Exam 1 - Version 112 – Exam 1 – Holcombe – (52460) 1...

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Unformatted text preview: Version 112 – Exam 1 – Holcombe – (52460) 1 This print-out should have 35 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. Helpful constants and conversions 1 atm = 760 torr = 760 mm Hg 0 degC = 273.15 K R = 8.314 J/K*mol R = 0.082 L*atm/K*mol 001 10.0 points Given that CO 2 (g) reacts with C(s) via the reaction C(s) + CO 2 (g) ⇀ ↽ 2 CO(g) and K p = 1.90 atm, what is the equilibrium partial pressure of CO 2 if 1.00 atm of CO 2 is placed in a vessel with PURE SOLID CAR- BON? (Note: There was no CO initially.) 1. 0.85 atm 2. 0.55 atm 3. 0.60 atm 4. 0.51 atm correct 5. 0.43 atm Explanation: K p = 1 . 9 atm C(s) + CO 2 (g) ⇀ ↽ 2 CO(g) ini, atm 1 Δ, atm- x 2 x eq, atm 1- x 2 x K p = P 2 CO P CO 2 = 1 . 9 (2 x ) 2 1- x = 1 . 9 4 x 2 + 1 . 9 x- 1 . 9 = 0 Therefore x =- 1 . 9 ± radicalbig (1 . 9) 2 + 16(1 . 9) 8 = 0 . 491476 P CO 2 = 1- x = 0 . 508524 atm 002 10.0 points Catalase (a liver enzyme) dissolves in water. A 14 mL solution containing 0 . 166 g of cata- lase exhibits an osmotic pressure of 1 . 2 Torr at 20 ◦ C. What is the molar mass of catalase? 1. 1 . 81 × 10 5 g / mol correct 2. 1 . 33 × 10 5 g / mol 3. 2 . 3 × 10 5 g / mol 4. 1 . 4 × 10 5 g / mol 5. 2 . 11 × 10 5 g / mol 6. 2 . 81 × 10 5 g / mol Explanation: V = 14 mL = 0 . 014 L m = 0 . 166 g T = 20 ◦ C + 273 = 293 K Π = 1 . 2 Torr R = 0 . 08206 L · atm / K / mol Π = n RT V = m MW RT V M = mRT V Π MW = (0 . 166 g) (0 . 08206 L · atm / K / mol) (0 . 014 L)(1 . 2 Torr) × (293 K) parenleftbigg 760 Torr 1 atm parenrightbigg = 1 . 80556 × 10 5 g / mol . 003 10.0 points For the process H 2 O(g) → H 2 O( ℓ ) Δ H would be 1. positive. 2. zero. 3. negative. correct Version 112 – Exam 1 – Holcombe – (52460) 2 Explanation: Condensation releases energy (is exother- mic), so Δ H will be negative. 004 10.0 points What is the form of the equilibrium constant for the process PbO(s) + CO 2 (g) ⇀ ↽ PbCO 3 (s) ? P species means the pressure of species. 1. K p = 1 P CO 2 correct 2. K p = [PbCO 3 ] [PbO] P CO 2 3. K c = [PbCO 3 ] 2 [CO 2 ] [PbO] 4. K p = P PbCO 3 P CO 2 5. No other choice is correct. Explanation: Solids are not included in the K expression. 005 10.0 points Suppose the reaction A ⇀ ↽ B has an equilibrium constant of 1.0 and the ini- tial concentrations of A and B are 0.5 M and 0.0 M, respectively. Which of the following is the correct value for the final concentration of A? 1. 1.00 M 2. None of these is correct. 3. 0.250 M correct 4. 0.500 M 5. 1.50 M Explanation: K = 1 . [A] ini = 0 . 5 M [B] ini = 0 M A ⇀ ↽ B ini, M 0.5 0.0 Δ, M- x x eq, M . 5- x x K = [B] [A] = 1 . x . 5- x = 1 ....
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This note was uploaded on 04/22/2011 for the course CHEM 302 taught by Professor Mccord during the Spring '10 term at University of Texas.

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Exam 1 - Version 112 – Exam 1 – Holcombe – (52460) 1...

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