Chapter 14
Chemical Kinetics
Overview:
• Reaction Rates
– Stoichiometry, Conditions, Concentration
• Rate Equations
–
–
–
–
–
Order
Initial Rate
Concentration vs. Time
First Order Rxns.
Second Order Rxns.
• Graphical Methods
Cont’d
• Molecular Theory
–
–
–
–
–
Activation Energy
Concentration
Molecular Orientation
Temperature
Arrhenius Equation
• Reaction Mechanisms
– Elementary Steps, Reaction Order, Intermediates
• Catalysts
Reaction Rates
• What Affects Rates of Reactions?
–
–
–
–
Concentration of the Reactants
Temperature of Reaction
Presence of a Catalyst
Surface Area of Solid or Liquid Reactants
Reaction Rates (graphical):
• Average Rate
=
[M]
D[M]
Dt
for reaction A  B
D[M]
Dt
time
Rates
• for
A  B
- D[A] = D[B]
Dt
Dt
Rate of the disappearance of A is equal in
magnitude but opposite in sign to the rate
of the appearance of B
• Average Rate--D mol (or concentration) over a
period of time, Dt
• Instantaneous Rate-- slope of the tangent at a
specific time, t
• Initial Rate-- instantaneous rate at t = 0
[M]
tangent at time, t
t
time
Average Rate =
[A]final time - [A]initial time
Dtfinal - Dtinitial
for A  B
Instantaneous Ratetime, t = slope of the tangent at time = t
[M]
tangent at time, t
t
time
Stoichiometry
4PH3
- 1D[PH3]
4 Dt
- D[PH3]
Dt
=>
P4
+ 6H2
= + 1 D[P4] =
1 Dt
= + 4 D[P4]
Dt
+ 1 D[H2]
6 Dt
= + 2 D[H2]
3 Dt
General Relationship
Rate =
- 1 D[A] = - 1 D[B] = + 1 D[C] = + 1 D[D]
aDt
b Dt
c Dt
d Dt
aA + bB  cC + dD
Conditions which affect rates
• Concentration
– concentration  rate
• Temperature
– temperature  rate
• Catalyst
– substance which increases rate but itself remains
unchanged
Rate Equations:
• aA + bB

xX
rate law
rate = k[A]m[B]n
• m, n are orders of the reactants
– extent to which rate depends on concentration
– m + n = overall rxn order
• k is the rate constant for the reaction
Examples:
• 2N2O5
=>
4NO2 +
rate = k[N2O5]
• 2NO
+
Cl2
1st order
=>
2NOCl
rate = k[NO]2[Cl2]
• 2NH3
=>
N2
O2
+
3rd order
3H2
rate = k[NH3]0 = k
0th order
Determination of Rate Equations:
Data for:
Expt. #
1
2
3
A
+
B
(0.10)2 M2
C
[A]
[B]
initial rate
0.10
0.10
0.20
0.10
0.20
0.10
4.0
4.0
16.0
rate = k[A]2[B]0 = k[A]2
k = 4.0 Ms-1
=>
= 400 M-1s-1
Exponent Values Relative to DRate
Exponent Value
0
1
2
3
4
[conc]
rate
double
double
double
double
double
same
double
x4
x8
x 16
Problem:
Data for:
2NO + H2 => N2O + H2O
Expt. #
1
2
3
[NO]
[H2]
rate
6.4x10-3
2.2x10-3
2.6x10-5
12.8x10-3
2.2x10-3
1.0x10-4
6.4x10-3
4.5x10-3
5.0x10-5
rate = k[NO]2[H2]
Units of Rate Constants
• units of rates
M/s
• units of rate constants will vary depending
on order of rxn
M = 1 (M)2 (M)
s
sM2
rate =
for
k [A]2 [B]
• rate constants are independent of the
concentration
Concentration vs. Time
1st and 2nd order integrated rate equations
• First Order: rate = - D[A] = k [A]
Dt
• ln [A]t = - kt
[A]0
A = reactant
• or ln [A]t - ln [A]0 = - kt
Conversion to base-10 logarithms:
ln [A]t = - kt
[A]0
to
log [A]t = - kt
[A]0
2.303
Problem:
The rate equation for the reaction of sucrose in
water is, rate = k[C12H22O11]. After 2.57 h at
27°C, 5.00 g/L of sucrose has decreased to 4.50
g/L. Find k.
C12H22O11(aq)
ln 4.50g/L
5.00g/L
+
H2O(l)
=>
= - k (2.57 h)
k = 0.0410 h-1
2C6H12O6
Concentration vs. Time
• Second Order:
rate = - D[A] = k[A]2
Dt
• 1 - 1 = kt
[A]t [A]0
• second order rxn with one reactant:
rate = k [A]2
Problem:
Ammonium cyanate, NH4NCO, rearranges in water to
give urea, (NH2)2CO. If the original concentration of
NH4NCO is 0.458 mol/L and k = 0.0113 L/mol min,
how much time elapses before the concentration is
reduced to 0.300 mol/L?
NH4NCO(aq) =>
1
(0.300)
-
(NH2)2CO(aq)
1
(0.458)
=
rate = k[NH4NCO]
(0.0113) t
t = 102 min
Graphical Methods
• Equation for a Straight Line
• y
=
bx
• ln[A]t
= - kt
• 1
[A]t
=
kt
+
+
a
ln[A]0
+
1
[A]0
b = slope
a = y intercept
x = time
1st order
2nd order
First Order: 2H2O2(aq) 
[H2O2]
time
2H2O(l)
+ O2(g)
First Order:
2H2O2(aq) 
2H2O(l)
slope, b = -1.06 x 10-3 min-1
ln [H2O2]
time
+ O2(g)
= -k
Second Order: 2NO2

2NO
1/[NO2]
slope, b = +k
time
+ O2
Half-Life of a
0.020 M
st
1
order process:
t1/2 = 0.693
k
[M]
0.010 M
0.005 M
t1/2
t1/2
time
Problem:
The decomposition of SO2Cl2 is first order in
SO2Cl2 and has a half-life of 4.1 hr. If you
begin with 1.6 x 10-3 mol of SO2Cl2 in a flask,
how many hours elapse before the quantity of
SO2Cl2 has decreased to 2.00 x 10-4 mol?
SO2Cl2(g)
=>
SO2(g)
+
Cl2(g)
Temperature Effects
• Rates typically increase with T increase
• Collisions between molecules increase
• Energy of collisions increase
• Even though only a small fraction of
collisions lead to reaction
• Minimum Energy necessary for reaction
is the Activation Energy
Molecular Theory
(Collision Theory)
Energy
Activation Energy, Ea
DH reaction
Ea forward rxn.
Ea reverse rxn.
Reactant
Product
Reaction Progress
Activation Energy
• Activation Energy varies greatly
– almost zero to hundreds of kJ
– size of Ea affects reaction rates
• Concentration
– more molecules, more collisions
• Molecular Orientation
– collisions must occur “sterically”
The Arrhenius Equation
• increase temperature, inc. reaction rates
• rxn rates are a to energy, collisions,
temp. & orient
• k = Ae-Ea/RT
k = rxn rate constant
A = frequency of collisions
-Ea/RT = fraction of molecules with energy necessary
for reaction
Graphical Determination of Ea
rearrange eqtn to give straight-line eqtn
y
=
bx
+ a
ln k = -Ea 1 + ln A
R T
ln k
slope = -Ea/R
1/T
Problem:
Data for the following rxn are listed in the
table. Calculate Ea graphically, calculate A
and find k at 311 K.
Mn(CO)5(CH3CN)+ + NC5H5 => Mn(CO)5(NC5H5)+ + CH3CN
ln k
k, min-1
T (K)
-3.20
-2.50
-1.85
0.0409
0.0818
0.157
298
308
318
1/T x 10-3
3.35
3.25
3.14
slope = -6373 = -Ea/R
-1.85
y intercept = 18.19 = ln A
A = 8.0 x 10 7
-2.50
k = 0.0985 min-1
-3.20
ln k
Ea = (-6373)(-8.31 x 10-3 kJ/K mol)
= 53.0 kJ
3.14 3.25 3.35
x 10-3
1/T
Problem:
The energy of activation for
C4H8(g) => 2C2H4(g)
is 260 kJ/mol at 800 K and k = 0.0315 sec
Find k at 850 K.
ln k2 = - Ea (1/T2 - 1/T1)
k1
R
k at 850 K = 0.314 sec-1
Reaction Mechanisms
• Elementary Step
– equation describing a single molecular event
• Molecularity
– unimolecular
– bimolecular
– termolecular
• 2O3
(1)
(2)
=>
3O2
O3
=>
O3 + O
O2 + O
=> 2 O2
unimolecular
bimolecular
Rate Equations
• Molecularity
unimolecular
bimolecular
bimolecular
termolecular
Rate Law
rate = k[A]
rate = k[A][B]
rate = k[A]2
rate = k[A]2[B]
– notice that molecularity for an elementary step is the
same as the order
2O3 =>
O3
3O2
=> O2 +
O3 + O =>
O
2O2
2O3 + O => 3O2 + O
O is an intermediate
rate = k[O3]
rate = k’[O3][O]
Problem:
• Write the rate equation and give the
molecularity of the following elementary
steps:
NO(g) + NO3(g) => 2NO2(g)
rate = k[NO][NO3]
bimolecular
(CH3)3CBr(aq) => (CH3)3C+(aq)
rate = k[(CH3)3CBr]
unimolecular
+ Br(aq)
Mechanisms and Rate Equations
rate determining step is the slow step -the overall rate is limited by the rate
determining step
step 1
step 2
overall
NO2 + F2 => FNO2 + F
k1
NO2 + F => FNO2
k2
rate = k1[NO2][F2]
slow
rate = k2[NO2][F]
fast
2NO2 + F2 => 2FNO2
rate = k1[NO2][F2]
Problem:
• Given the following reaction and rate law:
NO2(g) + CO(g) => CO2(g) + NO(g)
rate = k[NO2]2
– Does the reaction occur in a single step?
– Given the two mechanisms, which is most
likely:
NO2 + NO2 =>NO3 + NO
NO3 + CO => NO2 + CO2
NO2 => NO + O
CO + O => CO2
Reaction Mechanisms & Equilibria
2O3(g)
1:
2:
O3(g)
O(g)
3O2(g)
k1
k2
+
overall rxn
O2(g) + O(g)
equil.
O3(g)
k3
fast
rate1 = k1[O3]
rate2 = k2[O2][O]
2O2(g) slow
rate3 = k3[O][O3]
rate 3 includes the conc. of an intermediate and
the exptl. rate law will include only species that
are present in measurable quantities
Substitution Method
at equilibrium
k1[O3] = k2[O2][O]
rate3 =k3[O][O3]
substitute
rate3 = k3k1 [O3]2
k2 [O2]
overall rate
= k’ [O3]2
[O2]
[O] =
or
k1 [O3]
k2 [O2]
Problem:
Derive the rate law for the following
reaction given the mechanism step below:
OCl - (aq) + I -(aq)
OI -(aq) + Cl -(aq)
OCl - + H2O
I - + HOCl
HOI + OH -
k1
k3
k2
k4
HOCl + OH HOI + Cl H2O + OI -
fast
slow
fast
Cont’d
rate1 = k1 [OCl -][H2O] =
rate 2 = k2 [HOCl][OH -]
[HOCl] = k1[OCl -][H2O]
k2[OH -]
solvent
rate 3 = k3 [HOCl][I -]
rate 3 = k3k1[OCl -][H2O][I -]
k2 [OH -]
overall rate = k’ [OCl -][I -]
[OH -]
Catalyst
• Facilitates the progress of a reaction by
lowering the overall activation energy
– homogeneous
– heterogeneous
Ea
Energy
Ea
DHrxn
Reaction Progress
catalysts are used in an early rxn step but
regenerated in a later rxn step
Uncatalyzed Reaction:
O3(g) <=>
O(g)
O2(g) + O(g)
+ O3(g) => 2O2(g)
Catalyzed Reaction:
Step 1:
Cl(g) + O3(g) + O(g) =>
Step 2:
ClO(g) + O2(g) + O(g) => Cl(g) + 2O2(g)
Overall rxn:
ClO(g) + O2(g) + O(g)
O3(g) + O(g) => 2O2(g)
Ea uncatalyzed rxn
Ea catalyzed rxn
Cl + O3 + O
ClO + O2 + O
Cl + O2 + O2