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```							CHE 112 - MODULE 3

CHAPTER 15
LECTURE NOTES
Chemical Kinetics
   Chemical kinetics - study of the rates of
chemical reactions and is dependent on the
characteristics of the reactants

   Reaction mechanisms - detailed pathway that
atoms and molecules take as a chemical
reaction proceeds

   CD-ROM Screen 6.2 & 15.2
Effects on Chemical Kinetics
   Concentration of reactants – usually the rate
of a reaction increases with increased
concentration of reactants
   Concentration of a catalyst - catalyst speeds
up the reaction
   Temperature – rate will increase with an
increase in temperature (increased KE)
   Surface area – as the surface area increases,
the reaction will proceed at a faster rate
Rates of Chemical Reactions
   Rates - the change in concentration of a
chemical per unit of time ex. (M/sec)
– much like the speed of a car ex. (miles/hr)
– much like interest rates ex. (5.4%/year)
– much like sales taxes ex. (7cents/\$)

   Rates are just a ratio or fraction of one
thing changing with respect to another
Decomposition of N2O5
2 N2O5  4 NO2 + O2
   We look at the disappearance of N2O5

   Rate of Reaction = change in [N2O5]
change in time

   Rate of Reaction =  [N2O5]
t
Stoichiometry

2 N2O5  4 NO2 + O2
2moles : 4moles : 1mole
Therefore, for every 2 moles of N2O5
decomposed, you have 1 mole of O2
formed. The rate of formation of O2 is
equal to ½ the rate of decomposition of
N 2O 5
[O2]/t = -½ [N2O5]/ t
Decomposition of N2O5

   Therefore we can determine the rate of
decomposition by the rate of formation of
either NO2 or the O2 with consideration to
stoichiometry.
2 N2O5  4 NO2 + O2

Rf =  [NO2] = -2Rd      Rf =  [O2] = -½ Rd
t                     t
Calculations
2 N2O5  4 NO2 + O2
   Where ti = 600s and tf = 1200s and the
concentration of N2O5 decomposed from 1.24
x 10-2 M to 0.93 x 10-2 M
   Rate of decomposition = [N2O5 ] /t
= (0.93 x 10-2 M -1.24 x 10-2 M)
(1200s – 600s)
= -5.2 x 10-5 M/s
   Rate of formation O2 = -½ [-5.2 x 10-5 M/s]
= 2.6 x 10-5 M/s
Plot of Concentration vs. Time
   Figure 15.2 shows the graph of the
disappearance of N2O5.

   Average rate = change in concentration
over an interval in time (c/t)

   Instantaneous rate = concentration at
an instant in time; tangent to the curve
at a particular point in time (dc/dt)
Rate Law

   Rate Law – equation that relates the
rate of reaction to the concentration of
reactants raised to various powers

Rate = k [N2O5]x

– where k is the rate constant and x is the
order of the reaction
Determining Rate Law
We can perform an experiment to decompose
N2O5. If we doubled the concentration of the
reactant we can observe the change in rate
as follows:

Initial Conc.   Rate of decomposition
Exp.1 1x10-2 mol/l     4.8x10-6 mol/l•sec
Exp.2 2x10-2 mol/l     9.6x10-6 mol/l•sec

Observation: The rate of reaction doubled.
Determining Order
   When a reaction has this observable result, it
is said to be first order.

   As we double the concentration, the rate is 2x,
where x is the classification of the order of
that particular reaction.

   Or R2/R1 = 2x, where x=1, then we can see
that it is clearly first order.
Overall Rate Law for N2O5

We previously stated that:

Rate = k [N2O5]x

Where x = 1 determined experimentally,
therefore this reaction is a first order reaction
and follows the rules of first order kinetics.
Order of Reactions

Considering other reactions where we double
the initial reactant concentration, we can
observe the rate and order as follows:
Quadruple rate = 4 (2x=4, x=2) second order
Double rate       = 2 (2x=2, x=1) first order
No change         = 1 (2x=1, x=0) zero order
Half rate       = 1/2 (2x=1/2, x=-1) -1 order
1.4 the rate = 1.4 (2x=1.4, x=1/2) half order
Review of Logarithms
   Logs are the inverse function of an
exponential function (y = 3x)
   Algebraically it is written as x = 3y or y =
log3x where y is the exponent on 3 that
results in a value of x
– ex. 52 = 25 or log5 25 = 2
   Written as a function logs are expressed as
f(x) = logbx
   If b is not designated, b=10 is assumed.
   If b = e it is called the natural log (ln).
First Order Kinetics

2 N2O5  4 NO2 + O2
   Rate of decomposition = - [N2O5]
t
   Rate = kt

   Therefore - [N2O5] = -kt (by substitution)
t
   ln [N2O5]t = -kt + ln [N2O5]0 (derive by integration)
Integrated Rate Equations
   For a zero order reaction:
[A]t = -kt + [A]0

   For a first order reaction:
ln [A]t = -kt + ln [A]0

   For a second order reaction:
1/[A]t = kt + 1/[A]0
Iodine Clock Reaction Prep

Each solution is prepared in a 100ml
volumetric flask and qs with distilled
water. Accurately weigh the following:
2.998g of NaI = 0.2M
1.169g of NaCl = 0.2M
0.2482g of sodium thiosulfate = 0.01M
3.485g potassium sulfate = 0.2M
5.404g potassium persulfate = 0.2M
Half-life of Reaction

   Half-life (t½) = the time it takes for the
initial concentration of a reaction to
disintegrate to half its original
concentration

   First order reaction:   t½ = 0.693/k
Collision Theory

   Three conditions must be met for a
reaction to take place:
– The reacting molecules must collide with
one another
– Molecules must collide with sufficient
energy
– Molecules must collide with the proper
orientation
Activation Energy

   Ea = the minimum amount of energy
that must be absorbed by a system to
cause it to react where k = Ae -Ea/RT
   Activation energy can be determined
from the Arrhenius Equation:
ln k = ln A - Ea/R (1/T)

ln (k 2/k1) = - Ea/R [1/T2 - 1/T1]
Conditions Increasing Rates
   Temperature - increases the KE and
enough energy to overcome Ea

   Presence of a catalyst - provides
different pathways with lower Ea

   Surface area - increases the probability
of a collision with proper orientation
Reaction Mechanisms

   Reaction mechanisms = sequence of
steps that show the intermediates
formed ( bonds broken and bonds
formed) between the reactant side and
product side of any reaction
   See CD-ROM Screens 15.12 – 15.13
Rate Equations for Elementary Steps

   Elementary Step - a singular molecular
event classified by the # of reactant
molecules involved (molecularity)
– unimolecular: AP      where R = k[A]
– bimolecular: A + BP where R = k[A][B]
or A + AP where R = k[A]2
– termolecular: 2A +B P where R = k[A]2[B]

   See Table at the bottom of page 634
Rate-determining Step

Step 1:        A + B  X + M; k1 is slow
Step 2:        M + A  Y;     k2 is fast

Overall Rxn.   2A + B  X + Y

Because step 2 is fast, it does not
contribute to the overall rate, therefore:
R = k1[A][B]

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