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Hwk 11 Solutions

# Hwk 11 Solutions - kapur(rk7259 Homework 11 Lyon(52380 This...

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kapur (rk7259) – Homework 11 – Lyon – (52380) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. This HW assignment is due Tuesday, April 28, by 11PM. 001 10.0 points For the reaction 2 H 2 (g) + O 2 (g) 2 H 2 O(g) find the value for the work done at 300 K. 1. 7.5 kJ 2. - 2 . 5 kJ 3. 2.5 kJ correct 4. - 7 . 5 kJ Explanation: T = 300 K R = 8 . 314 J mol · K w = - P Δ V = - Δ nRT = - (2 - 3) mol parenleftbigg 8 . 314 J mol · K parenrightbigg (300 K) = 2494 . 2 J 002 10.0 points For the methanol combustion reaction 2 CH 3 OH( ) + 3 O 2 (g) -→ 2 CO 2 (g) + 4 H 2 O(g) estimate the amount of P Δ V work done and tell whether the work was done on or by the system. Assume a temperature of 27 C. 1. 7.5 kJ, work done on the system 2. 7.5 kJ, work done by the system correct 3. 2.5 kJ, work done by the system 4. 2.5 kJ, work done on the system 5. No work is done in this reaction. Explanation: T = 27 C + 273 = 300 K Considering only moles of gas, Δ n = n f - n i = (2 + 4) - 3 = 3 . w = - Δ nRT = - (3 mol) (8 . 314 J / mol · K) (300 K) = - 7500 J = - 7 . 5 kJ The system expands because Δ n is positive, so the system does the work on the surround- ings. Also, when w is negative, work is done by the system. 003 10.0 points Heat flow is considered positive when heat flows (into, out of) a system; work is consid- ered positive when work is done (by, on) a system. 1. out of; by 2. into; by 3. into; on correct 4. out of; on Explanation: Heat flow q is considered positive when heat flows into a system. Work w is considered positive when work is done on a system. 004 10.0 points Which one of the following statements is FALSE? For a reaction carried out at con- stant pressure in an open container 1. the heat absorbed by the system can be called Δ H . 2. the work done on the system can be set equal to - P Δ V . 3. the work done on the system can be set equal to - V Δ P . correct 4. the work done on the system can be set

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kapur (rk7259) – Homework 11 – Lyon – (52380) 2 equal to - n ) RT , where Δ n is the num- ber of moles of gaseous products minus the number of moles of gaseous reactants. Explanation: For P = const, w = - P Δ V = - n ) RT . 005 10.0 points The pressure-volume work done by an ideal gaseous system at constant volume is 1. - V Δ P 2. zero correct 3. - Δ P P 4. q 5. - Δ E Explanation: When V = constant, nothing moves through a distance and therefore no work is done: w = 0.
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