1. Lon-Capa 8th (final) HW assignment due Wednesday, November 30 by 5 pm.
Quiz #6 opens Friday (11/18) at 5 pm. Due Wednesday, November 30 by 10 pm.

2. Exam III
Thursday, December 1, 7:00-9:00 pm; rooms are on the website (same as before).
Conflict: (12/1) 4:30-6:30 pm in 165 Noyes Lab; sign up in 367 NL
Conflict with conflict? me right away.
Review sessions:
Monday, November 28: 217 Noyes Lab, 5-6 pm
Tuesday, November 29: 161 Noyes Lab, 7-8 pm

3. Clicker Question [A] Time 1.00 M 0.500 M 10 sec 0.250 M ?? sec
Determine the missing time given the following orders of reaction for
aA  Products
zero 1st 2nd a) b) c) d)
e) Not sure.
[A]
Time
1.00 M
0.500 M
10 sec
0.250 M
?? sec

4. Good to Know…

5. Using Half-Lives

6. Mechanisms Series of elementary steps. Must be chemically reasonable
Must have the correct stoichiometry. That is, the steps must add up to the overall reaction.
Must be chemically reasonable
How can we tell this?
Cannot prove but can support empirically.
Must agree with the overall rate law.

7. Rate Laws From Elementary Steps

8. Evaluating Mechanisms
The rate law comes from the rate-determining step (the “slow” step).
Fast equilibrium.
Rate forward = rate reverse.
Rate law cannot include intermediates.
Steady-state approximation. [VIDEO!]
[intermediate] = constant.
d[intermediate]/dt = 0

9. Rate Determining Step 2N2O(g)  2N2(g) + O2(g)
1. N2O(g) + N2O(g)  N4O2(g) (slow)
2. N4O2(g)  2N2(g) + O2(g) (fast)
vs.
1. N2O(g)  N2(g) + O(g) (slow)
2. N2O(g) + O(g)  N2(g) + O2(g) (fast)

10. 2NO(g) + 2H2(g)  N2(g) + 2H2O(g)
Fast Equilibrium Step
2NO(g) + 2H2(g)  N2(g) + 2H2O(g)
1. 2NO(g) N2O2(g) (fast equil.)
2. N2O2(g) + H2(g)  N2O(g) + H2O(g) (slow)
3. N2O(g) + H2(g)  N2(g) + H2O(g) (fast)

11. Steady-State Approximation
If we cannot determine a rate determining step.
Assumes that the concentration of an intermediate stays relatively constant during the course of the reaction.
Often used with enzyme-substrate studies in biochemistry
Video with an example different from the text.

12. Steady-State Approximation
2NO(g) + H2(g)  N2O(g) + H2O(g)
1. 2NO(g) N2O2(g)
2. N2O2(g) + H2(g)  N2O(g) + H2O(g)
  N2O2(g) is the intermediate

13. Clicker Question
The reaction 2A + B  C has the following proposed mechanism
A + B D (fast equil.)
D + B  E (slow)
E + A  C + B (fast)
The rate law from this mechanism is
a) rate = k[A][B] d) rate = k[A][B]2
b) rate = k[A]2[B] e) I don’t know
c) rate = k[A]2[B]2

14. Understanding Orders of Reaction: Second Order
NO2(g) + CO(g)  NO(g) + CO2(g)

15. Understanding Orders of Reaction: First Order
2N2O(g)  2N2(g) + O2(g)
1. N2O(g) + N2O(g) N2O*(g) + N2O(g) (fast)
2. N2O*(g)  N2(g) + O(g) (slow)
3. N2O(g) + O(g)  N2(g) + O2(g) (fast)

16. Understanding Orders of Reaction: Zero Order
C2H4(g) + H2(g)  C2H6(g)

17. Understanding Orders of Reaction: Zero Order
2N2O(g)  2N2(g) + O2(g)