Kinetics

Report
KINETICS
Chapter 14
14.1 Factors that Affect Reaction Rates
Chemical kinetics is the study of how
fast chemical reactions occur.
–Generally, the more frequently
collisions between reaction
particles occur, the faster the
reaction.
There are several important factors which
affect rates of reactions:
Physical state of the reactants
– Homogeneous mixtures react faster than
heterogeneous mixtures.
– When reactants are in different phases: more
surface area  faster reaction
Concentration of the reactants
– more reactant molecules so they collide more
frequently  faster reaction
Temperature of the reaction
– higher temperature molecules move faster and
collide more frequently  faster reaction
Presence or absence of a catalyst
– use a catalyst usually changes mechanism
14.2 Reaction Rates
• The speed of a reaction is defined as the
change that occurs per unit time.
• It is often determined by measuring the
change in concentration of a reactant or
product with time [M/sec] (could also use
change in moles but if volume is constant
[most reactions], then molarity and moles are
directly proportional – concentration is easier
to directly measure than moles. Pressure is
sometimes used for gases.)
• The speed of the chemical reaction is its
reaction rate.
•
•
•
•
•
•
Suppose A reacts to form B.
AB
Let us begin with [A] = 1.00 M
At t = 0 (time zero) there is 1.00 M A and no B present.
At t = 20 sec, there is 0.54 M A and 0.46 M B.
At t = 40 sec, there is 0.30 M A and 0.70 M B.
We can uses this information to find the average rate of the
reaction.
• For the reaction A  B, there are 2 ways of measuring rate:
• The rate of appearance of product B (i.e., change in
concentration of B per unit time) as in the preceding example.
Avg Rate =
Δ (Conc. B) (Conc. of B at t = 20s)-(Conc. of B at t = 0s)
=
Δt
20s - 0s
0.46M - 0.00 M
M
= 0.023
20 s -0 s
s
• The rate of disappearance of reactant A (i.e., the
change in concentration of A per unit time):
M
- Δ(conc. of A)
- (0.54 M - 1.00 M)
Avg Rate =

= 0.023
s
Δt
(20 - 0 s)
Avg Rate =
• Note the negative sign!
– Reaction rates are always positive!
Change of Rate with Time
•The average rate generally
decreases with time
•The rate at any instant in
time is called the
instantaneous rate.
• It is the slope of the
straight line tangent to the
curve at that instant.
• Instantaneous rate is
different from average rate.
• We usually call the
"instantaneous rate" the
rate.
Example:
Using the data in table above, calculate the
average rate of disappearance of C4H9Cl
over the time interval from 50.0 to 150.0
seconds. (b) Using the figure, estimate
the instantaneous rate of disappearance of
C4H9Cl at t = 0 (the initial rate).
[C4 H 9Cl ]
Average rate  t
(0.0741 - 0.0905) M
  1.64 x 10-4 M / s
(150.0 - 50.0) s
(b) the initial rate is given
by the slope of the
dashed line (y / x)
pick two points on the
tangent line:
(0, .100) and (200, .060)
(0.060 - 0.100) M
Rate   2.0 x 104 M / s
(200 - 0) s
Reaction Rates and Stoichiometry
• For the reaction:
C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)
• The rate of appearance of C4H9OH must equal
the rate of disappearance of C4H9Cl.
Δ C4H9Cl 
Δ C4H9OH
Rate = =
Δt
Δt
What if the stoichiometric relationships are not oneto-one?
For the reaction:
2HI(g)  H2(g) + I2(g)
The rate may be expressed as:
Δ H2
Δ I2
1 Δ HI
Rate = =
=
2 Δt
Δt
Δt
We can generalize this equation a bit.
For the reaction:
aA + bB  cC + dD
The rate may be expressed as:
Δ  A 
Δ C
Δ B 
Δ D
Rate = = =
=
aΔt
bΔt
cΔt
dΔt
The isomerization of methyl isonitrile CH3NC to acetonitrile, CH3CN, was studied
in the gas phase at 215oC and the following data was obtained:
Time (s)
0
2000
5000
8000
12000
15000
[CH3NC] (M) 0.0165
0.0110
0.00591
0.00314
0.00137
0.00074
(a) Calculate the average rate of reaction in M/s, for the time interval between each
measurement.
(a)
Time
(sec)
0
2000
5000
8000
12000
15000
Time Interval
(Δt) (sec)
2000
3000
3000
4000
3000
[CH3NC]
(M)
0.0165
0.0110
0.00591
0.00314
0.00137
0.00074
M
- 0.0055
- 0.0051
- 0.00277
- 0.00177
- 0.00063
Rate
(ΔM/ Δt)
2.8  10-6 - ( - 0.0055 M / 2000 s)
1.7  10-6
.923  10-6
.443  10-6
.21  10-6
Consider the hypothetical reaction: A(aq)  B(aq). A flask is charged with
0.065 mol of A and allowed to react to form B in a total volume of 100.0 mL.
The following data are collected:
Time (min) 0
10
20
30
40
Mol of A
0.065
0.051
0.042
0.036
0.031
(a) Calculate the number of moles of B at each time in the table, assuming there
are no molecules of B at time zero
(b) Calculate the average rate of disappearance of A for each 10 min interval, in
units of M/s.
(c) Between t = 10 min and t = 30 min, what is the average rate of appearance
of B in units of M/s? Assume the volume of the solution is constant.
Time(min)
0
10
20
30
40
(c)
Mol A
0.065
0.051
0.042
0.036
0.031
(a) Mol B
0
0.014
0.023
0.029
0.034
[A]
0.65
0.51
0.42
0.36
0.31
 [A]
(b) Rate -( [A]/time)
- 0.14
- 0.09
- 0.06
- 0.05
2.3  10-4
2  10-4
1  10-4
0.8  10-4
M B
(0.029 - 0.014 mol)/0.100 L 1 min
=
x
= 1.3 x 10-4 M/s
t
(30-10) min
60 s
- ( - 0.14 M / 600 s)
For each of the following gas-phase reactions, indicate how the rate
of disappearance of each reactant is related to the rate of
appearance of each product:
(a) H2O2(g)  H2(g) + O2(g)
(b) 2 N2O(g)  2 N2(g) + O2(g)
(c)
N2(g) + 3 H2(g)  2 NH3(g)
(a)
∆ 2 2
−
∆
(b)
∆ 2 
−
2∆
(c)
∆ 2
−
∆
=
∆ 2
∆
=
∆ 2
2∆
=
∆ 2
−
3∆
=
∆ 2
∆
=
∆ 2
∆
=
∆ 3
2∆
Consider the combustion of H2(g), 2 H2(g) + O2(g)  2 H2O(g). If hydrogen
is burning at the rate of 0.85 mol/s, what is the rate of consumption of
oxygen? What is the rate of formation of water vapor?
[H 2 O]
[H 2 ]
[O 2 ]
==2t
2t
t
- [H2 ]
H2 is burning,
= 0.85 mol/s
t
[O 2 ]
[H 2 ]
O 2 is consumed, = 
=
t
2t
H2O is produced,
1
x (0.85 mol/s) = 0.43 mol/s
2
1 [H2 O]
1 [H2 ]
= = 0.85 mol/s
2 t
2 t
The reaction 2 NO(g) + Cl2(g)  2 NOCl(g) is carried out in a closed
vessel. If the partial pressure of NO is decreasing at the rate of 23
torr/min, what is the rate of change of the total pressure in the vessel?
The change in total pressure is the sum of the changes of
each partial pressure. NO and Cl2 are disappearing and
NOCl is appearing.
P
NO = 23 /
−
t
−
PNO = − ∆ = ∆
∆
2∆
2t
ΔPNO/Δt = -23 torr/min (given)
ΔPCl /Δt = ½ (ΔPNO/Δt ) = ½ (-23) = -11.5 torr/min
ΔPNOCl /Δt = - ΔPNO/Δt = -(-23) = +23 torr/min
PTOT = -23 – 11.5 + 23 = -11.5 = -12 torr/min
Pressure is decreasing at a rate of 12 torr/min
14.3 Concentration and Rate
In general, rates:
• Increase when reactant concentration is increased.
• Decrease when reactant concentration is decreased.
• We often examine the effect of concentration on reaction rate
by measuring the way in which reaction rate at the beginning
of a reaction depends on different starting conditions.
• Consider the reaction:
NH4+(aq) + NO2– (aq)  N2(g) + 2H2O(l)
• We measure initial reaction rates.
• The initial rate is the instantaneous rate at time t = 0.
• We find this at various initial concentrations of each
reactant.
• As [NH4+] doubles with [NO2–] constant the rate
doubles.
– We conclude the rate is proportional to [NH4+].
• As [NO2–] doubles with [NH4+] constant the rate
doubles
– We conclude that the rate is proportional to [NO2–].
Rate Law
• The overall concentration dependence of reaction
rate is given in a rate law or rate expression.
• For a general reaction the rate law is:
Rate = k [A]m [B]n
The proportionality constant k is called the rate
constant.
Once we have determined the rate law and the rate
constant, we can use them to calculate initial
reaction rates under any set of initial
concentrations.
Exponents in the Rate Law
The exponents m and n are called reaction orders
The rate is “m” order in A
The rate is “n” order in B
We commonly encounter reaction orders of 0, 1 or 2.
But, fractional or negative values are possible.
The overall reaction order is the sum of the reaction orders.
The overall order of a reaction is m + n + ….
Note that reaction orders must be determined experimentally.
They DO NOT necessarily correspond to the stoichiometric
coefficients in the balanced chemical equation! (but they can and
sometimes do)
Using Initial Rates to Determine Rate Laws
• To determine the rate law, we observe the effect of changing
initial concentrations.
• A reaction is nth order if multiplying the concentration by x
causes a xn increase in rate.
x increase in concentration = xn increase in rate
(n is the order)
For our example – the concentration was multiplied by 2 and
the rate was also multiplied by 2 (21) so the rate is 1st order with
respect to both NH4+(aq) and NO2– (aq)
2 (increase in concentration) = 21 (increase in rate)
(1st order)
• Most Common Scenarios:
Another way to look at order of reaction vs. change in concentration:
Zero order rxn: change in concentration = no change in rate
1st order rxn: change in concentration = change in rate
2nd order rxn: change in concentration = change in rate is the
square of change in concentration
change in conc. of reactant
doubled
change in rate
no change - rate is multiplied by 1 or 20
doubled
doubled - rate is multiplied by 2 or 21
doubled
quadrupled - rate is multiplied by 4 or 22
tripled
no change - rate is multiplied by 1 or 30
tripled
tripled - rate is multiplied by 3 or 31
tripled
rate is multiplied by 9 or 32
order
0
1
2
0
1
2
• For the reaction in our example:
NH4+(aq) + NO2– (aq)  N2(g) + 2H2O(l)
The rate law is:
Rate = k [NH4+] [ NO2–]
• The reaction is said to be first order in [NH4+],
first order in [NO2–], and second order overall.
Units of Rate Constants
• Units of the rate constant depend on the overall
reaction order.
• Second order overall:
• Rate = k [A]2
Units of rate = (units of rate constant)(units of conc.)2
Ms
(Units of rate)
-1 -1
=
M
s
Units of rate constant =
=
2
2
M
(Units of conc.)
• First order overall:
• Rate = k [A]
Units of rate = (units of rate constant)(units of conc.)1
Ms
(Units of rate)
=
Units of rate constant =
=
1
1
M
(Units of conc.)
s-1
• Note that the rate, not the rate constant,
depends on concentration.
• The rate constant IS affected by
temperature and by the presence of a
catalyst (more on this later).
Examples of rate laws
What are the individual and overall reaction orders for the reactions
described in the following equations:
(1)
2 N2O5(g)  4 NO2(g) + O2(g)
Rate = k [N2O5]
(2)
CHCl3(g) + Cl2(g)  CCl4(g) + HCl (g)
Rate = k [CHCl3][Cl2]1/2
• Reaction (1) is first order in N2O5 and first order overall.
• Reaction (2) is first order in CHCl3 and ½ order in Cl2 and 3/2 order
overall (add the exponents).
A reaction A + B  C, obeys the following rate law:
rate = k [B]2. (a) if [A] is doubled, how will the rate
change? Will the rate constant change? (b) What are the
reaction orders for A and B? What is the overall reaction
order? (c) What are the units of the rate constant?
(a) if [A] is doubled, there will be no change in the rate or
the rate constant
(b) the reaction is zero order in A, second order in B and
second order overall
(c) rate = k [B]2
units of rate = (units of k)(conc. units) 2
rate units
units of k =
(conc. units) 2
M
1
s
units of k = 2 =
= M 1s-1
M
Ms
The decomposition of N2O5 in carbon tetrachloride
proceeds as follows: 2 N2O5  2 NO2 + O2. The rate
law is first order in N2O5. At 64oC, the rate constant is
4.82  10-3 s-1. (a) Write the rate law for the reaction.
(b) What is the rate of reaction when [N2O5] = 0.0240 M?
(c) What happens to the rate when the concentration of
N2O5 is doubled to 0.0480 M?
(a)
rate = k [N2O5] = (4.82  10-3 s-1) [N2O5]
(b) rate = (4.82  10-3 s-1) (0.0240 M) = 1.16  10-4 M/s
(c)
one way:
rate = 4.82  10-3 s-1 (0.0480 M) = 2.31  10-4 M/s
quicker way:
We know that the rxn is 1st order in N2O5, so doubling
the conc. will double the rate
(a)
(b)
(c)
Consider the following reaction:
CH3Br + OH-1  CH3OH + Br-1 . The rate law for this
reaction is first order in CH3Br and first order in OH-1.
When [CH3Br] is 0.0050 M and [OH-1] is 0.050 M, the
reaction rate at 298 K is 0.0432 M/s.
What is the value of the rate constant?
What are the units of the rate constant?
What would happen to the rate if the concentration of
OH-1 were tripled?
(a,b) rate = k [CH3Br]
at 298 K, k =
(c)
[OH-1]
rate
k =
[CH3 Br][OH-1 ]
0.0432 M/s
= 1.7 x 102 M-1s-1
(0.0050 M)(0.050 M)
Since the rate law is first order in [OH-1], if [OH-1] is
tripled, the rate triples.
The iodide ion reacts with hypochlorite ion (the active ingredient in bleach) in
the following way: OCl-1 + I-1  OI-1 + Cl-1. This rapid reaction gives the
following rate data:
Experiment
[OCl-1] (M)
[I-1] (M)
Rate (M/s)
1
0.0015
0.0015
1.36  10-4
2
0.0030
0.0015
2.72  10-4
3
0.0015
0.0030
2.72  10-4
(a) What is the rate law for the reaction?
(b) What is the value of the rate constant?
(c) Calculate the rate when [OCl-1] = 0.0020 M and [I-1] = 0.00050 M
(a)
from exp. 1&2, doubling [OCl-1] while keeping [I-1] constant doubles the
rate – rxn is 1st order in OCl-1
from exp. 1&3, doubling [I-1] while keeping [OCl-1] constant doubles the
rate – rxn is 1st order in I-1
rate = k [OCl-1][I-1]
(b)
solving the rate law for k and using the data of experiment 1:
rate
1.36 x 10-4 M/s
-1 -1
k=
=
=
60
M
s
-1
-1
[OCl ][I ] (0.0015 M)(0.0015 M)
(c)
using the rate law (from a) and the value of k (from b):
rate = (60 M-1s-1 ) (0.0020 M) (0.00050 M) = 6.0 x 10-5 M/s
(a)
(b)
(c)
(d)
(a)
(b)
Consider the gas-phase reaction between nitric oxide and bromine at 273oC:
2 NO + Br2  2 NOBr. The following data for the initial rate of NOBr
were obtained:
Experiment
[NO] (M)
[Br2] (M)
Initial Rate (M/s)
1
0.10
0.20
24
2
0.25
0.20
150
3
0.10
0.50
60
4
0.35
0.50
735
Determine the rate law.
Calculate the value of the rate constant for the appearance of NOBr.
How is the rate of appearance of NOBr related to the rate of disappearance
of Br2?
What is the rate of disappearance of Br2 when [NO] = 0.075 M and [Br2] =
0.25?
From exp. 1&2, multiplying [NO] by 2.5 multiplies the rate by 2.52 – rxn
is 2nd order in NO
From exp. 1&3, multiplying [Br2] by 2.5 multiplies the rate by 2.5 – rxn
is 1st order in Br2
The rate law for the appearance of NOBr is : rate = k [NO]2[Br2]
from experiment 1, k =
24 M/s
4
-2 -1
=
1.2
x
10
M
s
2
(0.10 M) (0.20 M)
(a)
(b)
(c)
(d)
Consider the gas-phase reaction between nitric oxide and bromine at 273oC:
2 NO + Br2  2 NOBr. The following data for the initial rate of NOBr
were obtained:
Experiment
[NO] (M)
[Br2] (M)
Initial Rate (M/s)
1
0.10
0.20
24
2
0.25
0.20
150
3
0.10
0.50
60
4
0.35
0.50
735
Determine the rate law.
Calculate the value of the rate constant for the appearance of NOBr.
How is the rate of appearance of NOBr related to the rate of disappearance
of Br2?
What is the rate of disappearance of Br2 when [NO] = 0.075 M and [Br2] =
0.25?
(c)
[Br2 ]
[NOBr]
=2 t
t
(d)
The rate law for the appearance of NOBr is : rate = k [NO]2[Br2]
The rate of disappearance of Br2 is ½ the rate
of appearance of NOBr.
[Br2 ]
1
= k [NO]2[Br2 ]
t
2
1
(1.2 x 104 M -2s-1 ) (0.075 M) 2 (0.25 M) = 8.4 M/s
2
Consider the reaction of peroxydisulfate ion, S2O8-2 , with iodide ion, I-1 , in aqueous
solution:
S2O8-2 + 3 I-1  2 SO4-2 + I3-1
At a particular temperature, the rate of disappearance of S2O8-2 varies with reactant
concentrations in the following manner:
Experiment
[S2O8-2] (M)
[I-1] (M)
Initial Rate (M/s)
1
0.018
0.036
2.6  10-6
2
0.027
0.036
3.9  10-6
3
0.036
0.054
7.8  10-6
4
0.050
0.072
1.4  10-5
(a) Determine the rate law for the reaction.
(a) from experiments 1 and 2, increasing [S2O8-2] by 1.5 increases the rate by 1.5
so [S2O8-2] is 1st order
Experiment
[S2O8-2] (M)
[I-1] (M)
Initial Rate (M/s)
1
0.018
0.036
2.6  10-6
2
0.027
0.036
3.9  10-6
3
0.036
0.054
7.8  10-6
4
0.050
0.072
1.4  10-5
ALTERNATE SOLUTION
In exp 1 and 2, keeping [I-1] constant and increasing [S2O8-2] by 1.5 increases the
rate by 1.5 meaning that the reaction is first order in [S2O8-2]
So we know that the rate law is: rate = k [S2O8-2] [I-1]x and we need to find x
Looking at exp 1 and 3, the rate inc by a factor of 3 or Rateexp 3 / Rateexp1 = 3
Rate 3
k [S2O8-2] [I-1]x
k [.036] [.054]x
2 [.054]x
-------- = -------------------- = -------------------- = ------------- = 3
Rate 1
k [S2O8-2] [I-1]x
k [.018] [.036]x
[.036]x
[.054]x
3
-------- = ----- (or 1.5)
[.036]x
2
(1.5)x = 1.5
x = 1, so the reaction is 1st order in [I-1]
rate = k [S2O8-2] [I-1]
Kinetics Part 2
14.4
• Goal: Convert the rate law into a
convenient equation that gives
concentration as a function of time.
First-Order Reactions
• Rate must depend on only one reactant and be raised to the 1st
power.
• For a first-order reaction, the rate doubles as the concentration of a
reactant doubles.
Rate = k[A]
  A
Rate  
t
  A
Rate  
 k  A
t
(slope= −k)
ln[A]t
After some calculus………
Time (s)
ln  A  kt  ln  A
t
y
= mx +
b
0
 A
ln  t
 A
 0
 kt
•A plot of ln[A]t versus t is a straight line with slope -k and intercept ln[A]0
Second-Order Reactions
• A second-order reaction is one whose rate depends on the
reactant concentration to the second power or on the
concentration of two reactants, each raised to the first power.
• For a 2nd order reaction with just one reactant:
  A
2


Rate  
 k  A
t
1
1
 kt 
 At
 A0
calculus
y
= mx + b
A plot of 1/[A]t versus t is a straight line with slope k and intercept 1/[A]0.
(slope= k)
1/[A]t
Time (s)
Half-life
• Half-life, t½ , is the time required for the
concentration of a reactant to decrease to half
its original value.
• That is, half life, t½, is the time taken for [A]0 to
reach ½ [A]0.
• Mathematically, the half life of a first-order
reaction is:
 At
ln
  kt
 A0
So, for t  t1 2 and  At  1 2  A 0
ln
1 2  A 0
 A
 0
 kt1 2
ln1 2 0.693
t1 2  

k
k
Radioactive decay
is 1st order.
Note that the half-life of a first-order reaction is independent of
the initial concentration of the reactant.
• We can show that the half-life of a second
order reaction is:
1
t1 2 
k  A 0
• Note that the half-life of a second-order
reaction is dependent on the initial
concentration of reactant.
Sample Exercise #1
• The first order rate constant for the
decomposition of a certain insecticide in water at
12oC is 1.45 yr-1. A quantity of this insecticide is
washed into a lake on June 1, leading to a
concentration of 5.0 x 10-7 g/cm3 of water.
Assume that the effective temperature of the lake
is 12oC. (a) What is the concentration of the
insecticide on June 1 of the following year? (b)
How long for the conc. of the insecticide to drop
to 3.0 x 10-7 g/cm3?
(a)
ln [A]t = - k t + ln [A]0
ln [A] t = - (1.45 yr-1)(1.00 yr) + ln(5.0 x 10-7)
eln [A] t = e-15.96
[A] t = 1.2 x 10-7 g/cm3
(note: conc. units for [A] and [A]0 must be the same)
(b)
ln [A]t = - k t + ln [A]0
ln (3.0  10-7) = - (1.45)(t) + ln (5.0  10-7)
t = 0.35 yr
• Using the figure to the
right, estimate the
half-life of the reaction
of C4H9Cl with water.
• The initial value of
[C3H9Cl] is 0.100 M.
• The half-life for this
reaction is the time for
[C3H9Cl] to be 0.050
M.
At the end of the second half-life, about 680 s, the
should decrease by another factor of 2
• This point occurs at concentration
to about 0.025 M.
approx. 340 s.
The graph indicates that this is true.
The reaction SO2Cl2  SO2 + Cl2 is first order in SO2Cl2. Using the
following kinetic data, determine the magnitude of the first order rate
constant:
Time (s)
Pressure SO2Cl2 (atm)
ln Pressure SO2Cl2
0
1.000
0
2500
0.947
- 0.0545
5000
0.895
- 0.111
7500
0.848
- 0.165
10000
0.803
- 0.219
Graph ln P vs. time (first order)
The graph is linear with slope of
about - 2.19  10-5 s-1
slope =
y
(-0.219 - 0)
=
= - 2.19 x 10 -5
x
(10000 - 0)
k = - (slope) = 2.19 x 10-5
Sample Exercise #4
The following data was obtained for the gas
phase decomposition of NO2(g) at 300oC:
2 NO2(g)  2 NO(g) + O2(g)
Time (s) [NO2] (M)
0.0
0.01000
50.0
0.00787
100.0
0.00649
200.0
0.00481
300.0
0.00380
What is the rate law for this reaction?
• To test whether the reaction is first or second
order, we can construct plots of ln [NO2] and
1 / [NO2] against time.
Time (s)
0.0
50.0
100.0
200.0
300.0
[NO2]
0.01000
0.00787
0.00649
0.00481
0.00380
ln [NO2]
1/[NO2]
-4.610
-4.845
-5.038
-5.337
-5.573
100
127
154
208
263
From the graphs above, only the plot of 1/[NO2] is a straight line. Thus, the
reaction is second order in NO2.
Rate = k [NO2]2
From the slope of the straight line graph, we can get:
k = 0.543 M-1 s-1
The decomposition of sulfuryl chloride (SO2Cl2) is a first-order process. The rate
constant for the decomposition at 660 K is 4.5  10-2 s-1. (a) if we begin with an
initial SO2Cl2 pressure of 375 torr, what is the pressure of this substance after 65 s?
(b) at what time will the pressure of SO2Cl2 decline to one-tenth its initial value?
(a)
ln Pt = - kt + ln Po
ln P65 = - 4.5  10-2 s-1 (65) + ln(375)
P65 = 20 torr
(b)
Pt = 0.10 P0 = (.10)(375) = 37.5 torr
ln Pt = - kt + ln Po
ln (37.5) = - (4.5 x 10-2 s-1 )t + ln (375)
t = 51 s
Consider the data presented
(a) By using appropriate
graphs, determine
whether the reaction
is first or second order.
(b) What is the value of the
rate constant for the
reaction?
(c) What is the half life for
the reaction?
1/[A]
(a) make both first and second order plots to see which is linear
the plot of 1/[A] vs. time is linear, so the reaction is second
order in [A]
(b) the slope of the line in the graph is the rate constant
slope =
y
(3.2 - 1.5)
=
=0.043 M  1min  1
x
(40 - 0)
(c) use the half-life formula for a 2nd order reaction:
t1/2 = 1 / k[A]o = 1 / [(0.043 M-1 min-1) (0.65 M)] = 36 min
14.5 Temperature and Rate
• Most reactions speed up as temperature
increases
• The rate law has no temperature term in it, so
the rate constant (k) must depend on
temperature.
– We see an approximate doubling of the rate with
each 10oC increase in temperature.
The Collision Model
• Rates of reactions are affected by concentration and
temperature.
• An explanation is provided by the collision model, based on
kinetic-molecular theory.
• In order for molecules to react they must collide.
• The greater the number of collisions the faster the rate.
- The more molecules present, the greater the probability
of collision faster rate
- The higher the temperature, the faster molecules move
and more often they collide  faster rate
However, not all collisions lead to products.
• In fact, only a small fraction of collisions lead to
products.
• In order for a reaction to occur the reactant molecules
must collide in the correct orientation AND with enough
energy to form products.
• The Orientation Factor
• The orientation of a molecule during collision can have a
profound effect on whether or not a reaction occurs.
• Consider the reaction between Cl and NOCl:
• If Cl collides with Cl of NOCl, the products are Cl2 and NO.
• If the Cl collides with the O of NOCl, no products are formed.
Cl
Cl-N=O
Cl
O=N-Cl
Activation Energy
• Arrhenius: molecules must posses a minimum amount of
energy to react. In order to form products, bonds must be
broken in the reactants.
• Bond breakage requires energy.
(ΔHrxn = bonds broken – bonds formed)
• Molecules moving too slowly, with too little kinetic
energy, don’t react when they collide.
• Activation energy, Ea is the minimum energy required to
initiate a chemical reaction (varies with the reaction).
• Energy is required to stretch
and break the bond between
the CH3 group and the NC
group
• The species at the top of the
barrier is called the activated
complex or transition state.
• Energy is released when the
C-C bond is formed.
• The change in energy for the
reaction is the difference in
energy between CH3NC and
CH3CN.
• The activation energy is the
difference in energy between
reactants, (CH3NC) and the
transition state.
The rate depends on the
magnitude of the Ea.
In general, the lower the Ea,
the faster the rate.
• Notice that if a forward reaction is exothermic (CH3NC 
CH3CN), then the reverse reaction is endothermic
(CH3CN  CH3NC).
How does this relate to temperature?
• At any particular temperature, the molecules present have an
average kinetic energy.
– Some molecules have less energy than the average while others have
more than the average value.
• Molecules that have an energy equal to or greater than Ea
have sufficient energy to react.
• The fraction of molecules with an energy equal to or greater
than Ea is given by:
f=e
-Ea
RT
(R is 8.314 J/mol-k , Temp in K, Ea in J)
At a higher temperature, more molecules have the
minimum Ea needed to react and the rate is faster.
The Arrhenius Equation
• Arrhenius discovered that most reaction-rate data obeyed an
equation based on three factors
1. The number of collisions per unit time.
2. The fraction of collisions that occur with the correct
orientation.
3. The fraction of the colliding molecules that have an
energy equal to or greater than Ea.
Arrhenius equation:
k=Ae
-Ea
RT
• Where k is the rate constant, Ea is the activation energy(J), R
is the ideal-gas constant (8.314 J/K•mol) and T is the
temperature (K).
• A is the frequency factor.
– It is related to the frequency of collisions and the probability that a
collision will have a favorable orientation.
More Useful Versions of the Arrhenius Equation
lnk = -
Ea
+ lnA
RT
y = mx + b
• A graph of ln k vs 1/T will be a line with slope of
–Ea/R and a y-intercept of ln A.
ln A
slope = -Ea / R
ln k
1/T
k1
Ea  1 1 
ln
=
 - 
k2
R  T2 T1 
• Useful when we have 2 different k values at 2
different temperatures
SAMPLE 1:
Consider a series of reactions having the following energy profiles:
Assuming that all three reactions have nearly the same frequency factors, rank
the reactions from slowest to fastest.
Solution
The lower the activation energy, the faster the reaction. The value
of ∆E does not affect the rate.
Hence the order is (2) < (3) < (1).
Calculate the fraction of atoms in a sample of argon gas at
400 K that have an activation energy of 10.0 kJ or greater.
f = e-Ea /RT
Ea = 10.0 kJ/mol = 10000 J/mol
10000 J/mol
- Ea /RT = = - 3.01
(8.314 J/mol K)(400 K)
f = e-3.01 = 0.049
T = 400 K
The gas phase reaction Cl(g) + HBr(g)  HCl(g) + Br(g) has an
overall enthalphy change of - 66 kJ. The activation energy for the
reaction is 7 kJ.
(a) Sketch the energy profile for the reaction and label Ea and ΔE.
(b) What is the activation energy for the reverse reaction?
(b) the reverse reaction has an Ea value of 73 kJ
A certain first order reaction has a rate constant of 2.75  10-2 s-1 at
20oC. What is the value of k at 60oC if Ea = 75.5 kJ/mol
T1 = 293 K
T2 = 333 K
k1
Ea  1 1 
ln
=
 - 
k2
R  T2 T1 
 2.75 x102  75.5 x 103 J/mol  1
1 
ln 
=


k
8.314
J/mol
333
293



2

 2.75 x102 
ln 
 = - 3.7
 k2

2.75 x102
= 0.024
k2
2.75 x102 s-1
k2 
= 1.1 s-1
0.024
The rate of the reaction
CH3COOC2H5 + OH-1  CH3COO-1 + C2H5OH
was measured at several temperatures and the following data were
collected. Using the data, construct a graph of ln k versus 1/T and
determine the value of Ea
slope =
y
(-2.955 - -1.103)
=
= - 5.6 x 103
x
(3.47 - 3.14)
the slope = -Ea / R
Ea = - (-5.6  103)(8.314 J/mol) = 47000 J/mol or 47 kJ/mol
The activation energy of a certain reaction is 65.7 kJ/mol. How many
times faster will the reaction occur at 50oC rather then 0oC assuming
equal initial concentrations?
Since we are considering one reaction and temp only affects k which directly
affects rate…the ratio of k values equals the ratio of rates at the two
temperatures.
T1 = 323 K ;
T2 = 273 K
 k 1  65.7 x 103 J/mol  1
1 
ln   =


k
8.314
J/mol
27
3
32
3


 2
 k1 
ln   = 4.48
 k2 
k1
= 88
k2
The reaction will occur about 88 times faster at 50oC
14.6 Reaction Mechanisms
• A chemical equation provides information about substances
present before and after a reaction.
• The reaction mechanism is the process of how the
reactants become products.
• Mechanisms provide a picture of which bonds are broken and
formed during the course of a reaction.
• Most reactions occur as a series of small steps or changes.
Elementary steps are processes that occur in a single step.
• The number of molecules present in an elementary step is
the molecularity of that elementary step.
•
Unimolecular: one molecule in the elementary step.
•
Bimolecular: two molecules in the elementary step.
•
Termolecular: three molecules in the elementary step.
Multistep Mechanisms
• consist of a sequence of elementary steps.
– The elementary steps must add together to give the
balanced chemical equation.
• Intermediates - these are species that appear in an
elementary step but are neither a reactant nor
product.
– They are formed in one elementary step and consumed in
another.
– They are NOT found in the balanced equation for
the overall reaction.
Rate Laws for Elementary Steps
• The rate laws of the elementary steps determine the
overall rate law of the reaction.
• For elementary processes, we DO NOT need to find
the reaction order by experimentation – it matches
the stoichiometry of the equation
• The rate law of an elementary step is determined by
its molecularity.
•
Unimolecular processes are first order.
•
Bimolecular processes are second order.
•
Termolecular processes are third order.
Rate Laws for Multistep Mechanisms
• When a reaction occurs by mechanisms with
more than one step, the slowest step limits
the overall reaction rate.
– This is called the rate-determining step
of the reaction.
– This step governs the overall rate law for
the overall reaction.
Consider the reaction:
NO2(g) + CO(g) → NO(g) + CO2(g)
Here is a proposed a mechanism for the reaction:
Step 1: NO2(g) + NO2(g)  NO3(g) + NO(g) slow step
Step 2: NO3(g) + CO(g)  NO2(g) + CO2(g) fast step
(Note that NO3 is an intermediate.)
The overall reaction rate will depend on the slowest step
Rate = k[NO2]2
If we do an experiment and find that the experimentally
derived rate law is: Rate = k[NO2]2 Then our theoretical rate
law is in agreement with the experimental rate law.
This supports (but does not prove) our mechanism.
The following mechanism has been proposed for the reaction of
methane gas with chlorine gas. All species are in the gas phase.
Step 1
Cl2
2 Cl
fast equilibrium
Step 2
CH4 + Cl  CH3 + HCl
slow
Step 3
CH3 + Cl2  CH3Cl + Cl
fast
Step 4
CH3Cl + Cl  CH2Cl2 + H
fast
Step 5
H + Cl  HCl
fast
(e) Identify the order of the reaction with respect to each of the
following according to the mechanism. In each case, justify your
answer.
(i)
CH4(g)
(ii) Cl2(g)
• The overall rate law depends on the rate law of the slow
step:
step 2: rate = k[CH4][Cl]
*The problem is we can’t use intermediates in overall rate
law…
Since step 1 is equilibrium the forward and reverse rates
are equal.
k[Cl2]=k[Cl]2
Rearranging:  =

[2 ]

= [2 ]
1
2
Plugging back into the rate law for step 2:
1
 =  4 [2 ] 2
1
 =  4 [2 ] 2
14.7 Catalysis
• A catalyst is a substance that changes the speed of a
chemical reaction without any permanent change to itself
– Works by providing an alternate mechanism for the reaction – one
that has a lower activation energy
• Lowering the activation energy allows for more molecules with Ea large enough to
react – more collisions that can make products – higher rxn rate (without changing
temperature)
– Very common in nature
– Much research to find new catalysts AND, much research to find
“inhibitors”(substances that slow down an undesired
reaction) Example: corrosion
• Homogeneous catalyst – catalyst is in the same phase as
reactants
• Heterogeneous catalyst – catalyst is in different phase than
reactants
• Enzymes – catalysts in living organisms
Effect of a catalyst on a reaction
A catalyst provides an alternate mechanism that has
a lower Ea than the original mechanism – this allows
the reaction to occur at a faster rate
What is the molecularity of each of the following elementary processes?
Write the rate law for each.
(a) Cl2  2 Cl
(b) OCl-1 + H2O  HOCl + OH-1
(c) NO + Cl2  NOCl2
(a) unimolecular rate = k [Cl2]
(b) bimolecular rate = k [OCl-1] [H2O]
(c) bimolecular
rate = k [NO] [Cl2]
(a)
(b)
(c)
(d)
Based on the following reaction profile,
how many intermediates are formed
in the reaction A  D?
How many transition states are there?
Which step is the fastest?
Is the reaction A  D
exothermic or endothermic?
This is a three-step mechanism, A  B, B  C, C  D
(a) 2 intermediates, B and C (A is reactant, D is product)
(b) 3 energy maxima, so there are 3 transition states
(c)Step C  D has the lowest activation energy, so it is the fastest
(d)The energy of D is greater than the energy of A, so the
overall reaction is endothermic
The following mechanism has been proposed for the gas-phase reaction of H2
with ICl:
H2 + ICl  HI + HCl
HI + ICl  I2 + HCl
(a) Write the balanced equation for the overall reaction
(b) Identify any intermediates
(c) Write rate laws for each elementary reaction in the mechanism
(d) If the first step is slow and the second one is fast, what rate law do you
expect to be observed for the overall reaction?
(e) If the second step is the rds, what rate law would you expect for the reaction
(a)
H2 + ICl  HI + HCl
HI + ICl  I2 + HCl
H2 + 2 ICl  I2 + 2 HCl
(b) HI
(c) first step: rate = k1 [H2] [ICl]
second step: rate = k2 [HI] [ICl]
(d) rate = k [H2] [ICl]
(e) rate = k [H2] [ICl]2
The reaction 2 NO + Cl2  2 NOCl obeys the rate law:
rate = k [NO]2 [Cl2]. The following mechanism has been proposed for
this reaction:
NO + Cl2  NOCl2
NOCl2 + NO  2 NOCl
(a) What would the rate law be if the first step were rate determining?
(b) Based on the observed rate law, what can we conclude about the
relative rate of the the two steps?
(a) rate = k [NO] [Cl2]
(b) Assuming the proposed mechanism is correct the 2nd step is
the r-d-s (slower step).
(The rate law based on this mechanism with the 2nd step as the r-ds matches the experimentally determined rate law.)
The oxidation of SO2 to SO3 is catalyzed by NO2. The reaction proceeds as
follows:
NO2(g) + SO2(g)  NO(g) + SO3(g)
2 NO(g) + O2(g)  2 NO2(g)
(a) Show that the two reactions can be summed to give the overall oxidation of
SO2 by O2 to give SO3. Hint: the top reaction must be multiplied by a factor
so the NO and NO2 cancel out. (b) Why do we consider NO2 a catalyst and
not an intermediate in this reaction? (c) Is this an example of homogeneous
catalysis or heterogeneous catalysis?
(a)
2 [NO2 + SO2  NO + SO3] = 2 NO2 + 2 SO2  2 NO + 2 SO3
2 NO + O2  2 NO2
__
2 SO2 + O2  2 SO3
(b) NO2 is a catalyst because it is present at the beginning of the reaction –
it is consumed and then reproduced in the reaction sequence.
(NO is an intermediate because it is produced and then consumed.)
(c) homogeneous catalysis

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