Sample1S - 1 Chem 6 Sample Exam 1 brief answers 1. a....

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

View Full Document Right Arrow Icon
1 Chem 6 Sample Exam 1 brief answers 1 . a. Consider the first 2 columns. Doubling [HI] quadruples the rate, so it must be second order in HI . Note that you get the same result by considering columns 2 and 3. So the rate law must be rate = k[HI] 2 . b. To get the rate constant, plug in some numbers. I used column 1: rate = 7.5x10 –4 Ms –1 = k(0.005M) 2 k = 30 M –1 s –1 c. Plug in [HI] = 0.0020M to get rate = 30M –1 s –1 (0.0020M) 2 = 1.2x10 –4 Ms –1 2. These sketches are not to scale, and my drawing program is not good at making curves, but you should be able to get the idea from these pictures. ..... reaction coordinate E reaction coordinate E (a) E = 10, E a = 25 (b) E = –10, E a = 50 (a) E = –50, E a = 50 E a E react. prod. E E a prod. react. reaction coordinate E E E a prod. react. Reaction (a) (the left-hand one) will be the fastest , since it has the lowest activation energy, and k = Ae –Ea/RT 3. Michaelis-Menten. From class discussion of the Lineweaver-Burk plot, we saw that the maximum possible rate is k 2 [E] o . So half of that is k 2 [E] o /2. Set this equal to the rate expression to get k 2 [E] o /2 = {k 2 [E] 0 [S]}/[S] + K M You can already see that this works when [S] = K M Or chug through the algebra: 1/2 = [S]/[S] + K M 1/2[S] + 1/2K M = [S] 1/2K M = 1/2[S] K M = [S]
Background image of page 1

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

View Full DocumentRight Arrow Icon
2 4 . (a) Here are the steps labeled as initiation, propagation and termination. (b) The rate is given by –d/dt[C 2 H 6 ] = k 2 [C 2 H 6 ][I] But [I] is an intermediate so we need to get rid of its concentration. Do this from steps 1 and 4, which define an equilibrium: [I] 2 /[I 2 ] = K eq = k 1 /k –1 So [I] = {[I 2 ]K eq } 1/2 Plug in to get Rate = k 2 [C 2 H 6 ][I] = k 2 [C 2 H 6 ]{[I 2 ]K eq } 1/2 Another approach to find [I] also works: doing the steady-state approximation on the
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 05/01/2008 for the course CHEM 5/6 taught by Professor Gribble during the Spring '08 term at Dartmouth.

Page1 / 4

Sample1S - 1 Chem 6 Sample Exam 1 brief answers 1. a....

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