Chapter 13
RATES OF REACTION
SUROVIEC
SPRING 2014
I. Reaction Rates
 Chemical Kinetics: the investigation of the rate at
which reactions occur
 Rate: how fast a quantity changes with time
I. Reaction Rates
Rate = change in quantity / time elapsed
A. Average Rate Vs. Instantaneous Rate
B. Writing Rate Reactions
 Consider the rate reaction:
2NO2 (g)  2NO (g) + O2 (g)
General form
aA + bB  cC + dD
Example
 Give the relative rates of disappearance of the
reactants and formation of products for each of the
following reactions:

2O3 (g)  3O2 (g)

2HOF (g)  2HF (g) + O2 (g)
Example
 A  2B
time
[B] (mol/L) Rate (D[B]/Dt)
0.00
0.000
10.0
0.326
20.0
0.572
30.0
0.750
40.0
0.890
 Rate increases or
decreases as there is less A
to react
 How is rate of
change of [A] related
to rate of change of
[B]?
 Find rate of change
of [A] for time
internal from 10.0 to
20.0 s
 What is the
instantaneous rate
when [B] = 0.750M
Example
N2 (g) + 3H2 (g)  2NH3 (g)
What are the relative rates?
If –D[H2]/Dt = 4.5X10-4 M/min,
What is –D[N2]/Dt?
What is D[NH3]/Dt?
II. Rate Laws
 Why look at initial rates?
 Ex. 2N2O5  4NO2 + O2
[N2O5]
Rate (M/s)
0.010
0.0180
0.020
0.0360
0.040
0.0720
0.060
0.1080
0.100
0.1800
General Form of the Rate Law
aA + bB  cC + dD
Rate = k[A]x[B]y
A. Method of Initial Rates
 Initial rate of reaction is measured with several
different starting amount of reactants
Example
The reaction between ozone and nitrogen dioxide at
231 K is first order in both [NO2] and [O3].
2 NO2 (g) + O3 (g)  N2O5 (g) + O2(g)
Write the rate equation for the reaction
2. If the concentration of NO3 is tripled, what is the
change in the reaction rates?
3. What is the effect on the reaction rate if the
concentration of the O3 is halved?
1.
Example
2NO(g) + O2 (g) 2NO2 (g)
1.
2.
[NO] (mol/L) [O2] (mol/L) Rate of [NO]
disappearing
(mol/L.s)
0.010
0.010
2.5 X 10-5
0.020
0.010
1.0 X 10-4
0.010
0.020
5.0 X 10-5
3.
4.
Determine the order
of the reaction for
each reactant
Write the rate
equation for the
reaction
Calculate the rate
constant
Calculate the rate in
M/s at the instant
when [NO] = 0.015 M
and [O2] = 0.0050M
Example
A+BC
What is the rate equation?
Run #
[A]i
[B]i
Initial Rate (M/s)
1
0.10
0.10
2.00X10-4
2
0.20
0.10
8.00X10-4
3
0.40
0.20
2.56X10-2
Base 10 log and Natural Log
 Base 10 log:
102 = 100  log 100 = _______
103 = 1000  log 1000 = _______
log 45 = 1.65  101.65 =_________
log x = y  10y =x
 Natural Log: base is e = 2.718281828…..
e1 = 2.71828
ln(2.71828) = 1
ln(e)=_______
ln(45) = 3.81  e3.81 =_________
ln x = y  ey =x
Important properties of both log and ln
 log ab = log a + log b
 log a/b = log a – log b
 log ab = b(log a)
Example
 Rate = k[A]2[B]x
III. Concentration-time relationships
 Useful to know
 How long a reaction is going to proceed to reach concentration
of interest for product or reactant
 After a given time what are the concentrations of the reactants
and the products?
A. 1st order reactions
 AB
 Rate = k[A]
 Useful for 3 reasons

 Integrated rate eqn for 1st order
reactions:


Can calculate k if we know
[A]/[A]o
if k and [A]o are known then
we can determine the
amount of material expected
after time t ([A])
If k is known than after time
t we can calculate the
fraction [A]/[A]o
B. 2nd order reactions
 Rate = k[A]2
 Integrated rate eqn for 2nd order reactions:
1/[A] = kt + 1/[A]o OR
1/[A] – 1[A]o = kt
Example
Ammonium cyanate, NH4NCO, rearranges in
water to give urea, (NH2)2CO.
NH4NCO (aq)  (NH2)2CO (aq)
The rate equation for this process is:
Rate = k [NH4NCO]2
where k = 0.0113 L/mol.min.
If the original concentration of NH4NCO in solution is
0.229M how long will it take for the concentration of
decrease to 0.180M?
Example
1st order graph
y = -0.0037x - 1.1488
2
R = 0.9998
-1
ln [sucrose]
Sucrose breaks down to form
glucose and fructose.
C12H22O11(aq) +H2O (l) 
2C6H12O6 (aq)
1. Plot ln sucrose versus time
and 1/[sucrose] versus time.
Find the order of the reaction
2. Write the rate constant for the
reaction and calculate k
3. Find concentration of sucrose
After 175 min
[sucrose]
ln [sucrose]
1/[sucrose]
0
0.316
-1.1520131
3.164557
39
0.274
-1.2946272
3.649635
80
0.238
-1.4354846
4.2016807
140
0.190
-1.6607312
5.2631579
210
0.146
-1.9241487
6.8493151
-1.4
-1.6
-1.8
-2
0
50
100
150
200
250
time (min)
+ 2.9817
2nd order graph y = 0.0175x
2
R = 0.9847
1/[sucrose]
time (min)
-1.2
7
6.5
6
5.5
5
4.5
4
3.5
3
0
50
100
150
time (min)
200
250
C. Zero Order Reactions
 Rate =k[A]0
 Integrated rate eqn for 0th order reactions:
[A]o – [A] = kt
Table 15-1, p.719
D. Half-Life
 Time required for half of a reactant to be consumed
1. 1st order
Fig. 15-9, p.720
Example
The rate equation for the decomposition of producing
NO2 and O2 is
–D[ N2O5]/Dt = k[N2O5]
The value of k = 5.0 X 10-4 s-1 for the reaction at a
known temperature.
1. Calculate the half-life of N2O5
2. How long does it take for the N2O5 concentration
to drop to 1/10th of its original value?
2. 2nd order and 0th order
 AB
 AB
 Rate = k[A]2
 Rate = k
IV. Activation Energy
NO2 (g) + F2 (g)  FNO2 (g) + F (g)
What makes this reaction occur?
What makes this reaction not occur?
A. Collision Theory
 Reactant Molecules must
collide with each other
 Reactant molecules must
collide with sufficient
energy
 Reactant molecules must
have proper orientation
2 questions:
 What makes a reaction
go faster?
 What does sufficient
energy mean?
A. What makes a reaction go faster?
1.
2.
B. What does sufficient energy mean?
RXN: AB + C  A + BC
Mechanism:
C. How can we measure Ea?
 From collision theory
Rate of reaction =
(collision freq)X(fraction with proper orientation)X (fraction of collisions with enough energy)
k = Ae-Ea/RT
Arrhenuis equation
Example
 When heated to a high temperature, cyclobutane,
C4H8, decomposes to ethylene.
C4H8 (g)  2C2H4 (g)
The activation energy, Ea for this reaction is 260
kJ/mol. At 800K the ate constant k = 0.0315 s-1.
Determine the value of k at 850K.
D. What is a catalyst?
 A substance added to a reaction to speed it up, but
not consumed in the reaction
MnO2 catalyzes decomposition of H2O2
2 H2O2 ---> 2 H2O + O2
Uncatalyzed reaction
Catalyzed reaction
V. Reaction mechanisms
AB + C  A + BC
Possible mechanisms
V. Reaction Mechanisms
 Molecularity
 Elementary step
A. Mechanisms and Rate Laws
1.
For net reactions, no
simple relationship can
be assumed between
reaction coefficients
and reactant orders
2. For each elementary
step reactant
coefficients DO equal
reactant orders
A. Mechanisms and Rate Laws
3.
The rate of a net reaction is essentially that of the
slowest step (the rate determining step)
A. Mechanisms and Rate Laws
4. When a fast step proceeds a slow step the fast step
can be assumed to be at equilibrium
RXN: H2 (g) + CO (g)  H2CO (g)
Step 1. H2 (g)  2 H (g) fast
Step 2. H (g)  HCO (g) slow
Step 3. H (g)  H2CO (g) fast
A. Mechanisms and Rate Laws
5. A catalyst can provide a new mechanistic pathway
for a reaction
2Ce4+ (aq) + Tl+ (aq)  2Ce3+ (aq) + Tl3+ (aq)
Mechanism
Rate equation
Add Mn2+ catalyst
In General
1.
Measure reaction rates
1.
2.
2.
3.
Through initial rates
Graphically concentraion vs. time
Formulate the rate law
Postulate the mechanism
1.
Cannot prove, can only disprove