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

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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?...
View Full Document

## This note was uploaded on 09/28/2009 for the course GE 206E taught by Professor staff during the Spring '09 term at University of Texas.

### Page1 / 8

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online