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Makeup Exam 1, Spring 2011

Makeup Exam 1, Spring 2011 - Version 025 Make up 1...

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Version 025 – Make up 1 – laude – (51635) 1 This print-out should have 30 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. Constants R = 8 . 314 JK 1 mol 1 Definitions Q [C] c [D] d [A] a [B] b G H - TS K w [H + ][OH ] pH ≡- log [H + ] pOH ≡- log [OH ] Equations Δ G = Δ H - T Δ S ln parenleftbigg P 2 P 1 parenrightbigg = Δ H vap R parenleftbigg 1 T 1 - 1 T 2 parenrightbigg Π = MRT Δ T b = iK b m Δ T f = iK f m P i = P i χ i P = summationdisplay i P i q = mc Δ T q = m Δ H Δ G = - RT ln K K = e Δ G RT ln K 2 K 1 = Δ H R parenleftbigg 1 T 1 - 1 T 2 parenrightbigg Q = K [at equil.] [H + ] = ( K a C a ) 1 / 2 [OH ] = ( K b C b ) 1 / 2 [H + ] = K a ( C a /C b ) pH = p K a + log( C b /C a ) [OH ] = K b ( C b /C a ) pOH = p K b + log( C a /C b ) 001 6.0 points In the derivation of the Clausius-Clapeyron equation, the Δ S vap term is present in early steps but isn’t present in the final equation. Why is this? 1. The Δ S vap term actually is present in the final equation. 2. We assume that Δ S vap is independent of temperature and are able to ignore it. 3. We assume ideal behavior for the gas, and the Δ S vap term reduces to a constant and is eliminated. correct 4. We assume that Δ S vap = Δ S vap , and we are able to eliminate the term altogether. Explanation: Because we assume that the gas behaves ideally, Δ S vap is the same regardless of the gas in question, and taking its base e expo- nent results in constant which can be readily eliminated. 002 6.0 points How many phase transitions occur by fol- lowing the steps described below? step 1: At 600 K going from 90 atm to 20 atm step 2: At 20 atm going from 600 K to 150 K step 3: At 150 K going from 20 atm to 65 atm 0 200 400 600 800 1000 0 20 40 60 80 100 Pressure, atm Temperature, K 1. 3 phase boundaries are crossed 2. 1 phase boundaries are crossed 3. 2 phase boundaries are crossed correct
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Version 025 – Make up 1 – laude – (51635) 2 4. 4 phase boundaries are crossed Explanation: 003 6.0 points What would be the pOH of a solution of Sr(OH) 2 (strontium hydroxide) prepared by dissolving 122 g of the base into 10 L of pure water (H 2 O)? 1. 13.3 2. 13 3. 0.7 correct 4. 1 Explanation: 122 g Sr(OH) 2 × 1 mol 122 g = 1 mol Sr(OH) 2 1 mol Sr(OH) 2 10 L H 2 O = 0 . 1 M Sr(OH) 2 Sr(OH) 2 (aq) Sr 2+ (aq) + 2 OH (aq) [OH ] = C b · 2 = 0 . 1M · 2 = 0 . 2M pOH = - log[OH ] = - log(0 . 2) = 0 . 7 004 6.0 points For a solution with an [OH ] of 10 8 M, what would be the value of [H + ] , pH and pOH, respectively? 1. 10 14 M, 8, 6 2. 10 6 M, 6, 14 3. 10 6 M, 6, 8 4. 10 14 M, 6, 8 5. 10 6 M, 8, 6 6. 10 6 M, 6, 8 correct Explanation: [H + ] = 10 14 [OH ] = 10 14 10 8 = 10 6 pH = - log 10 ([H + ]) pOH = - log 10 ([OH ]) = 14 - pH 005 6.0 points Molar Gibbs energy, G m Temperature, T pure liquid solvent solid Referring to the graph above, and using your knowledge of Δ H solution , complete the following statement concerning freezing point depression: The molar (enthalpy/entropy) of a pure liquid is always (greater/less) than that of a dilute solution. This lowers the value of G m for the solvent at all temperatures, thus lowering the freezing point.
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