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HW03-solutions

# HW03-solutions - kamalska(mk23835 HW03 Quigley(104001 This...

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kamalska (mk23835) – HW03 – Quigley – (104001) 1 This print-out should have 30 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points For the reaction A products , the following data were collected. time, s [A] t , M 0 1.00 1.00 0.430 2.00 0.270 3.00 0.200 4.00 0.160 What is the half-life for this reaction? 1. 1.08 s 2. 0.752 s correct 3. 0.922 s 4. 0.521 s Explanation: 002 10.0 points Consider the concentration-time dependence as shown below for two first order reactions. [A] 0 Molar Concentration of Reactant Concentration -→ Time -→ Which reaction has the greatest t 1 / 2 ? 1. the reaction represented by the lower curve 2. the reaction represented by the upper curve correct 3. Unable to determine Explanation: 003 10.0 points The reaction 2 HI H 2 + I 2 has rate constants k 1 = 9 . 7 × 10 6 M 1 s 1 and k 2 = 0 . 097 M 1 s 1 at T 1 = 326 . 85 C and T 2 = 526 . 85 C. What is the activation energy of this reaction? 1. 7 . 93 × 10 3 J 2. 2 . 86 × 10 4 J 3. 1 . 84 × 10 5 J correct 4. 6 . 59 × 10 4 J 5. 7 . 16 × 10 7 J Explanation: k 1 = 9 . 7 × 10 6 M 1 s 1 k 2 = 0 . 097 M 1 s 1 T 1 = 326 . 85 C + 273 . 15 = 600 K T 2 = 526 . 85 C + 273 . 15 = 800 K R = 8 . 314 J · mol 1 K 1 The Arrhenius equation gives ln K = ln A - E a RT , which is of the form y = c - mx , so - E a R is the slope of the line ln K vs 1 T . This slope can be estimated from - E a R = ln k 1 - ln k 2 1 T 1 - 1 T 2 = ln(9 . 7 × 10 6 ) - ln(0 . 097) 1 600 K - 1 800 K

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kamalska (mk23835) – HW03 – Quigley – (104001) 2 = - 22104 . 8 K E a = - R ( - 22104 . 8 K) = - (8 . 314 J · mol 1 K 1 ) ( - 22104 . 8 K) = - 1 . 83779 × 10 5 J / mol . 004 10.0 points We wish to find the specific rate constant for the reaction A products . After 5 minutes the amount of A has de- creased from 2 . 3 mol / L to 1 . 3 mol / L. Find k using minutes as the time scale. The reaction is second order in [A]. Correct answer: 0 . 0668896 L · mol 1 · min 1 . Explanation: [A] 0 = 2 . 3 mol / L [A] t = 1 . 3 mol / L t = 5 min Since time and concentration are involved, the integrated rate equation is used. For second order, 1 [A] t - 1 [A] 0 = akt k = 1 at parenleftbigg 1 [A] t - 1 [A] 0 parenrightbigg = 1 5 min parenleftbigg 1 1 . 3 mol / L - 1 2 . 3 mol / L parenrightbigg = 0 . 0668896 L · mol 1 · min 1 005 10.0 points The graph describes the energy profile of a reaction. A B Time Energy (kJ) 50 300 400 What are the values for Δ H and E a , respec- tively, for the reaction in the direction writ- ten? 1. 250 kJ, 350 kJ correct 2. - 250 kJ, 350 kJ 3. - 250 kJ, - 100 kJ 4. 250 kJ, 100 kJ 5. - 250 kJ, 100 kJ Explanation: E a Δ H A B Time Energy (kJ) 50 300 400 Δ H = 300 kJ - 50 kJ = 250 kJ E a = 400 kJ - 50 kJ = 350 kJ 006 10.0 points A non-steroidal anti-inflammatory drug is me- tabolized with a first-order rate constant of 3.25 day 1 . What is the half-life for the metabolism reaction? 1. 0.213 day correct 2. 2.25 day 3. 1.63 day 4. 0.308 day Explanation: 007 10.0 points Consider the multistep reaction that has the overall reaction 2 A + 2 B C + D . What is the rate law expression that would correspond to the following proposed mecha- nism?
kamalska (mk23835) – HW03 – Quigley – (104001) 3 A + B I (fast) I + B C + X (slow) X + A D (fast) 1. Rate = k [B] 2. Rate = k [A] 3. Rate = k [A] 2 [B] 4. Rate = k [A] [B] 5. Rate = k [A] 2 [B] 2 6. Rate = k [A] [B] 2 correct 7. Rate = k [A] [I] [B] 8. Rate = k [I] [B] 9. Rate = k

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HW03-solutions - kamalska(mk23835 HW03 Quigley(104001 This...

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