Homework 12-solutions

1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 1.0points The reaction NO 2 (g) + CO(g) → NO(g) + CO 2 (g) has been found to be second order with respect to NO 2 and zero order with re- spect to CO. At a certain temperature, the rate constant is found experimentally to be 3 . 0 × 10 − 5 L mol · s . What is the rate of formation of CO 2 at this temperature when the concen- tration of NO 2 is 6 . 7 mol / L, that of CO is 5 . 4 mol / L, that of NO is 3 . 6 mol / L, and that of CO 2 is 2 . 4 mol / L? 1. 0.0008748 2. 0.0003468 3. 0.0005547 4. 0.0013467 5. 0.0017787 6. 0.0014283 7. 0.00027 8. 0.0017328 9. 0.0013872 10. 0.0018252 Correct answer: 0 . 0013467 M · s − 1 . Explanation: [NO 2 ] = 6 . 7 mol / L k = 3 . 0 × 10 − 5 L mol · s [CO] = 5 . 4 mol / L [NO] = 3 . 6 mol / L [CO 2 ] = 2 . 4 mol / L Rate = Δ[CO 2 ] Δ t = k [NO 2 ] 2 = 3 . 0 × 10 − 5 L mol · s · (6 . 7 mol / L) 2 = 0 . 0013467 M · s − 1 002 1.0points HINT: Use the units of k to figure out the order of the reaction. For the reaction 2 NO 2 → 2 NO + O 2 Rate = 1 . 4 × 10 − 10 / M · s [NO 2 ] 2 at 25 ◦ C. If 1 mol of NO 2 are initially present in a sealed 1.00 liter vessel at 25.0 ◦ C, how many grams of NO 2 remain after 52 years? 1. 41.983 2. 52.5439 3. 67.9007 4. 48.0098 5. 29.1719 6. 23.4826 7. 36.3244 8. 31.5318 9. 40.7306 10. 43.1878 Correct answer: 31 . 5318 g. Explanation: a = 2 k = 1 . 4 × 10 − 10 M − 1 · s − 1 t = 52 y [NO 2 ] = 1 mol/L The rate law indicates a second order reac- tion. Since the relationship of interest involve [NO 2 ] and time, use the integrated rate law: 1 [NO 2 ] t − 1 [NO 2 ] 0 = akt 1 [NO 2 ] t = akt + 1 [NO 2 ] 0 = 2 ( 1 . 4 × 10 − 10 M − 1 · s − 1 ) × (52 years) × 365 d y × 24 h d × 3600 s h + 1 . 00 L 1 mol = 1 . 45916 M − 1 [NO 2 ] t = 0 . 685324 M = 0 . 685324 mol L × 1 . 00 L × 46 . 0055 g mol = 31 . 5318 g 003 1.0points The rate constant for the equation 2 C 2 F 4 → C 4 F 8

2 is 0.0440 M − 1 s − 1 . We start with 0.266 mol C 2 F 4 in a 5.00-liter container, with no C 4 F 8 initially present. What will be the concentra- tion of C 2 F 4 after 1.00 hour ? 1. 0.001417 2. 0.00298 3. 0.0008372 4. 0.001487 5. 0.0006875 6. 0.000996 7. 0.00249 8. 0.0007798 9. 0.0007934 10. 0.001552 Correct answer: 0 . 00298 M. Explanation: k = 0 . 0440 M − 1 s − 1 t = 1 . 00 hour a = 2 [C 2 F 4 ] ini = 0 . 266mol 5 . 00 L = 0.0532 M Based on the units of the rate constant, we know to use the second order integrated rate law, where a is the coefficient of the reactants: 1 [A] − 1 [A] 0 = akt 1 [A] = 1 [A] 0 + akt 1 [A] = 1 0 . 0532 M + bracketleftBig (2) (0 . 0440 M − 1 s − 1 ) × (3600 s) bracketrightBig [ A ] = 0 . 00298 M 004 1.0points Consider the first order reaction A → products where 25% of A disappears in 24 seconds. What is the half life of this reaction? 1. t 1 / 2 = 12 s 2. t 1 / 2 = 58 s correct 3. t 1 / 2 = 36 s 4. t 1 / 2 = 48 s 5. t 1 / 2 = 40 s 6. t 1 / 2 = 25 s Explanation: a = 1 t = 24 s [A] 0 = 100 [A] t = 75 ln parenleftbigg [A] 0 [A] t parenrightbigg = akt k = 1 at ln parenleftbigg [A] 0 [A] t parenrightbigg = 1 24 s ln parenleftbigg 100 75 parenrightbigg = 1 . 199 × 10 − 2 s − 1 t 1 / 2 = ln 2 k = 58 s 005 1.0points NOTE: If the concentration of A has fallen by a factor of 8, it means that A is now only 1/8 of its original concentration. A first order reaction, where A → products , has a rate constant of 1 . 56 × 10 7 s − 1 . At some time, a concentration of 1 . 06 × 10 − 6 M of species A is introduced into the reactor. How long does it take for the concentration of A to fall by a factor of 8? 1. t = 88 nsec = 8 . 8 × 10 − 8 sec 2. t = 5.5 nsec = 5 . 5 × 10 − 9 sec 3. t = 352 nsec = 3 . 52 × 10 − 7 sec 4. t = 133 nsec = 1 . 33 × 10 − 7 sec correct 5. t = 44 nsec = 4 . 4 × 10 − 8 sec Explanation: [A] 0 = 1.06 × 10 − 6 M [A] = 1 8 [A] 0