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.

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? Email me right away. Review sessions: Monday, November 28: 217 Noyes Lab, 5-6 pm Tuesday, November 29: 161 Noyes Lab, 7-8 pm

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) 20 20 20 b) 10 20 30 c) 30 20 15 d) 15 20 30 e) Not sure. [A] Time 1.00 M 0.500 M 10 sec 0.250 M ?? sec

Good to Know…

Using Half-Lives

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.

Rate Laws From Elementary Steps

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

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)

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)

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.

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

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

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

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)

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

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