Download Order and Half-life Equations

Notes – Unit 6 Kinetics
Vocabulary
Rate law
expresses how the rate depends on concentration of reaction
Chemical Kinetics
area of chemistry that concerns reaction rates
Reaction Rate
the change in concentration of a reactant per unit time
Rate Constant
the proportionality constant in the relationship between rxn rate and
reactant concentration
Order
the dependency of a rate on a particular reactant, appears in the rds
Overall Reaction Order
sum of the individual orders “m” and “n”
Enzyme
biological protein catalyst
Catalyst
a substance that increases the rate of a reaction without being consumed;
by lowering the activation energy, Ea
Heterogeneous Catalyst
exists in a different phase, usually a solid
Homogeneous Catalyst
present in a different phase, usually as the reacting molecules
Instantaneous Rate
the value of the rate at a particular time (because rate is not constant but
typically decreases as the amount of product increases)
Half Life
the time required for a reactant to reach half of its initial concentration
Reaction Mechanism
yhe series of chemical reaction steps that occur in a chemical reaction
Intermediate
a substance that is not a reactant or a product but is made and used during
the reaction mechanism
Rate determining step- (rds)
the slow step of a reaction mechanism, determines overall rate; A reaction
is only as fast as its’ slowest step
Collision model
a model based on the idea molecules must collide with proper angle and
energy to react; used to observe characteristics of reaction rates
Activation energy
the energy that must be added to start a chemical reaction
Notes – Unit 6 Kinetics
Reaction Order and Rate Laws

Order – the dependency of a rate on a particular reactant (never include products)
- there are three order: 0, 1st, and 2nd
Order
0
1
2

[react]
2x
2x
2x
Subsequent rate
2 so no change in the rate (goes away in rate law).
21 so 2x change in the rate
22 so 4x change in the rate
0
Rate Laws
- Math equations that govern the relationship between the rate and the reactant
concentrations. Rate is the change in concentration over time so it can be a derivative
d[x]/dt. However,…
Generally if A  B  C  D then, the format of the rate law is:
Rate  k [A]m  B 
n
where:
m and n are the orders for A and B respectively
mol  L1  s 1
unit for rate are mol
or
L s
k is the proportaionality constant known as the rate constant with units of
m  n 1
L
m  n 1
mol 
s
Note that the rate constant is temperature dependent due to the Arhenius
equation (listed later) and thus the rate constant k expresses why the speed of
a reaction changes with temperature.
“[ ]’s” mean concentration, and the unit is mol
L
Methods for determining the Rate Law
1. From the mechanism


Mechanism is a sequence of chemical steps for how the rxn happens
The order (and thus the rate law) is determined by how many times the reactant appears in the
Rate Determining Step (rds) also known as the slow step
fast step: A  B (no reverse reaction)
slow step: C  D (forward and reverse)
Example: Determine the rate law for each mechanism.
Step 1:
Step 2:
overall:
A  B  F  C (slow)
A  B  F  C (fast)
(fast)
FBD
FBD
A  2B  C  D
A  2B  C  D
note that “F” is an intermediate
rds is step 1 where A-once, B once
so, the order of A=1
Order of A = 1
Order of B = 1
Rate  k[ A][B]
(slow)
rds is step 2 where F-once, B-once
F is intmd’t from A + B
so, A-once, B-twice (1+1)
Order of A = 1
Order of B = 2
Rate  k[ A][B] 2
Notes – Unit 6 Kinetics
2. From ratio of initial rate data


You will be given a table of Expt’s that contain data of initial rates and conc’s
The order (and thus the rate law) is determined by analyzing (through 2 expt’s) how the
concentration of a reactant is changed and the subsequent effect on the rate
Example: Determine the rate law for the reaction:
Experiment
2CO2  NH 3  products
[CO2 ]0 mol
[NH 3 ]0 mol
L
1
2
3
L
0.10
0.10
0.20
0.05
0.10
0.10
mol
l
Initial Rate
s
1.2 x 10-3
2.4 x 10-3
9.6 x 10-3
Determining the Order:
CO 2 (compare experiments 2&3)
[CO2 ]  2 x
rate  4 x  2 2  2 nd order
NH 3 (compare 1&2)
[ NH 3 ]  2 x
rate  2 x 1st order
2
1
So the rate law = k[CO2 ] [ NH 3 ]
* The long way (how you should do it on the AP test)
m
n
rate3 k[CO2 ]3 [ NH 3 ]3

rate2 k[CO2 ]2 m [ NH 3]2 n
CO2:
(solve for m)
9.6 x10 3 (0.20) m (0.10) n

2.4 x10 3 (0.10) m (0.10) n
4  2m
NH3:
2m
rate2 k[CO2 ]2 m [ NH 3 ]2 n

rate1 k[CO2 ]1m [ NH 3]1n
(solve for n)
2.4 x103 (0.10) m (0.10) n

1.2 x103 (0.10) m (0.05) n
2  2m
1 m
3. From time and concentration data.



You will be given a table of data showing changes in concentration with time.
The order (and thus the rate law) is determined by linear regression analysis of three graphs.
The graph with the best straight line (r2 closest to 1) determines the order. Note that the
absolute value of the slope is the value of the rate constant.
The graphs (y vs x):
- Zeroeth Order: [A] vs. t
has an equation of [A] = -kt + [A]o
- First Order: ln[A] vs. t
has an equation of ln[A] = -kt + ln[A]o
- Second Order: [A]-1 vs. t
has an equation of [A] -1 = -kt + [A]o-1
Notes – Unit 6 Kinetics
Methods for Determining Half-life
1.
First order
[ A]o
 kT
[ A]
[ A]o
b. [ A] 
when t1 / 2  t
2
a.
2.
c. Half-life independent of initial concentration.
d. A plot of ln [A] vs. t is always a straight line.
Second Order
a. Half-life is dependent on [A] initial concentration
b.
3.
ln
Each successive half-life [A]o is halved so that t1 / 2 
1
k[ A]0
c. A plot of 1/[A] vs. t will produce a straight line where slope=k
Zero Order
a. Zero order has constant rate unlike 1st and 2nd it doesn’t change with concentration
b. A plot of [A] vs. t gives a straight line where the slope= -k
Order and Half-life Equations
Order
Zero
Rate Law:
Integrated Rate Law
Rate  k
[ A]  kt  [ A]o
t
First
Rate  k[ A]
ln[ A]  kt  ln[ A]o
t
Plot needed to give a straight
line
[A] versus
Relationship of rate constant
to the slope of straight line
Half-life
Slope  k
Slope  k
t 12 [ A]o
2k
t 12 
ln[A] versus
0.693
k
Second
Rate  k[ A] 2
1
1
 kt 
[ A]
Ao
1
versus t
[ A]
Slope  k
t 12 
1
k[ A]o
More on Reaction Mechanism
1.
2.
Terms
a.
b.
Elementary Step- a reaction whose rate law can be written from its’ molecularity
Molecularity- the number of species that must collide in the Rate determining step to
produce the reaction indicated.
c.
Unimolecular Step- reaction involving only one molecule (always 1st order)
d. Bimolecular Step- reaction involving collision of two species (always 2 nd order)
e.
Termolecular Step- reaction involving collision of three species. (Rare)
Reaction Mechanism (scientific definition): Series of elementary steps that must satisfy two
requirements.
a. Sum of elementary steps yields the overall equation for the reaction
b. Mechanism must agree with experimentally determined rate law.
Notes – Unit 6 Kinetics
3.
Overview:
Elementary Step
AProduct
2AProduct
A+BProduct
2A+BProduct
A+B+CProduct
Molecularity
Unimolecular
Bimolecular
Bimolecular
Termolecular
Termolecular
Rate Law
Rate=k [A]
Rate=k [A]2
Rate=k [A] [B]
Rate=k [A]2[B]
Rate=k [A] [B] [C]
How does a Chemist deduce a mechanism? First by determining the rate law then using rate law to
make reaction mechanism that follows the 2 rules. Through trial and error to eliminate false
mechanisms. (Note: A reaction Mechanism can never be proved absolutely)
Activation Energy: Propose by Svante Arrhenius during the 1880s.
a.
b.
c.
d.
Through collision models energy comes when kinetic energy in the molecules prior
to a collision converting into potential energy to break and make bonds.
Transition State- high point
The higher activation energy the slower the reaction is at a given temperature.
Higher temperature means more collisions thus more energy
 Ea
e.
f.

a.
b.
c.
RT
Number of collision with Ea  ( total # of collisions )e
To collide successfully
i. Collision must involve enough energy to produce the reaction collision
energy must be  activation energy
ii. Relative Orientation of the reactant must allow formation of any new bonds
necessary to produce products.
Arrhenius Equation
k  Ae
 Ea
RT
The fact that most rate constants obey the Arrhenius equation to a good approximation proves that
the collision model for chemical reaction is physically reasonable.
The Arrhenius Equation is a linear equation (y=mx+b)
ln( k ) 
Catalysts
-
 Ea 1
( )  ln( A)
RT T
Catalysts increase the rates of reactions, does not affect the energy difference, E,
between products and reactants
Catalysts speed up the reaction without being consumed, lowers activation energy