# Ch 12 ppt notes-new

```AP CHEMISTRY
CHAPTER 12
KINETICS
Chemical Kinetics
• Thermodynamics tells us if a reaction can occur
• Kinetics tells us how quickly the reaction occurs
– some reactions that are thermodynamically feasible are kinetically
so slow as to be imperceptible
C diamond  O 2g   CO 2g  G o298  396kJ
VERY SLOW
H +aq  + OH -aq   H 2 O l  G o298 = -79 kJ
INSTANTANE OUS
2
Reaction Rate- change in
concentration of a reactant or
product per unit time.
[A] = concentration in mol/L
Rate = [A]
t
If the rate expression involves a
reactant:
Rate = − [A]
t
(negative because [ ] decreases)
The above gives the average rate.
2NO2 2NO + O2
To get an instantaneous rate, we can
compute the slope of a line tangent
to the curve at that point.
Rate = −(slope of the tangent line)
The rate of a reaction is not
constant but changes with time
because concentrations change
with time.
We will only work with reaction
rates that are “initial rates”
(reverse reaction is negligible)
Differential Rate Law or Rate Equation
For the reaction aA+bB + … gG +hH+ …
Rate = k[A]m[B]n…
[A] & [B] represent molarities
The exponents are positive or
negative, integers or fractions.
usually positive integers (small whole
numbers)
k = rate constant
*value depends on reaction,
temperature and presence of a catalyst
*faster the reaction, larger the k value
The exponents determine the
order of the reactants. The sum
of the exponents is the order of
the reaction.
R = k[A][B]2 is first order in A,
second order in B and third order
overall.
The units of k can be calculated by
reaction orders and units of concentration and
rate. For example, if rate is in mol/Ls in the
above rate law, we can find the units for k as
follows:
Rate
= k so k = mol
[A][B]2
Ls
mol mol2
L
L2
This simplifies to: k = L2
mol2.s
The “shortcut” to determining units
of k is as follows:
.
x
x
k = L /(mol time).
The value of x will be one less than
the order of the reaction.
If the reaction is 3rd order,
.
2
2
k = L /mol time.
If the reaction is 2nd order,
k = L/mol.time
If the reaction is 1st order, k = 1/time
Determining Differential Rate
Laws from Experimental Data
If doubling the initial [ ] of a
reactant causes the initial rate to
double, the reaction is first order
in that reactant.
If doubling the initial [ ] of a
reactant causes the initial rate to
order in that reactant.
If doubling the initial [ ] of a
reactant does not change the
initial rate, the reaction is zero
order in that reactant and that
reactant is removed from the rate
law.
2A + B 2C
Ex.
[A]
[B]
1
0.1
0.2 2
1
2
0.1
0.4
1
2
3
0.2
0.4
Determine the rate law:
Rate = k[A]x[B]y
Rate = k[A]x[B]1
Rate = k[A]2[B]
Rate
0.10
0.20
0.80
Because 21=2, B is 1st order
Because 22 =4, A is 2nd order
2
4
AB
[A]
Rate
0.05 2 3 x 10-4 4
0.10 2 1.2 x 10-3
4
-3
0.20 4.8 x 10
Determine the rate law:
x
k[A]
Rate =
Rate = k[A]2
Because 22=4, A is 2nd order
A+BC
Exp [A]
[B]
Rate
1 2 0.10 1 0.20 1 1.0×105
2
0.202 0.20
1.0×105
1
2
3
0.20 0.40
2.0×105
Determine the rate law:
Rate = k[A]x[B]y
Rate = k[A]0[B]y Because 20=1, A is 0 order
Rate = k [B] Because 21=2, B is 1st order
A + 2B 2C
Ex.
[A]
[B]
Rate
1
0.1 2 0.1 1
0.10
2
0.2
0.1
0.20
2
2
3
0.4
0.2
0.40
Determine the rate law:
Rate = k[A]x[B]y
Rate = k[A]1[B]y
Rate = k[A]1[B]0
Rate = k[A]
Because 21=2, A is 1st order
No control in this example. Because
doubling A causes the rate to double,
doubling B must have had no effect. B is
2
2
NH4+ + NO2  N2 + 2H2
Exp
[NH4+] [NO2] Rate(mol/Lmin)
7
1
0.100
0.005
1.35
×
10
1
2
2
2
0.100
0.010
2.70 × 107
2
1
3
0.200
0.010 2 5.40 × 107
Determine the rate law:
Determine the value of k and its units:
Rate = k[NH4+]x[NO2]y
Rate = k[NH4+]x[NO2]1 Because 21=2, NH4+ is 1st order
Rate = k[NH4+] [NO2] orer
Because 21=2, NO2- is 1st order
1.35×107mol/Lmin = k(0.100mol/L)(0.005mol/L)
k = 2.7 ×104 L/mol.min
Experiment
[A]
1
10.1
2
3
2
19.8
40.2
[B]
2
4
Initial Rate
mol L-1s-1
2.01
3.99
2.00
16
1
5.00
80.0
80.0
Determine the rate law:
Rate = k[A]x[B]y
42=16 so [A] is 2nd order
Rate = k[A]2[B]y
Because A is 2nd order, doubling A will cause the rate to
quadruple. When both A and B were doubled, the rate was 16
times greater. This means that B is also 2nd order.
Rate = k[A]2[B]2
Determine the value of k and its units.
5.00 mol L-1s-1 = k(10.1mol/L)2(2.01mol/L)2
k = 0.0121 L3mol-3s-1
16
Integrated Rate Law
-expresses how the concentration
of the reactant depends on time
concentrations and using multiple
experiments, one experiment is
done and concentration changes
over time are measured.
1st order integrated rate law
 A0 
  kt
ln 
 At 
or
 At 
   kt
ln 
 A0 
A plot of ln[A] vs t always gives
a straight line for a 1st order
reaction.
The slope = -k
Ex. At 400oC, the 1st order conversion of cyclopropane
into propylene has a rate constant of 1.16 × 10-6s-1. If the
initial concentration of cyclopropane is 1.00 × 10-2 mol/L
at 400oC, what will its concentration be 24.0 hrs after the
reaction begins?
24 hrs 3600s = 86400s
1 hr
ln[A]t = −kt
[A]0
[A]t = 9.05×10−3
ln [A]t
= −1.16×10−6(86400)
1.00×10−2
Half-life (t1/2) is the length of time
required for the concentration of a
reactant to decrease to half of its
initial value.
t1/2 = 0.693/k
Where does 0.693 come from?
It is the natural log of 2.
A fast reaction with a short t1/2
has a large k.
A slow reaction with a long t1/2
has a small k.
A Plot of (N2O5) Versus Time for the Decomposition Reaction of N2O5
Example: The decomposition of SO2Cl and
Cl2 is a first order reaction with
k = 2.2×10−5s−1 at 320oC. Determine the halflife of this reaction.
t1/2 = 0.693/k
t1/2 = 0.693/(2.2×10−5s−1)
t1/2 = 31500 s or 525 min or 8.75 hours
Example: The decomposition of N2O5 dissolved in CCl4 is
a first order reaction. The chemical change is:
2N2O5  4NO2 + O2
At 45oC the reaction was begun with an initial N2O5
concentration of 1.00 mol/L. After 3.00 hours the N2O5
concentration had decreased to 1.21×10−3 mol/L. What is
the half-life of N2O5 expressed in minutes at 45oC?
ln [A]t = −kt
[A]0
ln 1.21 × 103 = −k(180 min)
1.00
k = 0.0373 min1
t1/2 = 0.693/k
t1/2 = 0.693/0.0373 min1 = 18.6 min
Second order integrated rate law
2
Rate = k[A]
Differential rate law
1 - 1 = kt
[A]t
[A]0
A plot of 1/[A]t versus t
produces a straight line with
slope k.
(a) A Plot of In(C4H6) Versus t
(b) A Plot of 1/(C4H6) Versus t
Zero Order
Rate = k
Differential rate law
formula sheet and will not have to be
calculated. Make sure that you
understand the graphing.
[A]t  [A]o = kt
A plot of [A] versus t produces
a straight line with slope -k.
A Plot of (A) Versus
t for a Zero-Order
Reaction
An easy way to keep the graphs straight is to remember CLR=0,1,2
C=concentration, L = ln concentration, R=reciprical of concentration
If the concentration vs time is straight, it is C (0 order).
If natural log of concentration vs time is straight it is L=1st order.
If reciprical of concentration vs time is straight it is R=2nd order.
CLR
012
What order is this reaction?
What order is this reaction?
Reaction Mechanisms
intermediate - a species that is
neither a product nor a reactant in
the overall equation
-is used up in a subsequent step
elementary step- a reaction whose
rate law can be written from its
molecularity (balanced equation)
molecularity- the number of species
that must collide to produce the
reaction represented by the
elementary step.
Unimolecular step- a reaction step
involving only one molecule
Bimolecular step- a reaction step
involving the collision of two
molecules (Rate law always 2nd
order)
Rate-determining step -slowest
step
Lucy clip
Elementary Reaction -agrees with
the balanced equation
Reaction mechanisms must:
1. Add up to the overall balanced
equation.
2. Agree with the rate law
We can’t prove a mechanism
absolutely. We can only come up
with a possible mechanism.
Look at this video where we can see
that the reaction is occuring through
a series of steps (mechanism). This
is called a “clock reaction”.
clock reaction
The iodine clock reaction uses a solution of iodate ion to which an
acidified solution of sodium bisulfite is added.
Iodide ion is generated by the following slow reaction between the
iodate and bisulfite:
IO3− (aq) + 3HSO3− (aq) → I− (aq) + 3HSO4−(aq)
This is the rate determining step. The iodate in excess will oxidize
the iodide generated above to form iodine:
IO3− (aq) + 5I− (aq) + 6H+ (aq) → 3I2 + 3H2O (l)
However, the iodine is reduced immediately back to iodide by the
bisulfite:
I2 (aq) + HSO3− (aq) + H2O (l) → 2I− (aq) + HSO4−(aq) + 2H+ (aq)
When the bisulfite is fully consumed, the iodine will survive (i.e.,
no reduction by the bisulfite) to form the dark blue complex with
starch.
Ex. Elementary Rxn :
NO + N2O  NO2+ N2
Rate Law: R = k[NO][N2O]
Ex. The reaction 2NO2(g) + F2(g)  2NO2F
is thought to proceed via the following twostep mechanism:
NO2 + F2  NO2F + F
slow
F + NO2  NO2F
fast
Rate law for the reaction:
Rate = k[NO2][F2]
When an intermediate is a
reactant in the rate-determining
step, the derivation of the rate
law is more difficult.
Ex. NO2 + CO  NO + CO2
k1
Mechanism: NO2 + NO2  NO3 + NO
k1
NO3 + CO  CO2 + NO2
Both fast w/
equal rates
slow
Ex. NO2 + CO  NO + CO2
k1
Mechanism: NO2 + NO2  NO3 + NO
k-1
NO3 + CO  CO2 + NO2
Both fast w/
equal rates
slow
Slow step determines rate law (rate-determining step)
Rate law: R = k[NO3][CO] But, NO3 was an intermediate.
We must come up with something equal to NO3 to substitute.
k[NO2]2 = k[NO3][NO]
[NO3] = [NO2]2
[NO]
R = k [NO2]2[CO]
[NO]
Ex. Cl2 + CHCl3  HCl + CCl4
Mechanism
k1
Cl2  2Cl
k-1
fast
Cl + CHCl3  HCl + CCl3 slow
Cl + CCl3  CCl4
fast
Rate Law:
Ex. Cl2 + CHCl3  HCl + CCl4
Mechanism
k1
Cl2  2Cl
k-1
fast
Cl + CHCl3  HCl + CCl3 slow
Cl + CCl3  CCl4
fast
Rate Law:
R = k[Cl][CHCl3]
Cl is an intermediate
k[Cl2] = k[Cl]2
[Cl2]1/2 = [Cl]
R = k[Cl2]1/2[CHCl3]
Increasing temperature increases
reaction speed. Rate and rate
constants often double for every 10o
increase in temperature. (Rule of
thumb)
Plot Showing the Number of Collisions with a Particular Energy at T1 and T2,
where T2>T1
Molecules must collide to react.
Only a small portion of collisions
produce a reaction.
Several Possible Orientations for a Collision Between Two BrNO Molecules
Activation energy (EA)
- energy that must be overcome
to produce a chemical reaction.
Rate of reaction depends on EA,
not E.
E has no effect on rate of
reaction. The higher the EA, the
slower the reaction at a given
temperature.
Molecules and collisions have varying
energies. As temperature increases, more
collisions will have sufficient energy to
overcome the activation energy. As
temperature doubles, the fraction of
effective collisions increases exponentially.
Reaction rate is smaller than would be
predicted from the number of collisions
having sufficient energy to react. This is
because of molecular orientations.
2 factors:
1. sufficient energy
2. proper orientation
Arrhenius Equation-can be used to calculate EA
This calculation is not required on the AP exam.
ln k = ln A - EA
RT
k = rate constant
R = 8.314 J/K mol
T = Kelvin temp
A = frequency factor
(constant as temperature changes)
EA = activation energy
As EA increases, k decreases.
The Arrhenius equation describes
a line.
We can plot 1/T vs ln k and get
a straight line whose slope is
equal to -EA/R.
Plot of In(k) Versus 1/T
for the Reaction
2N2O5(g) 2g) + O2(g)
Slope = -EA/R
A variation of the Arrhenius
equation can be used to calculate
EA or to find k at another
temperature if EA is known:
ln k1 = EA 1 - 1
k2 R T2 T1
This calculation is not required on the AP exam.
Catalysis
Catalysts, such as enzymes in a biological
system or the surfaces in an automobile's
catalytic converter, increase the rate of a
chemical reaction.
catalyst- substance that speeds up a
reaction without being consumed.
-produces a new reaction pathway with
a lower activation energy
-A catalyst lowers the EA for both the
forward and reverse reaction. Some
catalysts change the mechanism of the
reaction greatly and others simply
change the rate-determining step.
enzymes- biological catalysts
homogeneous catalyst- present
in the same phase as the reacting
molecules (usually liquid phase)
heterogeneous catalyst- exists in a
different phase
-usually involves gaseous reactants being
adsorbed on the surface of a solid catalyst
(such as a car’s catalytic converter)
Either a new reaction intermediate is formed
or the probability of successful collisions is
increased. This is sometimes called a
surface catalyst.
Absorption involves penetration.
Heterogeneous Catalysis of
the Hydrogenation of Ethylene
Catalytic Converter
How does the mechanism for the catalyzed
reaction differ from the mechanism of the
uncatalyzed reaction?
```
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