Determining Rate Law

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DETERMINING THE RATE LAW
If A B ? C
Then the rate law would ultimately be written as
R k Am Bn
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NH4 NO2- ? N2 2H2O R k NH4m NO2n

• k rate proportionality constant

• m reaction order for the ammonium ion

• n reaction order for the nitrite ion

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NH4 NO2- ? N2 2H2O
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R k NH4m NO2n
2x
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R k NH41 NO2n
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R k NH41 NO2n
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1
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R k NH41 NO21
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R k 0.02M1 NO2-1
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R k 0.02M10.200M1
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10.8 x 10-7M/s k 0.02M10.200M1
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10.8 x 10-7M/s k0.02M1 0.200M1
k 10.8 x 10-7 M/s /0.02M10.200M1
k 10.8 x 10-7 M/s / 0.004M2
k 2.7 x 10-4 1/M - s
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Suppose the concentrations for NH4 and NO2- are
0.15 M and 0.20 M, respectively
Since R kNH41 NO21 where k 2.7 x 10-4
1/M - s at 25? C
then R 2.7 x 10-4 1/M - s 0.15 M 0.20 M
R 8.1 x 10-6 M/s
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2NO(g) Br2(g) ? 2NOBr
1st order in Br2
R kNOm Br2n
k(0.0160)x (0.0120)1
3.24 x 10-4

6.42 x 10-4
k(0.0320)x (0.0060)1
1
2
(0.0160)x

2
(0.0320)x
1

.5x
4
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0 900 2,500 5,000 10,000 15,000 20,000 30,000
150 140 130 118 90 70 55 35
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ln At - k t ln A0
y mx b
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y mx b
3 r 0.92
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0 900 2,500 5,000 10,000 15,000 20,000 30,000
150 140 130 118 90 70 55 35
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Using Rate Law Equations to Determine Change in
Concentration Over Time
The first order rate constant for the
decomposition of a certain insecticide in
water at 12 C is 1.45 yr1-. A quantity of
insecticide is washed into a lake in June
leading to a concentration of 5.0 x 10-7 g/cm3.
Assume that the effective temperature of the lake
is 12 C. (a)What is the concentration of the
insecticide in June of the following year? (b)
How long will it take for the concentration of
the insecticide to drop to 3.0 x 10-7 g/cm3?
R - ?A kA
?t
lnAt - lnA0 ln
- kt
At
A0
ln At - k t ln A0
ln At - (1.45 yr.-1)(1.00 yr.) ln( 5.0 x
10-7 g/cm3)
ln At - 15.96, At e -15.96, At 1.2
x 10-7 g/cm3
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Using Rate Law Equations to Determine the
Half-Life of a Reaction
Lets begin with lnAt - lnA0 ln
let At ½ A0
given
½ A0
then
ln
- kt,
and ln ½ -kt ½
A0
t ½ -ln ½ 0.693
k
k
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Using Rate Law Equations to Determine the
Half-Life of a Reaction
The concentration of an insecticide
accidentally spilled into a lake was measured at
1.2 x 10-7 g/cm3. Records from the initial
accident show that the concentration of the
insecticide was 5.0 x 10-7 g/cm3. Calculate the
1/2 life of the insecticide.
1.2 x 10-7 g/cm3
ln
- k (1.00 yr.)
5.0 x 10-7 g/cm3
k 1.43 yr-1
t ½ 0.693 0.484 yr.
1.43 yr-1
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Examining the Relationship Between the Rate
Constant and Temperature
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Why does Temperature Have an Effect on Rate?
..its because of Collision Theory
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Its All About Activation Energy and Getting
over the HUMP!!
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1010 collisions/sec occur
1 in 1010 collisions succeeds
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...And What Do the Mathematicians Have to Say
About Collision Theory?
k Ae-Ea/RT
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Rewriting the Arrhenius Equation
k Ae-Ea/RT
ln k1 -Ea/RT1 ln Ae
ln k2 -Ea/RT2 ln Ae
lnk1 - ln k2 (-Ea/RT1 ln Ae) - (Ea/RT2 ln
Ae)

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Getting a Line on the Arrhenius Equation
But there is a simpler way to think about
the Arrhenius Equation
Slope -Ea/R
k Ae-Ea/RT
ln k
ln k1 -Ea/RT1 ln Ae
y m x b
1/T
R (ideal gas constant) 8.31 J/K-mol
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Mastering the Arrhenius Equation with your TI
The following table shows the rate constants
for the rearrangement of CH3CN (methyl
isonitrile) at various temperatures
Temperature (C) k (s-1)
2.52 x 10-5 5.25 x 10-5 6.30 x 10-4 3.16 x 10-3
189.7 198.9 230.3 251.2
From these data calculate the activation energy
of this reaction
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Mastering the Arrhenius Equation with your TI
k Ae-Ea/RT
L1 L2 L3 L4 L5
ln k1 -Ea/RT1 ln Ae
T(C) L1 273 1/L2 k ln L4
y m x b
Slope -Ea/R
Its as easy as L1, L2, L3
ln k (L5)
1/T (L3)
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Mastering the Arrhenius Equation with your TI
The following table shows the rate constants
for the following reaction at varying temperatures
CO(g) NO2(g) ? CO2(g) NO(g)
Temperature (C) k (M-1 s-1)
0.028 0.22 1.3 6 23
600 650 700 750 800
From these data calculate the activation energy
for this reaction
Now, you try it!
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Reaction Mechanisms The Shortest Distance
Between Two Points Often Requires More Steps
A Reaction Mechanism is the process which
describes in great detail the order in which
bonds are broken and reformed, changes in
orientation and the energies involved during
those re-bondings, and changes in orientations
NO(g) O3(g) ? NO2(g) O2(g)
A single reaction event is called an elementary
reaction
NO2(g) CO(g) ? NO(g) CO2(g)
While this reaction looks like an elementary
reaction, it actually takes place in a series of
steps
NO2(g) NO2(g)? NO3(g) NO(g)
NO3(g) CO(g) ? NO2(g) CO2(g)
NO2(g) CO(g) ? NO(g) CO2(g)
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Reaction Mechanisms The Shortest Distance
Between Two Points Often Requires More Steps
NO2(g) NO2(g)? NO3(g) NO(g)
Rate Determining step
Slow step
NO3(g) CO(g) ? NO2(g) CO2(g)
Fast step
NO2(g) CO(g) ? NO(g) CO2(g)
The rate law is constructed from the rate
determining step
Rate k1NO22
Yea, but what if the second step is the rate
determining step?
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Reaction Mechanisms The Shortest Distance
Between Two Points Often Requires More Steps
OK. Lets try this one experimentally 2NO(g)
Br2(g) ? 2NOBr(g) what we get is R kNO2Br2
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Reaction Mechanisms The Shortest Distance
Between Two Points Often Requires More Steps
k1
NO(g) Br2(g)? NOBr2(g)
Fast step
k-1
k2
Rate Determining step
NOBr2(g) NO(g) ? 2NOBr(g)
Slow step
2NO(g) Br2(g) ? 2NOBr(g)
1. We assume that reaction 1 is at equilibrium
and that Rf Rr
2. If the above is true, then k1NOBr2 k-1
NOBr2
3. If step 2 is the rate determining step, then R
kNOBr2Br2. Keep in mind, however, that one
cannot include an intermediate in the rate
determining step.
4. We can substitute for NOBr2 using step two,
however. Doing so would allow us to use NO(g) and
Br2(g) which are not intermediates
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Reaction Mechanisms The Shortest Distance
Between Two Points Often Requires More Steps
Let NOBr2 NOBr2
If R k NOBr2Br2.
Then by substitution, R k2 NOBr2
Br2.
We then let k k2
Finally, we get the rate law R kNOBr2Br2.
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Reaction Mechanisms The Shortest Distance
Between Two Points Often Requires More Steps
Prove that the following mechanism is consistent
with R kNO2Br2, the rate law which was
derived experimentally
k1
NO(g) NO(g) ? N2O2(g)
k-1
k2
N2O2(g) Br2(g) ? 2NOBr(g)
2NO(g) Br2(g) ? 2NOBr(g)
Now, you try it
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Catalysta substance that changes the speed of a
chemical reaction without it self undergoing a
permanent chemical change in the process
Br - is a homogeneous catalyst
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Catalysta substance that changes the speed of a
chemical reaction without it self undergoing a
permanent chemical change in the process
Ethylene
Ethane
This metallic substrate is a heterogeneous
catalyst