# Chapter 15 Powerpoint

```Chapter 15
Rates of Reaction
Overview


Reaction Rates
 Definition of Reaction Rates
 Experimental Determination of Rate
 Dependence of Rate on Concentration
 Change of Concentration with Time
 Temperature and Rate; Collision and TransitionState Theories.
 Arrhenius Equation
Reaction Mechanisms
 Elementary Reactions
 Rate Law and the Mechanism
 Catalysis
Definition of Reaction Rate
 Reaction
rate = increase in concentration
of product of a reaction as a function of
time or decrease in concentration of
reactant as a function of time
Rate Equation
 For
the reaction:
A+
2B  3C
conc change A [ A ]
Rate A 

time change
t
[ A ]2  [ A ]1

t 2  t1
Rate Equation
Concentration vs Reaction Time
A + 2B --&gt; 3C
Concentration, M
0.090
Init
Rate
0.045
Inst.
Rate
Ave.
Rate
0.000
0
250
500
Tim e, s

Rates are expressed as positive numbers.
For the reaction in the graph we have:
[ A ]
RA  
t
RB  
[B]
t
RC  
 [C ]
t
Reaction Rates and
Stoichiometry

A + B  C; RC = RA = RB.

A + 2B  3C;

For the general reaction: aA + bB  cC + dD
.
1
1
R A  RB  RC
2
3
a
a
a
R A  RB  R C  RD
b
c
d
Calculate the rate of decomposition of HI
in the reaction: 2HI(g)  H2(g) + I2(g).
Given: After a reaction time of 100 secs.
the concentration of HI decreased by
0.500 M.
the reaction 2A + 3B  4C + 2D;
determine the rates of B, C and D if the
rate of consumption of A is 0.100 M/s.
 For
Factors that Affect Reaction Rates





Nature of reactants
Concentrations of reactants
Temperature
Presence or absence of a catalyst a
substance that increases the rate of
reaction without being consumed in the
reaction.
Surface area for heterogeneous reactions.
Rate Laws and Reaction
Order

Rate Law – an equation that tells how the
reaction rate depends on the concentration of
each reaction.

Reaction order – the value of the exponents
of concentration terms in the rate law.
the reaction: aA + bB  cC + dD, the
initial rate of reaction is related to the
concentration of reactants.
 For
R
= k[A]m[B]n
 Varying
the initial concentration of one
reactant at a time produces rates, which
will lead to a determination of exponents.

The rate law describes this dependence: R =
k[A]m[B]n where k = rate constant and m and n
are the orders of A and B respectively.


m = 1 (A varied, B held constant) gives R = k’[A].
Rate is directly proportional to [A]. Doubling A
doubles R
m = 2 (A varied, B held constant) gives R = k’[A]2.
The rate is proportional to [A]2. Doubling A
Determine order of each reactant:
HCOOH(aq) + Br2(aq)  2H+(aq) +
2Br(aq) + CO2(g)
R = k[Br2]
The formation of HI gas has the following
rate law: R = k[H2][I2]. What is the order of
each reactant?
Problem
Determine the reaction orders for the reaction indicated
from the data provided.
A + 2B + C  Products.
[A]o
2.06
0.87
0.50
1.00
[B]o
3.05
3.05
0.50
0.50
[C]o
4.00
4.00
0.50
1.00
Ro
3.7
0.66
0.013
0.072
Problem
Determine the reaction order for each reactant
from the table.
BrO3-(aq)+5Br(aq)+6H+(aq)3Br2(aq)+3H2O(l)
[BrO 3 ]o
0.10
0.20
0.10
0.20
[Br]o
0.10
0.10
0.30
0.10
[H+]o
0.10
0.10
0.10
0.15
Ro
1.2
2.4
3.5
5.4
First–Order Reaction:
Integrated Rate Law

For a first order reaction, Rate = [A]/t = k[A]

Linear Graph, In [A] vs time

Equation of the line

Concentration vs time
ln[ A ]  kt  ln[ A ]o
ln
[A]
 kt
[ A ]o
Problem
Calculate the concentration of N2O
remaining after its decomposition
according to 2N2O(g)  2N2(g) + O2(g) if
it’s rate is first order and [N2O]o = 0.20M, k
= 3.4 s1 and T = 780&deg;C. Find its
concentration after 100 ms.
Problem
When cyclohexane (let's call it C) is
heated to 500 oC, it changes into propene.
Using the following data from one
experiment, determine the first order rate
constant.:
Half-Life: First Order Reaction

Half-life of First order reaction, t1/2 = 0.693/k.
the time required for the concentration of the
reactant to change to &frac12; of its initial value.
i.e. at t1/2 , [A] = &frac12; [A]o
1 / 2[ A]o 
ln 
  k  t1/ 2
 [ A]o 
1 
ln    k  t1/ 2
2
t1/ 2  0.693 / k
Problem
For the decomposition of N2O5 at 65 &deg;C,
the half-life was found to be 130 s.
Determine the rate constant for this
reaction.
Second–Order Reactions:
Integrated Rate Law
 Rate
law: R = k[A]2
 Plot
1
[ A ]t
of
vs. t gives a straight line with
a slope of k.
 Equation
 Half-life
of the Line:
is:
1
t1/ 2 
k  [ A ]o
1
1
 kt 
[ A ]t
[ A ]o
Problem
At 330&deg;C, the rate constant for the
decomposition of NO2 is 0.775 L/(mol*s).
If the reaction is second-order, what is the
concentration of NO2 after 2.5x102 s if the
starting of concentration was 0.050 M?
Zero–Order Reactions: Integrated
Rate Law
 Rate
law: R = k[A]0
 Plot
of [A] vs. t gives a straight line with a
slope of -k.
 Equation
 Half-life
of the Line: [A] = kt - [Ao]
is: t&frac12; = [Ao] / 2k
Reaction Mechanisms

Give insight into sequence of reaction events
 Each of the steps leading to product is called an
elementary reaction or elementary step.
 Consider the reaction of nitrogen dioxide with
carbon dioxide which is second order on NO2:
NO2(g) + CO(g)  NO(g) + CO2(g)
Rate = k[NO2]2.

Rate law suggests at least two steps.
 A proposed mechanism for this reaction involves
two steps.
Step 1
Step 2
Overall

2NO2(g)  NO3(g) + NO(g)
NO3(g) +CO(g) NO2(g) + CO2(g)
NO2 + CO  NO + CO2
NO3 is a reaction intermediate = a substance that is
produced and consumed in the reaction so that none
is detected when the reaction is finished.
 The
elementary reactions are often
described in terms of their molecularity.



Unimolecular One particle in elementary.
Bimolecular = 2 particles and
Termolecular = 3 particles


Rate Laws and Reaction
Mechanisms
Overall reaction order is often determined by the rate
determining step.
Use rate law of limiting step; No intermediates!
2NO2(g)  NO3(g) + NO(g),
NO3(g) +CO(g) NO2(g) + CO2(g)
NO2 + CO  NO + CO2
R1 = k1[NO2]2
Slow
R2 = k2[NO3][CO]
Fast
Robs = k[NO2]2
Problem

Determine the Rate Law for the mechanism
given below
k1
2*[N2O5(g) NO2 (g)  NO3 (g) ]
Fast
k2
NO3(g) +NO2(g) 
 NO 2 (g)  NO(g)  O 2 (g)
Slow
k 1
k3
NO3 + NO 
 2NO 2 (g)
k obs
2N2O5(g) 

 4NO 2 (g)  O 2 (g)
Fast
 Use steady state approximation. at “equilibrium” rates of forward and
reverse reactions are same. Use to eliminate intermediates from rate law
equations.
R1  R 1
k 1 [N2O5 ]  k 1[NO 2 ][NO3 ]
or
[NO3 ] 
k 1 [N2O5 ]
k 1 [NO 2 ]
Collision Theory
 Collision
theory assumes:
Reaction can only occur if collision takes
place.
 Colliding molecules must have correct
orientation and energy.
 Collision rate is directing proportional to the
concentration of colliding particles.
A + B  Products; Rc = Z[A][B]
2A + B  Products; Rc = Z[A]2[B], etc.

Only a fraction of the molecules, p (“steric
factor”), have correct orientation; multiply
collision rate by p.
 Particle must have enough energy. Fraction of
those with correct energy follows Boltzmann
equation f  e Ea / RT where Ea = activation
energy, R = gas constant and T = temp. (Kelvin
 This gives: k = Zpf

“Steric Factor”

Molecules must have the correct orientation
before a reaction can take place.
Transition State Theory

Explains the reaction resulting from the collision
of molecules to form an activated complex.
 Activated complex is unstable and can break to
form product.
Potential Energy Diagram

Endothermic reactions
 In endothermic reactions heat energy is taken in
from the surroundings and turned into potential
energy in the products. As a result, the enthalpy
of the products is greater than the enthalpy of
the reactants.

The potential energy (enthalpy) diagram for an
endothermic reaction is shown on the next slide
Potential Energy Diagram

Exothermic reactions
 In exothermic reactions potential energy in the
reactants is turned into heat energy and given
off to the surroundings. As a result, the enthalpy
of the products is less than the enthalpy of the
reactants.

The potential energy (enthalpy) diagram for an
exothermic reaction is shown on the next slide
Potential Energy Diagram
Reaction Rates and Temperature:
The Arrhenius Equation


Rate (rate constant) increases exponentially with
temperature.
Collision theory indicates collisions every 109s –
1010s at 25&deg;C and 1 atm.
i.e. only a small fraction of the colliding molecules
actually react.
f e
 Ea / RT
The Arrhenius Equation
Arrhenius Plot
-5
ln k
-6
-7
-8
0.00300
0.00305
0.00310
0.00315

1/T, K
 E 
k  A exp   a  where A = frequency factor.
 RT 

Summary:

Linear form: .

Plot ln k vs. 1/T; the slope gives Ea/R.

Two point equation sometimes used also:
E
ln k  ln A  a
RT
E
k
ln 2  a
k1
R
1
1
 

T
T
 1
2
Rate constant increases when T2&gt;T1
k, s1
4.8x104
8.8x104
1.6x103
2.8x103
Temp., &deg;C
45.0
50.0
55.0
60.0
Temp., K
318.15
323.15
328.15
333.15
Problem
Determine the activation energy for the
decomposition of N2O5 from the
temperature dependence of the rate
constant.
Problem
Determine the rate constant at 35&deg;C for
the hydrolysis of sucrose, given that at
37&deg;C it is 0.91mL/(mol*sec). The
activation energy of this reaction is 108
kJ/mol.
Catalysis

Catalysts a substance that increases the rate of
a reaction without being consumed in the
reaction.
 Catalyst provides an alternative pathway from
reactant to product which has a rate determining
step that has a lower activation energy than that
of the original pathway.
Potential Energy Diagram with
Catalyst
Homogeneous and
Heterogeneous Catalysts

Homogeneous catalyst: catalyst
existing in the same phase as the
reactants.

Heterogeneous catalysis: catalyst
existing in a different phase than the
reactants.

The catalytic hydrogenation of ethylene is an
example of a heterogeneous catalysis reaction:
Pt
H2C  CH2 (g)  H2 (g) 
 H3C  CH3(g)

ENZYMES (biological
catalysts)


They are proteins (large
organic molecules that are
composed of amino acids).
Slotlike active sites. The
molecule fits into this slot
and reaction proceeds.
Poisons can block active
site or reduce activity by
distorting the active site.
```