Chemical Kinetics - Fall River Public Schools

Report
AP Chem
 Area
of chemistry that deals
with rates or speeds at which a
reaction occurs
 The rate of these reactions are
affected by several factors
observable property
Rate 
time
Such as:
Concentration
Color
Bubbles
Temp
pH
Whatever is appropriate:
Hours
Minutes
Seconds

Candle Wax Example

Collisions cause reactions!
◦ Breaking of bonds directly linked to rate
◦ Must overcome repulsion of electron clouds
 Also correct orientation sometimes needed
◦ Example: Chalk dropping
 Concentration
of reactants
◦ With gases, pressure used instead
 Temperature
at which the
reaction occurs
 The presence of a catalyst
 Surface area of solid or liquid
reactants

Demo:
◦ Chalk + 1 M HCl / Chalk + 1 M Acetic Acid
 Prediction?
 Factor?
◦ Chalk + 1 M HCl / Chalk + 6 M HCl
 Prediction?
 Factor?
◦ Chalk + 1 M HCl (Room Temp) / Chalk + 1 M HCl
(Heated)
 Prediction?
 Factor?
 Speed
of reaction or reaction
rate is the time over which a
change occurs
 Consider the reaction A  B
 Reaction rate is a measure of
how quickly A is consumed or
B is produced

Average rate of reaction can be written:
 (moles B)
Average rate 
t

This is a measure of the average rate of
appearance of B

Average rate can also be written in terms of
A:
 (moles of A)
Average rate  
t


This is the rate of disappearance of A (equal
to B only negative)
Average Rates can only be positive

Start with one mole of A at time zero,
measure amounts of A and B at given time
intervals

Data for Reaction A  B


When mole ratios of equations are not 1:1
For the reaction:
aA + bB  cC + dD

Example
◦ If the rate of decomposition of N2O5 in a reaction
vessel is 4.2 x 10-7 M/s, what is the rate of
appearance of NO2 and O2
2 N2O5(g)  4 NO2(g) + O2(g)

Consider the reaction between butyl chloride
and water:
C4H9Cl(aq) + H2O(l) C4H9OH(aq) +HCl(aq)
C4 H 9Cl 
Average rate  
t


Using the curve
created from the
data, we can
determine the
instantaneous rate
for any given point
on the curve
Recall: slope is rise
over run!


Analogy: Distance between Fall River and
Norton is 29.7 mi along a certain route. It
takes Mr. N 30 minutes to get to school. His
average rate is 59.4 mph.
But at t = 15, Mr. N’s instantaneous rate is 95
mph, and at t = 1 Mr. N’s instantaneous rate
is 25 mph.
 Increasing
concentration of
reactants gives increasing rate
 Decreasing rates of reactions
over time is typical
◦ Due to decreasing concentration of
reactants


Rates of a reaction can be related to
concentrations with a rate constant (k)
For reaction:
aA + bB  cC + dD
Rate kA B
x

y
Rate laws are defined by reactant (not
product) concentrations

For the rate law expression:
Rate kA B
x

y
The overall order of reaction is the sum of the
powers (x + y)
◦ However, rate with respect to [A] is only x

In most rate laws reaction orders are 0, 1, or
2
◦ Can be fractional or negative at times
◦ Most commonly 1 or 2

Reaction orders are determined
experimentally, and do not necessarily relate
to coefficients of a balanced equation

Example:
What is the overall order of reaction for the
reaction below?
CHCl3(g) + Cl2(g) CCl4(g) + HCl(g)
Rate= k [CHCl3][Cl2]1/2
A.) ½
B.) 2
C.) 3/2
D.) 2/2
 Zero
order for a reactant means
concentration changes have no
effect on reaction rate
◦ Example: Drinking
 1st
order means concentration
changes give proportional
changes in reaction rate
◦ Double the concentration, double the rate
 2nd
order rate law, increasing
in concentration results in a
rate increase equal to the
concentration increased to the
second power
◦ Example:
 Double conc. = 22 = 4 (rate increase)
 Triple conc. = 32 = 9 (rate increase)

The units for the rate constant depend on the
order of the rate law
units of rate
Unitsof rateconstant
(unitsof concentrat
ion)Z
M /s
 units of rate constant
Z
M
Z = overall order of reaction
Rate Law
Overall Order
Units of Rate Constant
(k)
Rate = k
Zero
M/s (M s-1)
Rate = k [A]
First
1/s (s-1)
Rate = k [A][B]
Second
1/M s (M-1 s-1)
Rate = k [A]2[B]
Third
1/M2 s (M-2 s-1)

What is the unit for the rate constant for the
reaction below?
CHCl3(g) + Cl2(g) CCl4(g) + HCl(g)
Rate= k [CHCl3][Cl2]1/2
A.) M½/s
B.) M/s
C.) M2/s
D.) M-1/2/s
A particular reaction was found to depend on
the concentration of the hydrogen ion. The
initial rates varied as a function of [H+] as
follows:
[H+] (M)
0.0500
0.100
0.200
Initial rate(M/s) 6.4x10-7 3.2x10-7 1.6x10-7
 What is the order of the reaction in [H+]
 A.) 1
 B.) 2
 C.) -1
 D.) -1/2

Rate Data for the Reaction Between F2 and ClO2
[F2] (M)
[ClO2] (M)
Initial Rate (M/s)
0.10
0.010
1.2 x 10-3
0.10
0.040
4.8 x 10-3
0.20
0.010
2.4 x 10-3
What is the rate law expression for the reaction?


Rate law tells how rate changes with changing
concentrations at a particular temperature
We can derive equations that can give us the
concentrations of reactants or products at
any time during a reaction (instantaneous)
◦ These are known as integrated rate laws
Rate = k



Using calculus, the integrated rate law is:
[ A]t  kt  A0
[A]t is concentration of reactant at time t
[A]0 is initial concentration of reactant
[ A]t  kt  A0


y  m x  b
This has the same form as the general
equation for a straight line
Graphically, the slope is equal to -k


Separate equations can be derived relating to
time required for reactants to decrease to half
of initial concentration (aka half-life or t1/2)
When t = t1/2, [A]t is half of [A]0 ([A]t =[A]0/2)
Rate = k [A]

Using calculus, the integrated rate law
becomes:
lnAt  k  t  lnA0

This equation is of the general equation for a
straight line (like before)


Note that only the
second graph is
used so that the
slope can be
determined
Also, y-intercept is
ln [A]0

Example
2N2O5(g)  4NO2(g) + O2(g)

The decomposition of dinitrogen pentoxide is
a 1st order reaction with a rate constant of 5.1
x 10-4 s-1 at 45ºC. If initial conc. is 0.25M,
what is the concentration after 3.2 min.?

Example #2
2N2O5(g)  4NO2(g) + O2(g)
The decomposition of dinitrogen pentoxide is
a 1st order reaction with a rate constant of 5.1
x 10-4 s-1 at 45ºC. How long will it take for
the concentration of N2O5 to decrease from
0.25M to 0.15M?

For first order reactions:
t1/ 2


0.693

k
Note it is independent of concentration!
This is used to describe radioactive decay and
elimination of medications from the body
Rate = k [A]2

Using calculus, the integrated rate law
becomes:
1
1
 kt 
[ A]
[ A]0

Just like the previous two, this equation is of
the general equation for a straight line

Unlike first order, second order does depend
on initial concentrations:
1
k[ A]0
Ways to find Rate Constant or
Reaction Order
Conc. Vs. Time (Graphically)
k = slope
Rates vs. Conc. (Exp. Data/Rate Laws)
Half-life expressions
 Most
reactions increase in rate
with increasing temperature
 This is due to an increase in
the rate constant with
increasing temperature

Minimum amount of energy required to
initiate a chemical reaction
◦ Varies from reaction to reaction

This is the kinetic energy required by
colliding molecules in order to begin a
reaction
◦ Remember, even with sufficient KE, orientation is
still important

Activation energy must be enough to
overcome initial resistance for a reaction to
take place


Diagram can be used
to determine if
reaction is
exothermic (- ∆H) or
endothermic (+∆H)
Activated complex
(or transition state) is
the arrangement of
atoms at the peak of
the Ea barrier
◦ Unstable and only
appears briefly


The relationship between rate and
temperature was non-linear
Reaction rate obeyed an equation based on 3
factors:
1. Fraction of molecules that possess Ea
2. # of collisions per second
3. Fraction of collisions with proper orientation





k  Ae
 Ea / RT
k = the rate constant
R = gas constant (8.314 J/mol*K)
T = Absolute temperature (K)
Ea = the activation energy
A = frequency factor
◦ A is mostly constant with variations in temperature

Taking the natural log of both sides gives a
formula in straight line form:
Ea
ln k  
 ln A
RT

Graph of ln k versus 1/T will be a straight line
with a slope of –Ea/R and a y-intercept of ln A

In order to compare different rates at
different temperatures the equation can be
rearranged:
k1 Ea  1 1 
  
ln 
k2
R  T2 T1 

Example
◦ The rate constant of a first order reaction is 3.46 x
10-2 s-1 at 298 K. What is the rate constant at 350
K if the activation energy is 50.2 kJ/mol?
k1 = 3.46 x 10-2 s-1
T1= 298 K
k2 = ?
T2 = 350 K

Example
-2
1
-2
1
3.46x 10 s
50.2kJ K  mol  1
1 
ln




k2
mol 8.314J  350K 298K 
3.46x 10 s
ln
k2
50200J K  mol
4

 4.98 x10
mol 8.314J

-2
3.46x 10 s
ln
k2
1
 3.01

-2
3.46x 10 s
k2
-2
1
3.46x 10 s
.0493
 .0493
1
k2 = 0.702
 k2
-1
s
 The
process by which a
reaction is broken down into
multiple step reactions
◦ Chemical equations only show
beginning and ending substances
◦ Can show in detail bond breaking
and forming and structural changes
that occur during a reaction
 Elementary
steps
◦ Processes that occur in a single event or
step, are elementary processes.
 Can determine rate laws from elementary
steps, unlike overall reactions
◦ Particles collide with sufficient energy and
proper orientation for reaction to occur
◦ These are the small step reactions in
which an overall reaction occurs


Often times chemical reactions are a result of
multiple steps not shown by the overall
equation
NO2( g )  CO( g )  NO( g )  CO2( g )
The above rxn below 225 °C occurs as 2
elementary steps

1st step
◦ 2 NO2 molecules collide

NO
 NO2( g )  NO3( g )  NO( g )
nd 2( g )
2
step
◦ The NO3 then collides with CO and transfers an O
NO

CO

NO

CO
3(g)
(g)
2(g)
 The elementary steps must add to result2(g)
in
the overall chemical equation



Every reaction is made up of a series of
elementary steps
Rate laws reflect the relative speeds of these
steps
The rate law of an elementary step is directly
related to its molecularity
◦ This is the number of molecules that participate as
reactants
 Unimolecular
(Aproduct)
◦ Rate =k[A]
 Bimolecular
(A+B prod)
= 1st order
= 2nd order
◦ Rate= k[A][B]




Most reactions involve multiple steps
Often one step is much slower than the other
The Rate Determining Step (RDS) is the
slowest step in the reaction
The slowest step of a multi-step reaction
determines the overall rate of reaction!

Good things to know/determine:
◦ Steps must add up to overall reaction
◦ Identify Catalyst
 Consumed at first, regenerated later (not in overall)
◦ Identify Intermediates
 Produced and then consumed later
◦ Each step has a rate law
 Depends on number of reactants
◦ Rate Determining Step => slowest step

Example:
◦ What is the rate law of the following multi-step
reaction?
step1: NO2( g )  NO2( g )  NO3( g )  NO( g ) slow
k1
step2 : NO3(g)  CO(g)  NO2(g)  CO2(g)
k2
fast
Overall: NO2( g )  CO( g )  NO( g )  CO2( g )
 As
a general rule catalysts
change the rate of reaction by
lowering the Ea
 Usually this is done by giving
completely different
mechanism for the reaction
◦ This lowers the overall Ea

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