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Unformatted text preview: Panjwani, Sameer Test 4 Due: Aug 9 2007, 1:00 pm Inst: Fakhreddine/Lyon 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Consider the conversion of a substance from solid to liquid: SOLID LIQUID At one atmosphere pressure and at the melting point of the substance, 1. H = 0 for the process. 2. E = 0 for the process. 3. S = 0 for the process. 4. No answer is correct. 5. G = 0 for the process. correct Explanation: At standard phase change points G = 0 because the process is in equilibrium. 002 (part 1 of 1) 10 points Consider the equation NH 4 Br(s) NH 3 (g) + HBr(g) carefully, and think about the sign of S for the reaction it describes. H = +188.3 kJ. Which response describes the thermodynamic spontaneity of the reaction? 1. The reaction is not spontaneous at any temperatures. 2. The reaction is spontaneous only at rela- tively high temperatures. correct 3. The reaction is spontaneous at all tem- peratures. 4. All responses are correct. 5. The reaction is spontaneous only at rela- tively low temperatures. Explanation: Entropy ( S ) is high for systems with high degrees of freedom, disorder or randomness and low for systems with low degrees of free- dom, disorder or randomness. S(g) > S ( ) > S (s) . A reaction is spontaneous only when G is negative. H is positive for this reaction and S is positive. G = H- T S = (+)- T (+) = (+)- T G will be negative (spontaneous reactions) only at high values of T . 003 (part 1 of 1) 10 points 0.500 mole of N 2 (g) reacts with 1.50 moles H 2 (g) to produce NH 3 (g): N 2 (g) + 3H 2 (g) 2NH 3 (g) If this reaction is carried out in a system against a constant 0.75 atm pressure ( i.e. , a piston) at 0 C, calculate the magnitude of the P V work. R = 8.314 J/mol K. 1. 4 . 54 10 3 J 2. 22.7 J 3. 22 . 4 10 3 J 4. 22.4 J 5. 2 . 27 10 3 J correct Explanation: n N 2 = 0.5 mol n H 2 = 1.5 mol P = 0.75 atm T = 0 C = 273 K The equation states that N 2 (g) + 3H 2 (g) 2NH 3 (g) . Note, however, that in fact only 0.500 moles of N 2 and 1.50 moles of H 2 are used. Thus only Panjwani, Sameer Test 4 Due: Aug 9 2007, 1:00 pm Inst: Fakhreddine/Lyon 2 1 mol of NH 3 is produced in this particular reaction. On the reactant side of the reaction, we then have 2 moles of gas, but only 1 mole of gaseous product is formed. The amount of work that is done is given by w =- P V , but, in this case, since P V = n R T , w =- n R T . Substitute- 1 in for n (equal to the change in the number of moles of gas) and R and T to get the correct answer: w =- (- 1) 8 . 314 J mol K (273K) = 2269 . 72 J / mol Note that since only 1 mol of product was formed, the answer works out to 2269 . 72 J....
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This note was uploaded on 03/18/2008 for the course CH 301 taught by Professor Fakhreddine/lyon during the Fall '07 term at University of Texas at Austin.
- Fall '07