In the Binary Collision Model we made a good case for the

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In the Binary Collision Model we made a good case
for the rate expression
R?(?AB)2 lturelgt e -EA/RT (NA/V)(NB/V)
Often, NA/V and NB/V are concentrations in units
of molecules per ml. To get these in moles per
liter, just multiply by 1000/N0!
So the Binary Collision Model predicts RkRCACB
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Basically, the Binary Collision Model predicts a
reaction rate that is first order in A, first
order in B and second order overall.
B) First Order Reactions will need a model
later!
Assume this is first order to get ?
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Integrate both sides ? dCA/CA - ? kdt
Need to find g from initial conditions.
lnCAº - k (0) g
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lnCAo
? Slope -k
CA CA0 e-kt
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Equations for first order reactions are very
important. In the laboratory almost ALL reactions
can be made to APPEAR First Order.
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Bonus Bonus Bonus Bonus Bonus Bonus
Half Life or Half Time
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N2O5 ? 2NO2 1/2 O2
Rate (mol L-1 s-1)
3 x 10-5
Rate -dN2O5/dt kN2O5
2 x 10-5
1 x 10-5
0
1.0
2.0
N2O5 (mol L-1)
Example of a first order reaction.
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C) 2nd Order Kinetics A ? Products
- ? dCA/CA2 k ? dt
1/CA kt g
t t0 , CA CA0 Initial conditions
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1/CA 1/CA0 kt
At t1/2 , CA (CA0 / 2) and
t1/2 1 / kCA0
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2) Initial Rates Method
1st order reaction
c c0 - x Where c0 is the initial concentration
and x is a function of time, x x(t). x is
simply the amount reacted.
If xltltc0 ?dx/dt const kc0 (sure to be true
if t is small enough!)
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Initial part of curve will look like a straight
line if t is small!
CAo
?c
? Initial Slope -k CAo
?c/? t -k CAo
?t
CA
t
Blow up first one percent of CA vs t curve
Slope NOT Constant
dCA/dt -kCA
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Measure ?c vs. ? t for first 1 of reaction.
Here, c0gtgtx, ? c ? dx and ? t ? dt ?
Know c0 , measure ?c and ?t, ? obtain k
nth Order Reactions
A B C ? products
a initial conc of A b initial conc of B c
initial conc of C
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Bonus Bonus Bonus Bonus Bonus Bonus
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(Note, have kept b, c constant!)
(dx/dt)1 and (dx/dt)2 are measured in the
laboratory, while a1 and a2 are known
quantities.
Can do a similar trick for n2 , n3
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Mechanism Concept
1) Exponents in rate law do not depend on
stoichiometric coefficients in chemical
reactions.
2)What is the detailed way in which the reactants
are converted into products? This is not
described by the chemical equation, which just
accounts for mass balance.
3) Rate at which reaction proceeds and
equilibrium is achieved, depends on the Mechanism
by which reactants form products.
Elementary Reactions these are hypothetical
constructs, or our guess about how reactants are
converted to products.
The Mechanism is a set of Elementary Reactions!