m × 100% 4 . 03061 m = 27% 002 10.0 points The graph describes the energy profile of a reaction. A B Time Energy (kJ) 50 300 400 What are the values for Δ H and E a , respec- tively, for the reaction in the direction writ- ten? 1. 250 kJ, 100 kJ 2. - 250 kJ, - 100 kJ 3. 250 kJ, 350 kJ correct 4. - 250 kJ, 100 kJ 5. - 250 kJ, 350 kJ Explanation: E a Δ H A B Time Energy (kJ) 50 300 400 Δ H = 300 kJ - 50 kJ = 250 kJ E a = 400 kJ - 50 kJ = 350 kJ 003 10.0 points The following figure represents the progress of a given reaction at 298 K. G A B C D E rxn progress At point B on this figure, what is the relation- ship of Q to K ? 1. Cannot be determined 2. Q > K 3. Q = K 4. Q < K correct Explanation: Point B is on the reactants-heavy side of equilibrium, so Q is less than K . Note also that d G (slope) is negative here which means the reaction would be spontaneous in the for- ward direction. Spontaneous in a forward direction corresponds to Q being less than K . 004 10.0 points
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Version 074 – Final – Mccord – (52450) 3 The decomposition of cyclobutane is a first- order reaction. At a certain temperature, the half-life for this reaction is 137 seconds. What fraction of a sample of cyclobutane would be left after 685 seconds at this temperature?
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