Lecture A1a Reactivity and Chemical Kinetics
Textbook Chapter 13
Rate equals change in something with time
x
 velocity
t
 A
reaction rate =
t
velocity
A = reactant or product, [A] concentration of A in
For the reaction A  B , measure the rate two ways:
disappearance of reactant A
A
t
rate =

rate =
 B
.
t
or
appearance of product B
Self-tests 13.1 and 2
Physical factors that affect the rate of chemical reactivity
1. Phase and State of subdivision (greater surface area) within the phase
Solids: powders, grain size
Liquids: aerosols
2. Concentration
3. Temperature
(higher T - higher rate
Biological rates +10°C, rate doubles )
4. Presence of Catalysts
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A) Concentration Effects
Three requirements for a chemical reaction
1. Molecules must collide
2. Molecules must collide with sufficient energy (activated)
3. Molecules must collide with proper orientation (active site)
- Average and instantaneous rates
- Rates change over time since concentrations change over time.
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Rate Equations
How to find rate dependence on [ ]?
Through experiment!
Two major methods
First Method: vary initial rates by varying initial concentrations.
(Keep all Factors constant, but change initial concentration)
2A  B [specifically, 2 N2O5(g)  4 NO2(g) + O2(g) ]
As the initial concentration [A]0 is doubled, the initial rates are doubled:
Rate of disappearance of A = k [A]
The constant, k is called the “rate constant” and is the slope of the graph on the
right. Note that in this reaction, the rate is dependent on the concentration of A to
the first power, even though the coefficient of A in the chemical equation is 2.
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We call this reaction “First order in A”.
Now let’s look at the results for another 2A  B reaction
[specifically, 2 NO2(g)  2 NO2(g) + O2(g) ]
When the concentrations are plotted against time for several different starting
concentrations, we obtain a graph that is similar in appearance to the above graph
on the left. However, when we plot the initial rates to the initial concentrations, we
do not get a straight line as above on the right. We obtain a parabola, which
indicates a function like y = k∙x2. To test this, we plot rates against [A]2. When a
straight line is obtained we verify that the “reaction is second order in A”.
Rate of disappearance of A = k [A]2
Again, the “rate constant” k is the slope of the graph on the right.
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Now for another reaction where the coefficient of the reactant is 2.
2A  B [specifically, 2 NH3(g)  N2(g) + 3 H2(g), with catalyst]
At first, the experiment appears to go against principles in that the reaction is not
dependent on the concentration of A. (We will see later when we explore reactions
on the molecular level that the reaction depends on the number of catalyst atoms
available.)
The rate does not change over time, until the reactant is all used up and the rate
falls to zero.
Rate of disappearance of A = k [A]0 = k
We call this a reaction that is “zero order in A”.
Experimental determination of the order of reactants and of the rate
constant k
When there is more than one reactant, a series of rate experiments must be done
for each reactant
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Experiment: 2A + B  C
Initial rate of formation of C mol/L•s
Initial [A]
Initial [B]
mol/L
mol/L
1.
0.010
0.010
1.50
2.
0.020
0.010
3.00
3.
0.030
0.010
4.50
4.
0.030
0.020
18.0
Now set up rate expression: rate = [A]x[B]y
The object is to determine the value of x and y.
Compare experiments 1 and 2. [A] doubles and [B] is held constant.
rate  [A]x[B]y
rate  k[A]x[B]y k = "rate constant"
x, y determined from experiment (important!)
1.50 = k[0.010]x[B]y
3.00 = k[0.020]x[B]y
rate doubles as concentration of A doubles
rate doubles if x = 1
would quadruple if x = 2
4.50 = k[A]x[0.010]y
18.0 = k[A]x[0.020]y
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[0.010]y  [0.020]y
18.0
4
4.50
quadruples
doubles, so y = 2
4.50 = k[A]1[0.010]y
quad = quad
18.0 = k[A]1[0.020]2
"rate expression"
rate = k[A][B]2
“First order in A”, “second order in B”,
“Third order overall” (overall= sum of orders)
value of k ?
(take values from any one of the experiments in table)
1.50 mol/L•s = k[0.01][0.010]2
k = 1.50 mol/L•s /(0.010 mol/L)(0.010 mol/L)2
k = 1.50 x106 L2/mol2s1
General Formula for finding exponent x (when all other [] remain the same):
The formula works when the ratios are not integers.
1)
Look at any two experiments Rate1 = k [A]1x and Rate2 = k [A]2x
2)
Divide both equations
______
Rate1 = k [A]1x
________
Rate2 = k [A]2x
3) Separate the equations, (k’s cancel), then take ln of both sides
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(rate1/rate2) = ([A]1x / [A]2x) , ln(rate1/rate2) = ln([A]1x / [A]2x)
4) Remember ln (ax) = x∙ln (a) ln(rate1/rate2) = x∙ln([A]1 / [A]2)
x = ln(rate1/rate2) / ln([A]1 / [A]2)
Trying results for [B] from exps. 3 and 4:
x = ln(4.50/18.0) / ln (0.010/0.020) = ln 0.25 / ln 0.50 = -1.386/-0.693/- = 2
Self-tests 13.4
Why bother calculating k and rate expression?
Because rate depends on concentration, which changes over time.
The rate expression gives the rate for a particular set of concentrations of each
reactant.
Second Method to determine Rate Expression: “Integrated Rate Expressions”
(Only need to concentrations at a few points in time; the more common method)
I) First Order Reactions (Overall)
slope  d[A]/dt
[A]
time
-slope = -d[A]/dt = rate = k[A]
d[A] = -k∙[A] dt and [A]-1 d[A] = -k∙dt and ∫ [A]-1 d[A] = -k ∫ dt
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Calculus gives:
Integrated Rate Expression
ln ([A]/[A]0) = -kt
ln[A] - ln[A] 0 = -kt
[A]0 = initial conc.
[A] = conc. at time t
is a linear equation y=mx+b ln[A] = -kt + ln[A] 0

If you plot ln of [A] against time and there is a straight line,
then you know two things:
1) The rate is first order in A
2) The value of the slope is –k.
(natural) antilog ln([A]/[A]0) = (natural) antilog –kt
([A]/[A]0) = e-kt or [A] = [A]0 e-kt
Self-test 13.5,6,7
Half-Life,
t1
2
= time required for half of the original concentration of the limiting
reactant to be consumed.
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at
,
t1
2
, [A] = 1/2 [A0]
so for first order reactions in A
ln
[ A0 ]
[A ]
 kt1  ln 1 0  ln 2
2
[A]
2 [A0 ]
kt1  ln 2  0.693
2
0.693
k
t1 
2
First Order Reaction
(NB The sign of kt is positive and the ratio of concentrations was reversed in the
above equations. In effect, we have multiplied both sides of the previous equation
by -1.)
Remember: ln(a/b) = - ln(b/a)
II) Second order reactions
Rate = d[A]/dt = -k∙[A]2 or [A]-2 d[A] = -k dt or ∫ [A]-2 d[A] = -k ∫dt
The “Integrated Rate Expression” for a second order reaction is
1/[A] – 1/[A]0 = kt or
III)
1/[A] = kt + 1/[A]0 (y=mx+b form)
Zero order reactions
Rate = d[A]/dt = -k or d[A] = -k dt or ∫ d[A] = -k ∫dt
The “Integrated Rate Expression” for a second order reaction is
[A] – [A]0 = - kt or
[A] = - kt + [A]0 (y=mx+b form)
!!! Table 13.2 !!!
Zero Order
“rate law” Rate = k
First Order
Rate = k [A]
Second Order
Rate = k [A]2
“Integrated
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rate law” [A] = - kt + [A]0
ln[A] = -kt + ln[A] 0
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1/[A] = kt + 1/[A]0
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