Problem Set 1 answer key: posted on course website
Review Session: Tue, Aug 18th at 7:10 pm in Chem 179.
Midterm: Chapters 2.6-2.9, 9 and 10; Thu Aug 2oth
Steady State Kinetics Example
A + B k1 I
I k 2 P
d[ I ]
= k1 [ A][ B]
dt
d [ P]
= k 2 [I ]
rate =
dt

Thermodynamics vs Kinetics
Chemical Reaction or Biological Process:
kf
AB
krev
Keq = [B]/[A] = kf/krev
kf = forward rate constant
(Thermo, 107A)
(Kinetics, 107B)
Thermodynamics predicts how far a rxn proceeds (Keq).
Keq depends on stability of product vs

Rate Depends on Collision Frequency (Z11)
and Boltzmann Factor (e-Ea/RT)
A + B Product
Ea
P
Rate of rxn = (# of collisions per second) X (# of collisions having E > Ea)
N
Z11 = Z1
V
N
Z1 = d c
V
2
exp(-Ea/RT)
m c2
2
Collision Frequency (Z1)
For one mo

University of California at Davis
CHEM 107 B
Physical Chemistry for the Life Sciences
Summer 2015
Practice Midterm 1 Exam
Student Name: _
Student ID: _
Instruction:
This is a CLOSED BOOK EXAM.
DO NOT OPEN the exam until instructed to do so.
Useful equa

Practice Midterm: see course website.
Review Session: Tue Aug 18th at 7pm in 179 Chem.
Midterm: Chapters 2.6-2.9, 9 and 10; Thu Aug 20th
Rate Constant (k) vs. Temperature
k
A Products
Normal
rate = -d[A]/dt = k[A]
Enzyme-catalyzed
A I + heat
IP
AP
E
a
Z11

Problem Set #1: due at end of class on Tue Aug 11th.
First-Order Kinetics
A products
What is [A] vs time?
First order reaction: rate = k[A]1
rate = [A]t
[A]0
Recall calculus:
ln
[A]t
[A]0
= -kt
Integrated
Rate law
d[A ]
dt
= k [A]
t
d[A ]
[A]
= - k dt = -

Chem 107B
Summer 2015
Ames
Problem Set 1 (due Tue Aug 11, 2015).
1. The diagram below shows the Maxwell speed distribution curves for an ideal gas at two different
temperatures (T1 and T2).
(a) What is the mean speed ( c ) of the gas at 298.0 K?
c
8RT 4

Problem Set #1: due at end of class today.
Practice Midterm: see course website.
Midterm: Chapters 2.6-2.9, 9 and 10; Thu Aug 20th
Review Session: Tue Aug 18?
Reaction Mechanism
NO2(g) + CO(g) NO(g) + CO2(g)
Overall rxn
rate = k[NO2]2
Step (1) NO2(g) + NO