Lecture 6 Kinetika Kesetimbangan Kimia
Rate Laws
• Rate laws are always determined experimentally.
• Reaction order is always defined in terms of reactant
(not product) concentrations.
• The order of a reactant is not related to the
stoichiometric coefficient of the reactant in the
balanced chemical equation.
F2 (g) + 2ClO2 (g)
2FClO2 (g)
rate = k [F2][ClO2] 1
Determine the rate law and calculate the rate constant
for the following reaction from the following data:
2SO42- (aq) + I3- (aq)
S2O82- (aq) + 3I- (aq)
Experiment
[S2O82-]
[I-]
Initial Rate
(M/s)
1
0.08
0.034
2.2 x 10-4
2
0.08
0.017
1.1 x 10-4
3
0.16
0.017
2.2 x 10-4
rate = k [S2O82-]x[I-]y
y=1
x=1
rate = k [S2O82-][I-]
Double [I-], rate doubles (experiment 1 and 2)
Double [S2O82-], rate doubles (experiment 2 and 3)
2.2 x 10-4 M/s
rate
k=
=
= 0.08/M•s
2[S2O8 ][I ] (0.08 M)(0.034 M)
First-Order Reactions
[A]
rate = t
A produk
rate = k [A]
k=-
D[A]
[A] Dt
k=
-
= k [A]
-
ln[A] - ln[A]0 = - kt
= det-1
= k dt
[A] is the concentration of A at any time
[A]0 is the concentration of A at time t=0
ln
= - kt
[A] = [A]0e-kt
2N2O5 4NO2 + O2
(det)
[N2O5]
ln[N2O5]
0
0.91
-0.094
300
0.75
-0.29
600
0.64
-0.45
1200
0.44
-0.82
3000
0.16
-1.83
Decomposition of N2O5
k = 5,7 x 10-4 detik-1
13.3
The reaction 2A
B is first order in A with a rate
constant of 2.8 x 10-2 s-1 at 80C. How long will it take for
A to decrease from 0.88 M to 0.14 M ?
[A] = [A]0e-kt
[A]0 = 0.88 M
ln[A] - ln[A]0 = - kt
[A] = 0.14 M
ln[A]0 - ln[A] = kt
ln[A]0 – ln[A]
=
t=
k
ln
[A]0
[A]
k
ln
=
0.88 M
0.14 M
2.8 x 10-2 s-1
= 66 s
13.3
First-Order Reactions
half-life, t½, is the time required for the concentration of a reactant to decrease to half of
initial concentration (waktu yang diperlukan agar konsentrasi reaktan turun menjadi setengah
i konsentrasi awalnya)
t½ = t when [A] = [A]0/2
ln
t½ =
[A]0
[A]0/2
k
0.693
ln 2
=
=
k
k
What is the half-life of N2O5 if it decomposes with a rate
constant of 5.7 x 10-4 s-1?
0.693
t½ = ln 2 =
= 1200 s = 20 minutes
-4
-1
k
5.7 x 10 s
How do you know decomposition is first order?
units of k (s-1)
First-order reaction
A
product
half-life
[A] = [A]0/n
1
2
2
4
3
8
4
16
Second-Order Reactions
D[A]
rate = Dt
rate/[A]2
M/dt
= .dt
M
M
rate = k [A]2
[A] is the concentration of A at any time t
[A]0 is the concentration of A at time t=0
1
1
= kt
[A]0
[A]
Half life for second order
t½ = t when [A] = [A]0/2
1
t½ =
k[A]0
Zero-Order Reactions
D[A]
rate = Dt
[A] - [A]0 = kt
rate = k [A]0 = k
[A] is the concentration of A at any time t
[A]0 is the concentration of A at time t=0
Half life for zero order
t½ = t when [A] = [A]0/2
[A]0
t½ =
2k
Summary of the Kinetics of Zero-Order, First-Order
and Second-Order Reactions
Order
0
Rate Law
rate = k
1
rate = k [A]
2
[A]2
rate = k
Concentration-Time
Equation
[A] - [A]0 = - kt
Half-Life
t½ =
[A]0
2k
ln[A] - ln[A]0 = - kt
Ln 2
t½ =
k
1
1
= kt
[A]0
[A]
1
t½ =
k[A]0
A+B
C+D
Exothermic Reaction
Endothermic Reacti
produk lebih stabil
dingkan reaktan, maka
si akan diiringi dengan
asan kalor (eksotermik).
liknya, jika produk kurang
l dibandingkan reaktan,
kalor akan diserap dari
ungan oleh campuran yang
aksi dan reaksinya bersifat
termik.
activation energy (Ea) is the minimum amount of energy required to initiate a chemical
tion.
si yang terbentuk sementara oleh molekul reaktansebagai akibat tumbukkan sebel
bentuk produk dinamakan kompleks teraktifasi (activated complex).
Temperature Dependence of the Rate Constant
k = A • exp -Ea/RT
(Arrhenius equation)
Ea is the activation energy (J/mol)
R is the gas constant (8.314 J/K•mol)
T is the absolute temperature
A is the frequency factor
-Ea 1
ln k = + lnA
R T
ln k = ln A -
(3) Reaction Mechanisms
The overall progress of a chemical reaction can be represented at the
molecular level by a series of simple elementary steps or elementary
reactions.
The sequence of elementary steps that leads to product formation is the
reaction mechanism (tahap elementer yang mengarah pada pembentukkan
produk disebut mekanisme reaksi.
2NO (g) + O2 (g)
2NO2 (g)
N2O2 is detected during the reaction!
Elementary step:
NO + NO
N2O2
+ Elementary step:
N2O2 + O2
2NO2
Overall reaction:
2NO + O2
2NO2
Reaction Intermediates
Intermediates are species that appear in a reaction mechanism
but not in the overall balanced equation.
An intermediate is always formed in an early elementary step and
consumed in a later elementary step.
Elementary step:
NO + NO
N2O2
+ Elementary step:
N2O2 + O2
2NO2
Overall reaction:
2NO + O2
2NO2
Rate Laws and Rate Determining Steps
Writing plausible reaction mechanisms:
• The sum of the elementary steps must give the overall
balanced equation for the reaction.
• The rate-determining step should predict the same rate
law that is determined experimentally.
The rate-determining step is the slowest step
in the sequence of steps leading to product
formation.
Rate Laws and Elementary Steps
Unimolecular reaction
A
products
rate = k [A]
Bimolecular reaction
A+B
products
rate = k [A][B]
Bimolecular reaction
A+A
products
rate = k [A]2
anyaknya molekul yang bereaksi dalam tahap elementer menentukan molekularitas reaksi
olecularity of reaction). Setiap tahap elementer yang baru dibahas disebut reaksi bimolecul
imolecular reaction), yaitu tahap elementer yang melibatkan dua molekul. Reak
nimolekular (unimolecular reaction) ialah reaksi yang tahap elementernya hanya melibatka
atu molekul yang bereaksi.
A catalyst is a substance that increases the rate of a
chemical reaction without itself being consumed.
Ea
uncatalyzed
k
catalyzed
ratecatalyzed > rateuncatalyzed
13.6
Energy Diagrams
Exothermic
Endothermic
(a) Activation energy (Ea) for the forward
reaction
50 kJ/mol
300 kJ/mol
150 kJ/mol
100 kJ/mol
(b) Activation energy (Ea) for the reverse
reaction
-100 kJ/mol
+200 kJ/mol
(c) Delta H
The experimental rate law for the reaction between NO2
and CO to produce NO and CO2 is rate = k[NO2]2. The
reaction is believed to occur via two steps:
Step 1:
NO2 + NO2
NO + NO3
Step 2:
NO3 + CO
NO2 + CO2
What is the equation for the overall reaction?
NO2+ CO
NO + CO2
What is the intermediate? Catalyst?
NO3
NO2
What can you say about the relative rates of steps 1 and 2?
rate = k[NO2]2 is the rate law for step 1, so
step 1 must be slower than step 2
Write the rate law for this reaction.
Rate = k [HBr] [O2]
List all intermediates in this reaction. HOOBr, HOBr
List all catalysts in this reaction.
None
Ostwald Process
4NH3 (g) + 5O2 (g)
Pt catalyst
2NO (g) + O2 (g)
2NO2 (g) + H2O (l)
4NO (g) + 6H2O (g)
2NO2 (g)
HNO2 (aq) + HNO3 (aq)
Pt-Rh catalysts used
in Ostwald process
Hot Pt wire
over NH3 solution
Catalytic Converters
CO + Unburned Hydrocarbons + O2
2NO + 2NO2
catalytic
converter
catalytic
converter
CO2 + H2O
2N2 + 3O2
13.6
Enzyme Catalysis
13.6
Type of reactions
rous reactions
s) + 2H2O(l) 2KOH(aq) + H2(g)
Very rapid reactions
Formation of insoluble salts
Ag+(aq) + Cl−(aq) AgCl(s)
Very rapid reactions
Potassium reacts with
water vigorously
Formation of insoluble bases
Fe3+(aq) + 3OH−(aq) Fe(OH)
Slow reactions
ery rapid reactions
id-alkali neutralization reactions
+(aq)
Fermentation of glucose
C6H12O6(aq) 2C2H5OH(aq) + 2CO2(g)
+ OH−(aq) H2O(l)
pid or moderate reactions
placement reactions of metals:
(s) + 2Ag+(aq) Zn2+(aq) + 2Ag(s)
Very slow reactions
Rusting of iron
4Fe(s) + 3O2(g) + 2nH2O(l) 2Fe2O3 · nH2O(s)
placement reactions of halogens:
aq) + 2Br(aq) 2Cl(aq) + Br2(aq)
Extremely slow reactions
CaCO3(s) + 2H+(aq) Ca2+(aq) + CO2(g) + H2O(l)
Before corrosion
After corrosion
Amount is usually expressed in
erage rate of reaction
Total change in amount of a product or a reactant
Total time taken for the change to occur
Concentration
Mass
Volume
Pressure
mol dm−3
g
cm3 or dm3
atm
The Concept of Equilibrium
Chemical equilibrium occurs when a reaction and
its reverse reaction proceed at the same rate.
• As a system approaches
equilibrium, both the
forward and reverse
reactions are occurring.
• At equilibrium, the
forward and reverse
reactions are
proceeding at the same
rate.
A System at Equilibrium
Once equilibrium is
achieved, the amount of
each reactant and
product remains
constant.
A System at Equilibrium
Rates become equal
Concentrations become constant
a system at equilibrium, both the forward and reverse reactions are runnin
ultaneously. We write the chemical equation with a double arrow.
The Equilibrium Constant
Forward reaction:
Rate law
Reverse reaction:
Rate law
The Equilibrium Constant
At equilibrium
The ratio of the rate constants is a constant (as long as T is constant).
The expression becomes:
To generalize, the reaction:
has the equilibrium expression:
Kc =
is depending upon the temperature (T)
a K >1 kesetimbangan akan terletak disebelah kanan, cenderung ke arah prod
a K
• Rate laws are always determined experimentally.
• Reaction order is always defined in terms of reactant
(not product) concentrations.
• The order of a reactant is not related to the
stoichiometric coefficient of the reactant in the
balanced chemical equation.
F2 (g) + 2ClO2 (g)
2FClO2 (g)
rate = k [F2][ClO2] 1
Determine the rate law and calculate the rate constant
for the following reaction from the following data:
2SO42- (aq) + I3- (aq)
S2O82- (aq) + 3I- (aq)
Experiment
[S2O82-]
[I-]
Initial Rate
(M/s)
1
0.08
0.034
2.2 x 10-4
2
0.08
0.017
1.1 x 10-4
3
0.16
0.017
2.2 x 10-4
rate = k [S2O82-]x[I-]y
y=1
x=1
rate = k [S2O82-][I-]
Double [I-], rate doubles (experiment 1 and 2)
Double [S2O82-], rate doubles (experiment 2 and 3)
2.2 x 10-4 M/s
rate
k=
=
= 0.08/M•s
2[S2O8 ][I ] (0.08 M)(0.034 M)
First-Order Reactions
[A]
rate = t
A produk
rate = k [A]
k=-
D[A]
[A] Dt
k=
-
= k [A]
-
ln[A] - ln[A]0 = - kt
= det-1
= k dt
[A] is the concentration of A at any time
[A]0 is the concentration of A at time t=0
ln
= - kt
[A] = [A]0e-kt
2N2O5 4NO2 + O2
(det)
[N2O5]
ln[N2O5]
0
0.91
-0.094
300
0.75
-0.29
600
0.64
-0.45
1200
0.44
-0.82
3000
0.16
-1.83
Decomposition of N2O5
k = 5,7 x 10-4 detik-1
13.3
The reaction 2A
B is first order in A with a rate
constant of 2.8 x 10-2 s-1 at 80C. How long will it take for
A to decrease from 0.88 M to 0.14 M ?
[A] = [A]0e-kt
[A]0 = 0.88 M
ln[A] - ln[A]0 = - kt
[A] = 0.14 M
ln[A]0 - ln[A] = kt
ln[A]0 – ln[A]
=
t=
k
ln
[A]0
[A]
k
ln
=
0.88 M
0.14 M
2.8 x 10-2 s-1
= 66 s
13.3
First-Order Reactions
half-life, t½, is the time required for the concentration of a reactant to decrease to half of
initial concentration (waktu yang diperlukan agar konsentrasi reaktan turun menjadi setengah
i konsentrasi awalnya)
t½ = t when [A] = [A]0/2
ln
t½ =
[A]0
[A]0/2
k
0.693
ln 2
=
=
k
k
What is the half-life of N2O5 if it decomposes with a rate
constant of 5.7 x 10-4 s-1?
0.693
t½ = ln 2 =
= 1200 s = 20 minutes
-4
-1
k
5.7 x 10 s
How do you know decomposition is first order?
units of k (s-1)
First-order reaction
A
product
half-life
[A] = [A]0/n
1
2
2
4
3
8
4
16
Second-Order Reactions
D[A]
rate = Dt
rate/[A]2
M/dt
= .dt
M
M
rate = k [A]2
[A] is the concentration of A at any time t
[A]0 is the concentration of A at time t=0
1
1
= kt
[A]0
[A]
Half life for second order
t½ = t when [A] = [A]0/2
1
t½ =
k[A]0
Zero-Order Reactions
D[A]
rate = Dt
[A] - [A]0 = kt
rate = k [A]0 = k
[A] is the concentration of A at any time t
[A]0 is the concentration of A at time t=0
Half life for zero order
t½ = t when [A] = [A]0/2
[A]0
t½ =
2k
Summary of the Kinetics of Zero-Order, First-Order
and Second-Order Reactions
Order
0
Rate Law
rate = k
1
rate = k [A]
2
[A]2
rate = k
Concentration-Time
Equation
[A] - [A]0 = - kt
Half-Life
t½ =
[A]0
2k
ln[A] - ln[A]0 = - kt
Ln 2
t½ =
k
1
1
= kt
[A]0
[A]
1
t½ =
k[A]0
A+B
C+D
Exothermic Reaction
Endothermic Reacti
produk lebih stabil
dingkan reaktan, maka
si akan diiringi dengan
asan kalor (eksotermik).
liknya, jika produk kurang
l dibandingkan reaktan,
kalor akan diserap dari
ungan oleh campuran yang
aksi dan reaksinya bersifat
termik.
activation energy (Ea) is the minimum amount of energy required to initiate a chemical
tion.
si yang terbentuk sementara oleh molekul reaktansebagai akibat tumbukkan sebel
bentuk produk dinamakan kompleks teraktifasi (activated complex).
Temperature Dependence of the Rate Constant
k = A • exp -Ea/RT
(Arrhenius equation)
Ea is the activation energy (J/mol)
R is the gas constant (8.314 J/K•mol)
T is the absolute temperature
A is the frequency factor
-Ea 1
ln k = + lnA
R T
ln k = ln A -
(3) Reaction Mechanisms
The overall progress of a chemical reaction can be represented at the
molecular level by a series of simple elementary steps or elementary
reactions.
The sequence of elementary steps that leads to product formation is the
reaction mechanism (tahap elementer yang mengarah pada pembentukkan
produk disebut mekanisme reaksi.
2NO (g) + O2 (g)
2NO2 (g)
N2O2 is detected during the reaction!
Elementary step:
NO + NO
N2O2
+ Elementary step:
N2O2 + O2
2NO2
Overall reaction:
2NO + O2
2NO2
Reaction Intermediates
Intermediates are species that appear in a reaction mechanism
but not in the overall balanced equation.
An intermediate is always formed in an early elementary step and
consumed in a later elementary step.
Elementary step:
NO + NO
N2O2
+ Elementary step:
N2O2 + O2
2NO2
Overall reaction:
2NO + O2
2NO2
Rate Laws and Rate Determining Steps
Writing plausible reaction mechanisms:
• The sum of the elementary steps must give the overall
balanced equation for the reaction.
• The rate-determining step should predict the same rate
law that is determined experimentally.
The rate-determining step is the slowest step
in the sequence of steps leading to product
formation.
Rate Laws and Elementary Steps
Unimolecular reaction
A
products
rate = k [A]
Bimolecular reaction
A+B
products
rate = k [A][B]
Bimolecular reaction
A+A
products
rate = k [A]2
anyaknya molekul yang bereaksi dalam tahap elementer menentukan molekularitas reaksi
olecularity of reaction). Setiap tahap elementer yang baru dibahas disebut reaksi bimolecul
imolecular reaction), yaitu tahap elementer yang melibatkan dua molekul. Reak
nimolekular (unimolecular reaction) ialah reaksi yang tahap elementernya hanya melibatka
atu molekul yang bereaksi.
A catalyst is a substance that increases the rate of a
chemical reaction without itself being consumed.
Ea
uncatalyzed
k
catalyzed
ratecatalyzed > rateuncatalyzed
13.6
Energy Diagrams
Exothermic
Endothermic
(a) Activation energy (Ea) for the forward
reaction
50 kJ/mol
300 kJ/mol
150 kJ/mol
100 kJ/mol
(b) Activation energy (Ea) for the reverse
reaction
-100 kJ/mol
+200 kJ/mol
(c) Delta H
The experimental rate law for the reaction between NO2
and CO to produce NO and CO2 is rate = k[NO2]2. The
reaction is believed to occur via two steps:
Step 1:
NO2 + NO2
NO + NO3
Step 2:
NO3 + CO
NO2 + CO2
What is the equation for the overall reaction?
NO2+ CO
NO + CO2
What is the intermediate? Catalyst?
NO3
NO2
What can you say about the relative rates of steps 1 and 2?
rate = k[NO2]2 is the rate law for step 1, so
step 1 must be slower than step 2
Write the rate law for this reaction.
Rate = k [HBr] [O2]
List all intermediates in this reaction. HOOBr, HOBr
List all catalysts in this reaction.
None
Ostwald Process
4NH3 (g) + 5O2 (g)
Pt catalyst
2NO (g) + O2 (g)
2NO2 (g) + H2O (l)
4NO (g) + 6H2O (g)
2NO2 (g)
HNO2 (aq) + HNO3 (aq)
Pt-Rh catalysts used
in Ostwald process
Hot Pt wire
over NH3 solution
Catalytic Converters
CO + Unburned Hydrocarbons + O2
2NO + 2NO2
catalytic
converter
catalytic
converter
CO2 + H2O
2N2 + 3O2
13.6
Enzyme Catalysis
13.6
Type of reactions
rous reactions
s) + 2H2O(l) 2KOH(aq) + H2(g)
Very rapid reactions
Formation of insoluble salts
Ag+(aq) + Cl−(aq) AgCl(s)
Very rapid reactions
Potassium reacts with
water vigorously
Formation of insoluble bases
Fe3+(aq) + 3OH−(aq) Fe(OH)
Slow reactions
ery rapid reactions
id-alkali neutralization reactions
+(aq)
Fermentation of glucose
C6H12O6(aq) 2C2H5OH(aq) + 2CO2(g)
+ OH−(aq) H2O(l)
pid or moderate reactions
placement reactions of metals:
(s) + 2Ag+(aq) Zn2+(aq) + 2Ag(s)
Very slow reactions
Rusting of iron
4Fe(s) + 3O2(g) + 2nH2O(l) 2Fe2O3 · nH2O(s)
placement reactions of halogens:
aq) + 2Br(aq) 2Cl(aq) + Br2(aq)
Extremely slow reactions
CaCO3(s) + 2H+(aq) Ca2+(aq) + CO2(g) + H2O(l)
Before corrosion
After corrosion
Amount is usually expressed in
erage rate of reaction
Total change in amount of a product or a reactant
Total time taken for the change to occur
Concentration
Mass
Volume
Pressure
mol dm−3
g
cm3 or dm3
atm
The Concept of Equilibrium
Chemical equilibrium occurs when a reaction and
its reverse reaction proceed at the same rate.
• As a system approaches
equilibrium, both the
forward and reverse
reactions are occurring.
• At equilibrium, the
forward and reverse
reactions are
proceeding at the same
rate.
A System at Equilibrium
Once equilibrium is
achieved, the amount of
each reactant and
product remains
constant.
A System at Equilibrium
Rates become equal
Concentrations become constant
a system at equilibrium, both the forward and reverse reactions are runnin
ultaneously. We write the chemical equation with a double arrow.
The Equilibrium Constant
Forward reaction:
Rate law
Reverse reaction:
Rate law
The Equilibrium Constant
At equilibrium
The ratio of the rate constants is a constant (as long as T is constant).
The expression becomes:
To generalize, the reaction:
has the equilibrium expression:
Kc =
is depending upon the temperature (T)
a K >1 kesetimbangan akan terletak disebelah kanan, cenderung ke arah prod
a K