Kinetics
Chapter 15
1
The study of rxn rates
• Rxn rate =
•
concentration/time
Example:
2N2O5  4NO2 + O2
• As reactant concentration
decreases  products’
concentrations increase
-[N 2 O5 ]
rate of rxn =
t
[NO 2 ]
rate of rxn =
t
[O 2 ]
rate of rxn =
t
2
Let’s do some calculations
• From 300s to 400s,
•
•
•
-[N2O5]/t =
-.0019M/100s
What are [NO2]/t &
[O2]/t during the same
time interval?
Let’s work this out…
Is rxn rate constant?
– Let’s look at [NO2]/t
at different time
intervals
3
More…
• The previous rates were
•
•
•
average rates
Different from
instantaneous rates
(@ single pt)
Ex:[N2O5]=-0.0019M
(@ 200s)
Car speed is analogous
4
Continued
• We can also calculate
changes
mathematically based
on stoichiometric
relationships
• Let’s see
• http://wps.prenhall.co
m/wps/media/objects
/167/172009/Decomp
ositionofN2O5.html
5
Writing rate expression
•
2N2O5  4NO2 + O2
To equate rates of
disappearance or
[N 2 O5 ]
[NO 2 ]
[O 2 ]
appearance
1
1

=+
=+
2
4
– Divide the
stoichiometric
coefficient in the
balanced equation
t
t
t
6
Let’s work on this
• a) Give four related rate expressions for
the rate of the following reaction:
2H2CO(g) + O2(g)  2CO(g) + 2H2O(g)
• b) Give three related rate expressions for
the rate of the following reaction:
N2(g) + 3H2(g)  2NH3(g)
7
Let’s also work on this:
•
•
•
Consider the reaction:
A + 2B  C
Give the three related
rate expressions for the
rate of the reaction.
Using the following
data, determine the
average rate of the
reaction, and the rate
between 30 and 40
seconds.
• Time (s): 0.0, 10.0, 20.0,
•
30.0, & 40.0
[A] (M): 1.000, 0.833,
0.714, 0.625, & 0.555,
respectively
8
Reaction conditions and rates
• Higher temp  faster rxn rate
– Raise in temp by 10°C  double reaction rate
• Higher concentration  faster rxn rate
• Catalysts speed up rxn rates
– They don’t react
– They lower the activation energy
• Enzymes (proteins) in organisms
• Metals, salts, etc. in chemical rxns
• Question: What does acid rain do to enzymes?
9
Concentration:
2N2O5  4NO2 + O2
• If you double [N2O5], you double the rxn
rate
– Rate of rxn  [N2O5]
• Other rxns have different relationships
w/concentration-rxn rates
10
Rate equations
• Describe relationship between reactant concentration
•
•
•
•
and rxn rate
For 2N2O5  4NO2 + O2
– Rate of rxn = k[N2O5]
k = rate constant
– Is the rate of rxn constant too?
Generic expression for rate eq: aA + bB  xX
So rate = k[A]m[B]n
– Where m & n need not equal stoichiometric ratios!
– They can be zero, fractions, even negative #’s
– Empirically verified values
11
Order of rxn:
Exponent of its concentration term
for each item
• 2NO(g) + Cl2(g)  2NOCl(g)
• Rxn rate for NO is second order
• Rxn rate for Cl2 is first order
•
– Thus, rxn rate = k[NO]2[Cl2]
Total order of rxn = summation of exponents of
all concentration items
– Therefore, overall rxn rate is third order
• Remember, m & n need not equal stoichiometric
ratios!
12
Reaction mechanisms
• Step by step chemical equations that, when
•
summed, give net rxn
For ex:
Br2(g) + 2NO(g)  2BrNO(g)
1) Br2(g) + NO(g)  Br2NO(g)
2) Br2NO(g) + NO(g)  2BrNO(g)
• Each step:
– Elementary step
13
Molecularity
• Elementary steps
•
•
classified according to #
of reactant molecules
– Molecularity
Unimolecular:
– A  products
– Rate expression: k[A]
Bimolecular:
– A + B  products
– Rate expression:
• Termolecular:
– A + B + C  products
• (Or 3A or 2A + B, etc.)
– Rate expression: k[A][B][C]
(or k[A]3 or k[A]2[B], etc.)
• In elementary steps,
stoichiometry defines rate
equation!
– K different for each step
k[A][B]
14
Rate-determining step
• The slowest elementary step is the ratelimiting step
– Overall rxn follows this rate-limiting step
• However, intermediate not included in final rate
law
15
An example
•
•
•
•
•
•
•
2NO(g) + O2(g) → 2NO2(g)
NO(g) + NO(g)  N2O2(g) (fast, eq.)
N2O2(g) + O2(g) → NO2(g) (slow)
Using slow, rate-determining, step
– Rate = k2[N2O2][O2]
Since fast-step at eq.
– k1[NO]2 = k-1[N2O2]
– Thus, K = k1/k-1 = [N2O2]/[NO]2
Intermediate ([N2O2]) = K[NO]2
– Since rate = k2[N2O2][O2]
• Rate = k2 K[NO]2[O2]
Let’s say k2 K = k’
 rate = k’[NO]2[O2]
16
Another problem
2H2(g) + 2NO(g)  2H2O(g) + N2(g)
• 2NO(g)  N2O2(g); fast
• H2(g) + N2O2(g)  H2O(g) + N2O(g); slow
• N2O(g) + H2(g)  N2(g) + H2O(g); fast
17
Solution
2NO(g)  N 2 O 2(g) ; fast
H 2(g) + N 2 O 2(g)  H 2 O(g) + N 2 O(g) ; slow
N 2 O(g) + H 2(g)  N 2(g) + H 2 O(g) ; fast
2
fast: ratefwd = k1[NO] , rate rev = k -1[N 2O 2 ]
Since ratefwd  rate rev , k1[NO]2  k -1[N 2O 2 ]
k1[NO]2
Thus, [N 2O 2 ]=
 K'[NO]2
k -1
slow: rate = k 2 [H 2 ][N 2 O 2 ]=k 2 [H 2 ]  K'[NO]2
 rate = K"[H 2 ][NO]2
18
Rate = change
• 2NH3  N2 + 3H2
• Rate = k[NH3]0 = k
– Zero order rxn rate
• Any tinkering of
•
concentration will not
change rxn rate of
species
k = mol/(L  time)
19
Rate = k[NO]2[Cl2]
• Rxn rate for Cl2 is
•
•
first order
 rate doubled when
[Cl2] doubled
k = time-1
20
Rate = k[NO]2[Cl2]
• Rxn rate for NO is
•
•
second order
Doubling [NO] 
quadruples (x4) rxn
rate
k = L/(mol  time)
21
Let’s do this:
• Nitrosyl bromide, NOBr, is formed from NO and
Br2.
•
2NO(g) + Br2(g)  2NOBr(g)
Experiment show that the reaction is first order
in Br2 and second order in NO.
–
Write the rate law for the reaction.
• If the concentration of Br2 is tripled, how will
•
the reaction rate change?
What happens to the reaction rate when the
concentration of NO is doubled?
22
Determining rate equations
• Done in the lab
• Generally, calculated after 1-2% of limiting
reactant consumed
– Lessens chance of side-rxns throwing things
off
23
Let’s work on this:
•
The rate for the oxidation of iron (II) by cerium (IV) is
measured at several different initial concentrations of
the two reactants:
Ce4+(aq) + Fe2+(aq)  Ce3+(aq) + Fe3+(aq)
•
•
•
[Ce4+] (M): 1.1 x 10-5, 1.1 x 10-5, 3.4 x 10-5
[Fe2+] (M): 1.8 x 10-5, 2.8 x 10-5, 2.8 x 10-5
Rate (M/s): 2.0 x 10-7, 3.1 x 10-7, 9.5 x 10-7
– Write the rate law for this reaction, determining the
orders of the reaction with respect to Ce (IV) and
Fe (II). What is the overall order of the reaction?
– Calculate the rate constant, k, and give its units.
– Predict the reaction rate for a solution in which
[Ce4+] is 2.6 x 10-5 M and [Fe2+] is 1.3 x 10-5 M. 24
Concentration-time relationships:
Integrated rate laws
• Zero-order rxns
• -([R]/t) = k[R]0
• Using integral
calculus, integrated
rate equation:
[R]i - [R]f = kt
k = concentration/time
25
Concentration-time relationships:
Integrated rate laws
• First-order rxns
• -[R]/t = k[R]
[R]f
ln
= -kt
[R]i
k = time-1
26
Integrated rate laws
• Second-order rxns
• -[R]/t = k[R]2
1
1
= kt
[R]f [R]i
k=1/(concentrationtime)
27
Using graphs to solve rxn order and
rate constants
• Zero-order:
• [R]i – [R]f = kt
• Rearrange this in the form:
y = mx + b
28
Using graphs to solve rxn order and
rate constants
• First-order:
[R]f
ln
= -kt
[R]i
• Rearrange this in the form:
y = mx + b
• Solve the following:
•
•
2H2O2(aq)  2H2O(l) + O2(g)
The reaction is first order in peroxide, and the rate constant,
k, for this reaction is 1.06 x 10-3/min.
If the initial peroxide concentration is 0.020 M, what is the
concentration after 135 min?
29
Solution
[R]f
ln
= -kt
[R]i
-3
[R]f
ln
= -1.06 10
 (135 mins)
min
[0.020 M]
ln[R]f  ln[0.020 M]  -1.06 10
[R]f  e
4.06
-3
min
 (135 mins)
 .017M
30
Using graphs to solve rxn order and
rate constants
• Second-order:
• Rearrange this in the form:
1
1
= kt
[R]f [R]i
y = mx + b
• The gas phase decomposition of HI into hydrogen and
•
•
iodine is second order in HI.
The rate constant for the reaction is 30/Mmin.
How long must one wait for the concentration of HI to
decrease from 0.010 M to 0.0050 M?
31
Solution
1
1
= kt
[R]f [R]i
1
1
 30 M
t
min
[0.0050M]
[0.010M]
t=3.3 min
32
Half-life
• Time required for
reactant
concentration to
decrease to ½ initial
value
– Longer half-life
means slower rxn
– Usually used for
1st-order rxns (like
radioactive decay)
[R]f = 1 [R]i
2
1  [R]f
2 [R]
i
[R]f
1
ln( )  ln(
)   kt 1
2
2
[R]
i
ln( 1 )  0.693  kt 1
2
2
0.693
t1 =
2
k
33
Half-life
• For 1st-order rxns, t1/2
is independent of
conc
34
Half-life: 2nd order rxns
• Half-life increases
with decreasing
concentration
1
t1 =
2
k[R]i
• Derive this!
35
Half-life: zero-order rxns
• Half-life decreases
with decreasing
concentration
t1
2
[R]i

2k
36
Problem
• Sucrose decomposes to fructose and
glucose in acid solution with the rate law
of:
Rate = k[sucrose]; k=0.208 hr-1 @ 25°C
• What amount of time is required for
87.5% of the initial concentration of
sucrose to decompose?
37
Solution
[12.5M]
ln
= -0.208  t
hr
[100.0M]
t=1.00hr
38
Collision Theory
1. Reacting molecules must collide with one
another.
2. Reacting molecules must collide with
sufficient energy to break bonds.
3. Molecules must collide in correct
orientation to form new species.
39
Temperature
• Higher temperatures
•
lead to fast reactions
Molecules have
varying temperatures
– At higher
temperatures, more
molecules have higher
energies
40
Activation energy and catalysts
• Activation energy:
– Energy required to form
products
• Methane + oxygen gas
– Low activation energy
• Xenon + oxygen gas
– Humongous activation
energy
• Catalysts dramatically
reduce activation energy
w/out being consumed
41
Arrhenius equation
• Includes tenets of collision
•
•
•
•
theory; i.e., collision
frequency, temp (energy),
and correct orientation
k = Ae-Ea/RT
A = frequency factor
(L/mols)
– Empirically derived
relationship between
reaction rate and temp
e-Ea/RT = fraction of
molecules w/min energy
needed for rxn (R = gas
constant = 8.314 J/molK)
Can put equation in
“y=mx+b” form
– How?
-E a 1
ln k =
 + ln A
R T
42
The Arrhenius equation
43
Problem
• Given:
y = -24371x + 28.204
• Solve for Ea
44
Solution
y = -24371x + 28.204
Ea
Ea
-24371K = 
J
R
8.314
mol  K
5 J
E a  2.0262 10
mol
45
If given two points…
-E a
k2
1 1
ln( ) =
( - )
k1
R
T2 T1
46