CHEMICAL KINETICS
Goal of kinetics experiment is to measure concentration of a species
at particular time during a rxn so a rate law can be determined
rates are obtained from concen. vs fct of time
Rate of Rxn: describes how fast reactants used up & pdts formed
Chem Kinetics: 1)study of rates,
2) factors that affect rxn rates, &
3) mechanisms (steps) by which rxns occur
From a chem eqn, rate can be determined by following the Dconcen
of any subst that is quantitatively detected
Order of rxn cannot be deduced from chemical eqn. of rxn
Rate law expressions
- calculate rate of rxn from rate constant & reactant concen
- convert into eqn to determine concen of reactants @ any time
4 factors that affect chem rxns
1) nature of reactants
2) concen of reactants
3) temp
4) catalyst present
Write rate law for rxn to describe how rate depends on concen.
Rate Law is deduced experimentally from how its rate varies w/ concen
Order of rxn cannot be deduced from chemical eqn. of rxn
For rxn: A + B -----> pdts
general form: rate = k[A]x[B]y
exponents x & y
- usually integers
- value of x is the order of rxn
w/ respect to A
- y??
Values for k, x, & y have no relation to coeff of balanced chem eqn.,
remember, must be determined experimentally
Exponent
Define
rate not depend on [reacts]
0
rate is directly proportional to [reacts]
1
rate is directly proportional to square of concen; [reacts]2
2
overall order of rxn = x + y
Sum of orders of reacts
Examples of observed rate laws for following rxns
3NO(g) ------> N2O(g) + NO2 (g)
rate = k[NO]2
order in NO: 2nd
overall order:2nd
2NO2(g) + F2(g) ------> 2NO2F (g)
rate = k[NO2][F2]
order in NO2:1st order in F2: 1st
overall order: 2nd
quick summary
- order in rate law may not match coeff. in balanced eqn
- no way to predict rxn orders overall from balanced eqn
- orders must be determined experimentally
rate law can be determined by 2 methods:
1) Method of Initial Rates (if time)
2) using Integrated Rate Eqn
ZERO ORDER
Has a rate which is independent of concentration of reactant(s),
therefore, increasing concen. of rxning species not speed up rate
A -----> pdts
Rate is:
rate  k[A]O
k
r-
d A 
k
dt
Rate is a CONSTANT
integration gives eqn called integrated zero-order rate law
[A] = -kt + [A]o
concentr of chemical
@ particular time
Initial concentr
[A] = -kt + [A]o
eqn line:
y = mx + b
calculate k from plot of graph;
straight line plot of [A] vs time,t; slope = -k
Determine units:
[A]
time, t
k-
d [A] M

dT
s
half-life describes time needed for half of reactant to be depleted
t1
2

A O

2k
FIRST ORDER
Depends on concentration of only 1 reactant, if other reactants present but
each will be zero-order
1st order rate constant,
units of 1/time
eqn for first-order reaction A -----> pdts
rate is: rate 
d [A]
dt
know: rate  k[A]
-
d [A]
 k[A]
dt
integration gives eqn called integrated first-order rate law
ln
d [A]O
 kt
[A]
ln[A] = -kt + ln[A]o
eqn line:
y= mx + b
calculate k from plot of graph;
plot of ln[A] vs time,t; gives straight line
slope = -k
Determine units: k  -
d [A] 1
M
1


d [A] dt M  s s
ln[A]
[A]
time, t
time, t
half-life describes time needed for half of reactant to be depleted
t1 
2
Ln 2
k
0.693

k
SECOND ORDER
second-order
rate law
A. A ------ > pdts
depends on concentration of 2nd-order reactant
rate is: rate  k[A]2

d [A]
 k[A] 2
dt
d [A]
 k dt
2
[A]
1
1
[A]O @ t = 0 & [A] @ t: kt 

[ A ] [A] O
integrated in the form:
d [A]
d [B]
B. or, A + B = pdts


 k[A][B]
st
dt
dt
two 1 -order reactants:
Another way to represents rate laws, take ln of both sides:
ln r = ln k + 2 ln[A]
1
1
 kt 
[A]
[A]O
eqn line:
Plot 1/[A] vs time,t;
y= mx + b
slope = 2nd-order rate constant; +k
1/[A]
time, t
half-life for
2nd
order dependent on one
Determine units: k  -
2nd
order reactant: t 1
d [A] 1
M
1


[A] 2 dt M 2  s M  s
2
1

kA O
FIRST ORDER REACTION
2 N2O5 (aq) --------> 4 NO2 (aq) + O2 (g)
DATA
Time, s
[N2O5], M
0
600
1200
1800
2400
3000
3600
0.0365
0.0274
0.0206
0.0157
0.0117
0.00860
0.00640
ln[N2O5]
- 3.310
- 3.597
- 3.882
- 4.154
- 4.448
- 4.756
- 5.051
1/[N2O5], M-1
27.4
36.5
48.5
63.7
85.5
116
156
ln[N2O5]
[N2O5]
time,s
1/[N2O5]
time,s
rate = k[N2O5]
time,s
ln[N2O5]
time,s
ln[ A]3000  ln[ A]0 (-4.756) - (-3.310)

 - 4.820 *10-4 /s
slope =
t 3000  t 0
3000 s
SECOND ORDER REACTION
2 NO2 (g) --------> 2 NO (g) + O2 (g)
DATA
Time, s
0
60
120
180
240
300
360
[N2O5], M
0.0100
0.00683
0.00518
0.00418
0.0350
0.00301
0.00264
ln[N2O5]
- 4.605
- 4.986
- 5.263
- 5.477
- 5.655
- 5.806
- 5.937
1/[N2O5], M-1
100
146
193
239
286
332
379
ln[NO2]
[NO2]
time,s
1/[NO2]
time,s
rate = k[NO2]2
time,s
1/[NO2]
time,s
1   1 
slope =  [A] 300  [A] 0  (332) - (100)  0.773 L/mol  s
t 300  t 0
300 s
ZERO ORDER REACTION
Can occur if:
1) rate limited by [catalyst]
2) photochemical rxn if rate determined
by light intensity
3) most often occur when subst as a metal
surface or enzyme required for rxn to occur
2 N2O (g) --------> 2 N2 (g) + O2 (g)
N2O
N2O
N2O
N2O
N2O
N2O
N2O N2O N2O
N2O
N2O
N2O
N2O
N2
O
N2O
N2O
N2O
N2
O
N2O
N2O
N2O
N2O
Pt metal
surface
N2O
N2O
N2O
N2O
N2O
N2O
N2O
N2O N2O N2O
N2O
N2O
N2O
N2O
N2O
N2O
N2O
N2O
N2O
N2O
N2O
N2O
N2O
N2O
N2O
N2O
N2O
N2O
N2O
N2O
N2O
N2O N2O N2O
N2O
N2O
N2O
N2O
N2O
N2O N2O N2O
Describe what is happening
Rxn occurs on a hot Pt surface, when surface completely
covered w/ N2O molecules, an increase of [N2O] has no
effect on rate, since only N2O molecules on the surface
are reacting.
Therefore, the rate is constant because rsn is controlled
by what happens on Pt surface rather than total [N2O].
[N2O]
time,s
rate = k[N2O]0
Determine:
0 ORDER
1ST
ORDER
2ND ORDER
UNITS
d [A] mol
k
dT
Ls
d [A] 1
M
1
k

d [A] dt M  s s
k-
d [A] 1
M
L


[A]2 dt M 2  s mol  s
HALF-LIFE
t1
2
t1
t1
2
2

A O

2k
.693

k
1

kA O
Summary for reaction orders 0, 1, 2, & n
Zero-Order
First-Order
Second-Order
nth-Order
Rate Law
d A 
k
dt
d [A]
 k[A]
dt
d [A]
 k[A] 2
dt
-
Integrated
Rate Law
[A] = [A]Oe-kt
[A] = [A]O - kt ln[A] = ln[A] - kt
O
d [A]
 k[A] n
dt
1
1
1
1


 kt [A]n 1 [A]n -1  (n -1)kt
O
[A] [A]O
[Except1st order]
Units of
Rate
Constant (k)
Linear Plot
to determine
k
y-intercept
Half-life
mol
Ls
[A] vs t
-k
ln[A] vs t
-k
ln[A]O
[A]O
t1 
2
A O
2k
L
mol  s
1
s
t1 
2
.693
k
1
vs t
[A]
-k
1
[ A ]O
t1 
2
1
kA O
1
 mol 


L


n 1
1
vs t
[A]n1
[Except1st order]
s
NOTES
Rate Rxn
- describe rate rxn must determine concen of react/pdt at various times as rxn proceeds
- devising methods is challenge for chemists
-spectroscopic method: if 1 subst colored measure inc/dec in intensity of color
4 Factors: help control rates
METHOD OF INITIAL RATES
Experiment
1
2
3
[A]O
deduce rate law from experimental rate data
[B]O
1.0 * 10-2 M
1.0 * 10-2 M
2.0 * 10-2 M
1.0 * 10-2 M
2.0 * 10-2 M
1.0 * 10-2 M
initial rate
1.5 * 10-6 M.s-1
3.0 * 10-6 M.s-1
6.0 * 10-6 M.s-1
describing same rxn in each experiment, same rate law, form:
rate = k[A]x[B]y
Notice, [A]O same in #1 & #2, what would affect the rxn rate?
Des in rxn rate due to diff initial concen of B
Comparing the 2 experiments, [B] is Ded by factor of:
2.0 *102
 2.0  [B] ratio
2
1.0 *10
rate Des by factor of:
3.0 *106
 2.0  rate ratio
6
1.5 *10
Exponent y deduced from:
rate ratio = ([B])y
2.0 = (2.0)y solving, y = 1
What order is rxn order in [B]?
rate = k[A]x[B]1
Experiments 1 & 3 show [B]O same but [A]O different
[A] is Ded by factor of:
rate Des by factor of:
2.0 *102
 2.0  [A] ratio
2
1.0 *10
6.0 *106
 4.0  rate ratio
6
1.5 *10
Exponent x deduced from:
rate ratio = ([A])x
4.0 = (2.0)x
What order is rxn order in [A]?
solving, x = 2
rate = k[A]2[B]1
Rate constant, k, substitute data from any set of 3 sets into rate-law expression
rate1 = k[A]12[B]11
or, rate = 1.5 M-2.s-1[A]2[B]