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Homework 10

# Homework 10 - ycs73 Homework 10 Fakhreddine(53605 This...

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ycs73 – Homework 10 – Fakhreddine – (53605) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Under certain conditions, the reaction 3 A + 2 B 4 C was observed to proceed at a rate of 0 . 00223 M · s 1 . What was the corresponding rate of change in reactant A? 1. 0 . 001115 M · s 1 2. - 0 . 00446 M · s 1 3. 0 . 00446 M · s 1 4. 0 . 000743333 M · s 1 5. - 0 . 001115 M · s 1 6. - 0 . 00223 M · s 1 7. 0 . 00223 M · s 1 8. - 0 . 000743333 M · s 1 9. - 0 . 00669 M · s 1 correct 10. 0 . 00669 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 10.0 points The reaction N 2 + 3 H 2 2 NH 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. 2 . 25 × 10 1 moles/sec 2. The question cannot be answered from the information given; we need to know the rate law for the reaction. 3. 7 . 5 × 10 3 moles/sec 4. 1 . 125 × 10 2 moles/sec correct 5. The question cannot be answered from the information given; we need to know the volume of the container. 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 · 0 . 150 mol 20 s = 0 . 00375 mol / s The rate of disappearance of H 2 is Rate = - 1 3 · Δ[H 2 ] Δ t 0 . 00375 mol / s = - 1 3 · Δ[H 2 ] Δ t Δ[H 2 ] Δ t = - 0 . 01125 mol / s Thus H 2 disappears at a rate of 1 . 125 × 10 2 mol/s. 003 10.0 points The following data were collected for the re- action 2 A + B 2 + C D Initial Initial Initial Initial [A] [B 2 ] [C] rate (M) (M) (M) (M/s) 1 0 . 01 0 . 01 0 . 01 1 . 250 × 10 3 2 0 . 02 0 . 01 0 . 01 5 . 000 × 10 3 3 0 . 03 0 . 01 0 . 05 1 . 125 × 10 4 4 0 . 04 0 . 02 0 . 01 8 . 000 × 10 4 Find the rate law.

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ycs73 – Homework 10 – Fakhreddine – (53605) 2 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] 2 [B 2 ] 2 4. Rate = (1 . 25 × 10 7 ) [A] [B 2 ] [C] 2 5. Rate = (1 . 25 × 10 11 ) [A] 2 [B 2 ] 2 correct 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 = parenleftbigg 0 . 02 0 . 01 parenrightbigg x parenleftbigg 0 . 01 0 . 01 parenrightbigg y parenleftbigg 0 . 01 0 . 01 parenrightbigg 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 = parenleftbigg 0 . 04 0 . 01 parenrightbigg 2 parenleftbigg 0 . 02 0 . 01 parenrightbigg y parenleftbigg 0 . 01 0 . 01 parenrightbigg z 64 = 4 2 · 2 y 2 y = 64 16 = 4 = 2 2 y = 2 Rate 3 Rate 1 = k [A] 2 3 [B 2 ] y 3 [C] z 3 k [A] 2 1 [B 2 ] y 1 [C] z 1 1 . 125 × 10 4 1 . 250 × 10 3 = parenleftbigg 0 . 03 0 . 01 parenrightbigg 2 parenleftbigg 0 . 01 0 . 01 parenrightbigg y parenleftbigg 0 . 05 0 . 01 parenrightbigg z 9 = 9(5) z 5 z = 5 0 = 1 z = 0 k = Rate [A] 2 [B] 2 [C] 0 = 1 . 250 × 10 3 M / s (0 . 01) 4 M 4 = 1 . 250 × 10 11 M 3 · s 1 Rate = 1.250 × 10 11 [A] 2 [B] 2 004 10.0 points What is the rate of the reaction 4 NH 3 + 5 O 2 4 NO + 6 H 2 O when 0.0500 mol/L of NH 3 is being consumed per second?
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