Physical Chemistry 20130403 week 1 Wednesday April 3, 2013 page 1
2N2O5 ⇌ 2NO2 + 2NO3
s=2 (may be omitted)
NO2 + NO3 → NO + O2 + NO2
NO + NO3 → 2NO2
s=1
s=1
sum of those 3 equations: 2N2O5 → 4NO2 + O2
The s=2 for the first equation is because that reaction has to happen twice.
s is stoichiometric number
ν (nu) is stoichiometric coefficient
The 2, 4, and 1 in the bottom equation are ν.
Each of those top 3 equations is an elementary stop or a single step.
ethene (CH2=CH2) + 1,3 butadiene (CH2=CH-CH=CH2) → cyclohexene (C6H10)
That reaction happens in one step.
In thermodynamics we just need to know 2 states: initial state and final state, time doesn’t matter
pseudo-order reaction
C12H22O11 + H2O → C6H12O6 + C6H12O6 (sucrose + water = glucose + fructose)
r=k”[C12H22O11][H2O]6[H+] so order = 8
Keep [H2O] constant to simplify, since the experiment is hard to do. Then:
r=k’[C12H22O11][H+]
Keep [H+] constant to simplify. Then:
r=k[C12H22O11] so this reaction is pseudo first order.
Measurement of reaction rates
1. chemical methods
can stop reaction by diluting
use GC or MS
less preferable method
2. physical methods
more accurate and quicker than chemical methods
Integration of rate laws
1. zero order reaction
very uncommon
example is enzyme catalysis
aA → product(s)
r=k[A]0 = k
-1 d[A]
=k=r t is time
a dt
-d[A]=akdt then integrate
[A]
∫
t=t
-d[A]=ak ∫
[A]0
ak=k a
t=0
[A]
t=t
-[A] |
=k A t |
[A]0
t=0
-[A]+[A]0 = kA(t-0)
[A]0-[A]=kAt
integrated rate law
[A] = -kAt + [A]0
The rate r of a zero order reaction is a constant then. Graphing [A]0-[A] on the vertical axis and t on the
horizontal axis gives a straight line of slope kA. This line comes from the origin, indicating it’s zero order.
Graphing [A] on the vertical axis and t on the horizontal axis gives a line of slope –kA. A reaction that
happens on the surface of a solid catalyst is often zero order.
First order reaction
First order reactions are most important because they’re very common. Nuclear decay is always first
order.
aA → product(s)
r=
-1 d[A]
=k[A]
a dt
d[A]
= -ak[A]
dt
d[A]
= -k A [A]
dt
ak=k A
separate variables
d[A]
= -k A dt
[A]
then integrate
[A]
t
d[A]
= -k A ∫ dt
[A]0 [A]
0
∫
[A]
t
ln[A] |
=-k A t |
[A]0
0
ln (
[A]
) = -k A t
[A]0
first order
[A]= [A]0 e-k At
ln[A]= -k A t+ln[A]0
Graphing ln[A] on the vertical axis and t on the horizontal axis gives a straight line with slope = -kA. The y
intercept is ln[A]0. Graphing [A] on the vertical axis and t on the horizontal axis gives a curve that goes
asymptotically to 0 as t→∞. Each distance for t1/2 (half life) is the same amount of time.
half-life
when t= t 1⁄
2
[A]=
[A]0
2
so [A]0 =2[A]
Remember, [A] is what’s left is the system, not how much the system has lost.
ln (
t 1⁄ =
2
.693
kA
[A]
) = -k A t 1⁄
2
2[A]
the [A]' s cancel
first order reaction
constant
=constant
constant
The first order equations don’t apply to pseudo-first order reactions.
second order reactions
Second order reactions are more common than first order reactions. First order reactions can be of the
type aA → product(s) or A + B → product(s). Assume the type aA → product(s) for now since it’s easier.
d[A]
= -k A [A]2 integrate
dt
[A]
t
d[A]
∫
= -k A ∫ dt
2
[A]0 [A]
0
-1 [A]
t
|
=-k A t |
[A] [A]0
0
-1
1
+
=k A t
[A] [A]0
1
1
=k A t
[A] [A]0
second order
Graphing 1/[A] on the vertical axis and t on the horizontal axis gives a line with slope kA.
t 1⁄ =
2
1
[A]0 k A
second order
Graphing [A]/[A]0 on the vertical axis and t on the horizontal axis gives a curve coming down. Each t1/2 is
a different width on the graph (length of time).