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Unformatted text preview: Martin, Robert – Homework 10 – Due: Apr 18 2007, 11:00 pm – Inst: Deb Walker 1 This printout should have 18 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. The due time is Central time. Read all of the options before choosing one. There may be cases in which more than one response appears to be true. PICK THE RE SPONSE THAT BEST answers the question asked. 001 (part 1 of 1) 3 points Under certain conditions, the reaction 3A + 2B → 4C was observed to proceed at a rate of . 00235 M · s 1 . What was the corresponding rate of change in reactant A? 1. . 000783333 M · s 1 2. . 001175 M · s 1 3. . 0047 M · s 1 4. . 0047 M · s 1 5. . 00705 M · s 1 6. . 000783333 M · s 1 7. . 001175 M · s 1 8. . 00235 M · s 1 9. . 00705 M · s 1 correct 10. . 00235 M · s 1 Explanation: Reactant A is reacting at 3 times the stated rate of the overall reaction. A is also a reac tant, so the concentration is DECREASING with time, which corresponds to a NEGA TIVE rate. 002 (part 1 of 1) 3 points The reaction N 2 + 3H 2 → 2NH 3 is proceeding under conditions that 0.150 moles of NH 3 are being formed every 20 sec onds. What is the rate of disappearance of H 2 ? 1. 7 . 5 × 10 3 moles/sec 2. 2 . 25 × 10 1 moles/sec 3. The question cannot be answered from the information given; we need to know the rate law for the reaction. 4. The question cannot be answered from the information given; we need to know the volume of the container. 5. 1 . 125 × 10 2 moles/sec correct Explanation: n NH 3 = 0.150 mol t = 20 s The rate of appearance of NH 3 is Rate = 1 2 Δ[NH 3 ] Δ t We can consider this to be Rate = 1 2 · . 150 mol 20 s = 0 . 00375 mol / s The rate of disappearance of H 2 is Rate = 1 3 · Δ[H 2 ] Δ t . 00375 mol / s = 1 3 · Δ[H 2 ] Δ t Δ[H 2 ] Δ t = . 01125 mol / s Thus H 2 disappears at a rate of 1 . 125 × 10 2 mol/s. 003 (part 1 of 1) 3 points The following data were collected for the re action 2 A + B 2 + C → D Martin, Robert – Homework 10 – Due: Apr 18 2007, 11:00 pm – Inst: Deb Walker 2 Initial Initial Initial Initial [A] [B 2 ] [C] rate (M) (M) (M) (M/s) 1 . 01 . 01 . 01 1 . 250 × 10 3 2 . 02 . 01 . 01 5 . 000 × 10 3 3 . 03 . 01 . 05 1 . 125 × 10 4 4 . 04 . 02 . 01 8 . 000 × 10 4 Find the rate law. 1. Rate = (1 . 25 × 10 9 ) [A] 2 [B 2 ] 2. Rate = (1 . 25 × 10 9 ) [A] [B 2 ] 2 3. Rate = (1 . 25 × 10 7 ) [A] [B 2 ] [C] 2 4. Rate = (1 . 25 × 10 11 ) [A] 2 [B 2 ] 2 correct 5. Rate = (1 . 25 × 10 7 ) [A] 2 [B 2 ] 2 Explanation: Rate = k [A] x [B 2 ] y [C] z Rate 2 Rate 1 = k [A] x 2 [B 2 ] y 2 [C] z 2 k [A] x 1 [B 2 ] y 1 [C] z 1 5 . 000 × 10 3 1 . 250 × 10 3 = µ . 02 . 01 ¶ x µ . 01 . 01 ¶ y µ . 01 . 01 ¶ z 4 = 2 2 = 2 x x = 2 Rate 4 Rate 1 = k [A] 2 4 [B 2 ] y 4 [C] z 4 k [A] 2 1 [B 2 ] y 1 [C] z 1 8 . 000 × 10 4 1 . 250 × 10 3 = µ . 04 . 01 ¶ 2 µ . 02 . 01 ¶ y µ . 01 . 01 ¶ z 64 = 4 2 · 2 y 2 y = 64 16 = 4 = 2 2 y = 2...
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 Spring '08
 LIU
 Rate equation, Reaction rate constant, Deb Walker

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