HW12SOL - Moore Robert – Homework 12 – Due midnight –...

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Unformatted text preview: Moore, Robert – Homework 12 – Due: Apr 29 2007, midnight – Inst: McCord 1 This print-out should have 25 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 The first order rate constant for A → products is 6 . 6 s- 1 . If [A] = 1 . 3 M at t = 0, what is the value of [A] at t = 0 . 6 s? 1. . 00495641 M 2. 6 . 19551 × 10- 6 M 3. 1 . 2391 × 10- 5 M 4. . 495641 M 5. . 024782 M correct Explanation: [A] = 1 . 3 M t = 0 . 6 s k = 6 . 6 s- 1 If the reaction is first order, the integrated rate equation is ln µ [A] [A] ¶ = a k t ln[ A ]- ln[ A ] = a k t ln[ A ] = ln[ A ]- a k t = ln(1 . 3 M)- (1) ( 6 . 6 s- 1 ) (0 . 6 s) =- 3 . 69764 M [ A ] = 0 . 024782 M 002 (part 1 of 1) 10 points A reaction A → products is observed toobey first-order kinetics. Which of the following plots should give a straight line? 1. ln[A] vs k 2. [A] vs 1 t 3. [A] vs 1 k 4. [A] vs t 5. ln[A] vs t correct 6. [A] vs k 7. ln[A] vs 1 k 8. ln[A] vs 1 t Explanation: The first order integrated rate equation is ln µ [A] [A] ¶ = a k t ln[A]- ln[A] = a k t ln[A] =- a k t + ln[A] Since a , k , and ln[A] are constants, this equa- tion is in the form y = mx + b , where y = ln [A] and x = t . A straight line is produced by y = mx + b . 003 (part 1 of 1) 10 points Consider the following data for the reaction A → products. Initial Trial [A] Half-life (M) (s) 1 . 2 30 2 . 5 12 3 . 1 60 The reaction 1. is first order. 2. is second order. correct 3. No choice is correct. Moore, Robert – Homework 12 – Due: Apr 29 2007, midnight – Inst: McCord 2 4. is zero order. 5. has an order that cannot be determined from the data given. Explanation: 004 (part 1 of 1) 10 points A first order elementary reaction A → products has a rate constant of 3 . 16 × 10 8 sec- 1 . At an instant in time, a concentration of 3 . 16 × 10- 6 M of species A is created. How long does it take for the concentration to fall by a factor of 4? 1. 4 . 2 × 10- 6 sec 2. 4 . 4 × 10- 9 sec correct 3. 1 . 28 × 10- 8 sec 4. 1 . 6 × 10- 9 sec 5. 6 . 4 × 10- 9 sec Explanation: a = 1 k = 3 . 16 × 10 8 s- 1 [A] = 3 . 16 × 10- 6 M [A] t = 1 4 [A] ln µ [A ] [A] ¶ = a k t Since the concentration of A is decreasing by a factor of 4, ln(4) = a k t ln(4) = (1) ( 3 . 16 × 10 8 s- 1 ) t t = ln(4) 3 . 16 × 10 8 s- 1 = 4 . 38 × 10- 9 s 005 (part 1 of 1) 10 points 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?...
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HW12SOL - Moore Robert – Homework 12 – Due midnight –...

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