HW #9 - Chemistry 132, Winter 2008 Solutions to Homework...

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Chemistry 132, Winter 2008 Solutions to Homework No. 9 Problem 1. In a chain reaction, the chain carriers are reactive intermedi- ates that occur in chain propagation steps (Levine p. 565). A step that generates chain carriers from relatively unreactive species is an initiation step. A termination step consumes chain carriers without generating new ones. A branching step is one that generates more chain carriers than it con- sumes (Levine p. 566). Although not clearly stated in Levine p. 565, it can be inferred that a propagation step generates as many chain carriers as it consumes. The chain carriers in the given mechanism are O, H, and OH, as mentioned in the problem statement. Based on the above definitions, the steps of the given mechanism can be identified as follows: step type reason 1 initiation generates H 2 initiation generates O 3 initiation generates H 4 branching consumes H, generates OH and O 5 branching consumes O, generates OH and H 6 propagation consumes OH, generates H 7 termination consumes H 8 termination consumes O 9 termination consumes H Problem 4. The Michaelis constant K M is a parameter of the Michaelis- Menten equation (Levine eq. 17.125), which gives the initial rate r 0 (or v 0 ) of an enzymatic reaction as a function of initial substrate concentration S 0 . Taking the reciprocal of both sides of the Michaelis-Menten equation gives (Levine eq. 17.126): 1 r 0 = K M k 2 E 0 1 S 0 + 1 k 2 E 0 , where k 2 is the rate constant for the second step of the as- sumed mechanism (Levine eq. 17.120) and E 0 is the ini- tial concentration of the enzyme. Then, a plot of y = 1 /r 0 against x = 1 /S 0 should yield a straight line with slope a = K M /k 2 E 0 and intercept b = 1 /k 2 E 0 , so that K M = a/b . Figure 1 shows that performing linear regression on the given data points and taking the ratio of slope to intercept gives K M = 2 . 2 × 10 - 3 M. Problem 7. See Levine problem 22.8. Solve this problem for argon gas ( M = 39 . 948 × 10 - 3 kg / mol) in a 20-cm 3 box at 300 T. The number n of available translational states with energy less Figure 1: Excel spreadsheet for problem 4 1 2 3 4 5 6 7 8 9 10 11 12 13 A B C D E F y =1/ r 0 x =1/ S 0 / (10 3 M −1 ) 2.8 0.7 3.2 0.9 4.2 1.3 6.2 2.2 9 3.3 a 1.08125 b / (10 -3 M) 2.3802083 b/a / (10 -3 M) 2.2013487 cell formula F7 =INTERCEPT(E2:E6,F2:F6) F8 =SLOPE(E2:E6,F2:F6) F9 =F8/F7 y = 2.3802x + 1.0813 0 2 4 6 8 10 0 1 2 3 4 x y than 3 kT is given by (Levine p. 836) n 60 ± mkT h 2 ² 3 / 2 V = 60 ± MkT N A h 2 ² 3 / 2 V = 60 ³ (39 . 948 × 10 - 3 kg / mol)(1 . 38065 × 10 - 23 J / K) (6 . 022142 × 10 23 mol - 1 ) × (300 K) (6 . 62607 × 10 - 34 J s) 2 ´ 3 / 2 (20 × 10 - 6 m 3 ) = 1 . 879 × 10 28 . Problem 10. See Levine problem 22.15. Solve this problem for O 2 in a 10- cm 3 cubic box. The kinetic energy associated with motion in the x direction is given by (Levine eq. 18.42) ± tr,x ( n x ) = n 2 x h 2 8 ma 2 , where n x = 1 , 2 ,... is a quantum number and a = V 1 / 3 is the length of the side of the cubic box. The difference between
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This note was uploaded on 05/22/2008 for the course CHEM 132 taught by Professor Lindenberg during the Winter '08 term at UCSD.

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HW #9 - Chemistry 132, Winter 2008 Solutions to Homework...

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