Exam 4- - Version 116 Exam 4 Holcombe (52460) 1 This...

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

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Version 116 Exam 4 Holcombe (52460) 1 This print-out should have 28 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points Consider the data below: [CH 4 ] [O 2 ] initial rate M M M s 1 Exp 1 . 5 1 . 3 2 . 5 10 3 Exp 2 . 5 2 . 6 1 . 10 2 Exp 3 1 . 1 . 3 5 . 10 3 Which of the following is a correct rate law for the reaction? 1. k [CH 4 ] 1 [O 2 ] 2. k [CH 4 ][O 2 ] 3. k [CH 4 ] 1 [O 2 ] 2 4. k [CH 4 ][O 2 ] 2 correct 5. k [CH 4 ] Explanation: rate 1 rate 2 = [O 2 ] 1 [O 2 ] 2 2 . 5 10 3 1 . 10 2 = parenleftbigg 1 . 3 2 . 6 parenrightbigg x x = 2 rate 3 rate 1 = [CH 4 ] 3 [CH 4 ] 1 5 . 10 3 2 . 5 10 3 = parenleftbigg 1 . . 5 parenrightbigg y y = 1 002 10.0 points Biochemical reactions are most often cat- alyzed by 1. isomers 2. prostaglandins 3. heat energy 4. monomers 5. NSAIDs 6. enzymes correct Explanation: Enzymes are biological catalysts. 003 10.0 points Consider the concentration-time dependence graph for a first-order reaction. [A] Molar Concentration of Reactant A B C Concentration- Time- At which point on the curve is the reaction fastest? 1. B 2. A correct 3. The rates are the same at all points. 4. A + t 1 / 2 5. C Explanation: Steeper portions of the graph define faster reactions. 004 10.0 points Consider the reaction 2 NOCl(g) 2 NO(g) + Cl 2 (g) with rate constant 0.0480 M 1 s 1 when con- ducted at 200 C. The initial concentration of Version 116 Exam 4 Holcombe (52460) 2 NOCl was 0.521 M. What is the concentration of NOCl after 0.268 minutes and at 200 C? 1. 0.112 M 2. 0.200 M 3. 0.289 M correct 4. 0.372 M 5. 0.514 M 6. 0.111 M Explanation: k = 0 . 0480 M 1 s 1 t = 0 . 268 min [A] = 0.521 M The units on k indicates a second order reaction: 1 [A] t- 1 [A] = a k t 1 [A] t = 1 [A] + a k t = 1 . 521 M +2 ( . 0480 M 1 s 1 ) (0 . 268 min) 60 . 0 s min = 3 . 463 M 1 [A] t = 0 . 289 M 005 10.0 points Consider the elementary reaction: CaCO 3 ( s )- CO 2 ( g ) + CaO( s ) If k = 1 . 03 10 2 M s 1 , and there is ini- tially 0 M CO 2 , what is the [CO 2 ] after 10 minutes have passed? 1. 6 . 18 M correct 2. . 10 M 3. 1 . 03 M 4. . 62 M Explanation: [O 2 ] = [O 2 ] + kt = 0 + ( 1 . 03 10 2 ) (600) = 6 . 18 M 006 (part 1 of 4) 10.0 points Consider the following potential energy dia- gram. e f c d a b A+B X+Y Reaction progress Potentialenergy Which arrow represents the potential en- ergy of the reactants?...
View Full Document

Page1 / 8

Exam 4- - Version 116 Exam 4 Holcombe (52460) 1 This...

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

View Full Document Right Arrow Icon
Ask a homework question - tutors are online