Intro to kinetics

Kinetics: Zumdahl Chapter
12
L. Allen
Stonington High School

Chapter objectives (lots!)
 Identify factors that influence reaction rates
 Calculate rate of consumption of a reactant or of
creation of a product from stoichiometry
 Write the rate law for a reaction
 Determine the rate law of a reaction from data
 Determine the order of a reaction from time/rate
data
 Solve for the rate constant from time/rate data
 Determine the instantaneous rate of a reaction
 Create graphs to determine order of a reaction
 Use the integrated rate law to determine
concentration at time t
 Determine half life of a reaction
 Determine the activation energy for a reaction

Didn’t we already figure out if a
reaction would run
spontaneously?
 Remember the rule from
thermochemistry – spontaneous does
not mean fast! Just because a
reaction is energetically favored, does
not mean it will run in a reasonable
amount of time. Thermochemistry
takes into account the energy of the
products and of the reactants, not any
activation energy required to run a
reaction.

Factors that influence reaction
rates
 Concentration – the more particles
bump into each other, the more likely
they are to react
 Temperature – the more particles
move, the more they bump into each
other
 Increasing surface area – powdered
aluminum is much more interesting
than a chunk of aluminum
 Addition of a catalyst – determined
experimentally

Effect of a catalyst on activation
energy
Reactions can be
endothermic or
exothermic – either
way, a catalyst can
facilitate overcoming
the activation energy
“bump”

Simple stoichiometry
 In the reaction: N2 + 3 H2 2
NH3, what is the relationship between
the rate of production of NH3 and the
rate of consumption of H2?
 In any given amount of time, twice as
much NH3 is created as nitrogen gas
disappears. Its RATE of generation is
twice as great. Its rate of generation
is 2/3 that of hydrogen’s
disappearance.

Chapter objectives (lots!)
 Identify factors that influence reaction rates
DONE
creation of a product from stoichiometry DONE
data
 Determine the activation energy for a reaction

Determine the rate law of a
reaction
First,
Notice that Zumdahl
only selects examples
of reactions that have
a good reason not to
run both ways.
Usually, it is that one
of the products is a
gas that escapes.
Thus, only reactants
are considered in
elementary kinetics.
Next,
Notice that we are
beginning with
decomposition
reactions, in which
there is only one
reactant. This means
we are only looking at
the rate of change of
one substance – the
product concentrations
are stoichiometrically
dependent on the
reactant anyway.

2NO2(g)  2NO(g) + O2(g)
 See page 551 for table of
concentrations as a function of time
 The rate of change is going to be
defined as the change in
concentration over the elapsed time.
The instantaneous rate is graphically
the slope of the line tangent to the
curve. You will NOT have to calculate
the slope of that tangent yourself
when the graph is a curve.

“WRITE THE RATE LAW FOR THIS REACTION”
RATE = K[NO2]N
For the reaction
2NO2(g)  2NO(g) + O2(g)
 This k is NOT the same k as we used in
equilibrium. Sorry!
 N is an exponent – this tells us whether the
disappearance of NO2 is directly related to
concentration (n=1), exponentially related
(n=2) or not based on concentration at all
(n=0)
 The concentrations of the products are not
included because the reverse rate of the
reaction is not significant in this case – in
fact, you will never need to consider an
equilibrium situation for these problems.

2N2O5 (aq)  4NO2(aq) + O2(g)
 Notice in this
problem, the oxygen
leaves – no worries
about the reverse
reaction
 On page 557, Z has
cut to the chase – the
concentration of N2O5
and the instantaneous
rate are presented
together
 As the concentration
doubles, the
instantaneous rate
doubles. No more
math required; you
know that the value of
the exponent is 1
 rate = k[N2O5]n
 Now plug in n=1, any
concentration of
N2O5, and solve for k
 You can now find the
rate at any
concentration of N2O5

Sneaky sneaky Zumdahl…
 Notice that on page 552, you get a big
gorgeous graph of the concentrations of all
the substances in this reaction as a function
of time. The tangents at specific points are
calculated for you
 In a tiny table on page 553, you get the
rate/time data, but even that is not
instantaneous.
 If you graphed the change of the
instantaneous velocities, you would get a
straight line – this means it is a first order
reaction, although Zumdahl never tells you
that.
 In fact, he moves to a different equation

Chapter objectives (making
progress!) Identify factors that influence reaction rates
DONE
 Write the rate law for a reaction DONE
 Determine the rate law of a reaction from
dataDONE
data

Reactions with two reactants
 NH4
+
(aq) + NO2
-
(aq)  N2(g) + 2H2O(l)
 Notice that the N2(g) leaves.
 Z provides initial instantaneous rates
from 3 different experiments on page
559.
 Spend a minute reading the table –
see if you can reach any conclusions
without a lot of math about the
relationships between concentration
and rate for each reactant.

Need a hand?
 Look at the initial concentration of
ammonium ions. Do you see how
experiment 2 and 3 are related?
 Notice that the NO2- does not change
between experiment 2 and 3. This
means the change in the rate is
exclusively dependent on the ammonium
concentrations
 Notice the difference between
concentrations in experiments 1 and 2.
 Can you establish the exponent for each
reactant in the rate law now?

rate = k[NH4
+ ]n[NO2
-]m
 The rate doubles as the concentration
of NH4
+ doubles. Therefore, n=1
 The rate doubles as the concentration
of NO2
- doubles. Therefore, m=1
 Now you can plug in any of the values
for one experiment and calculate k.
 Calculating k from different initial
concentrations should NOT yield
different k values.

WHAT DOES IT MEAN
WHEN THEY SAY….?
Dealing with the words
rate = k[NH4
+ ]n[NO2
-]m
data
rate = k[NH4
+ ]1[NO2
-]1
 Determine the order of a reaction from
time/rate data
1 + 1 = 2 so the overall order of a reaction is 2
 Determine the instantaneous rate of a
reaction
 Solve for the rate constant from time/rate

WHAT DOES IT MEAN
WHEN THEY SAY….?
Dealing with the words
rate = k[NH4
+ ]n[NO2
-]m
rate = k[NH4
+ ]1[NO2
-]1
 Determine the order of a reaction from time/rate data
1 + 1 = 2 so the overall order of a reaction is 2
 Solve for the rate constant from time/rate
data
 k = rate/ [NH4
+ ][NO2
-]
 Determine the instantaneous rate of a
reaction
◦ Plug in your new k into the reaction
◦ rate = k[NH4
+ ] [NO2
-] for any concentrations of the
reactants

progress!) Identify factors that influence reaction rates DONE
dataDONE
data DONE
DONE
DONE
 Create graphs to determine order of a
reaction

HOMEWORK
 End of chapter 12, numbers
21, 24, 25, 29, 30, 36, 39, 40, 46, 66
 Due next class – Friday April 26
 Also, read to the end of chapter
12, and come prepared to ask
questions or to solve problems!

Which data am I being given?
Initial concentrations and
rates
Concentrations and
time
 In this case, look for
patterns that will tell you
if the exponent in the rate
law is a zero, 1, or 2
 No additional calculated
values are necessary to
find the rate order
 In this case, add 2 more
columns of data
 Include 1/[A] and ln[A]
columns of data when
presented with
concentration and time

2C4H6C8H12
Given
[C4H6] time
Calculated
Time 1/ [C4H6] ln[C4H6]
 .01000 0
 .00625 1000
 .00476 1800
 .00370 2800
 .00313 3600
 .00270 4400
 .00241 5200
 .00208 6200
 0 100 -4.605
 1000 160 -5.075
 1800 210 -5.348
 2800 270 -5.599
 3600 320 -5.767
 4400 370 -5.915
 5200 415 -6.028
 6200 481 -6.175

Using the inverse and the ln
 A graph of [A] over time will be a
straight line for zero order reactions.
 One other data set you will need is the
natural log of the concentration.
 A graph of ln[A] over time will be a
straight line for first order reactions.
 The third data set you will need is the
reciprocal of the concentration.
 A graph of 1/[A] vs. time will be a
straight line for second order
reactions.

Graph each data set with time on
the x axis
 The graph of [C4H6] is not a straight line.
 This is not a zero order reaction.
 The graph of the ln [C4H6] is not a
straight line. The reaction is not first
order for [C4H6]
 REJECT the data set of ln [C4H6].
 The graph of 1/[C4H6] is a straight line.
Therefore, the reaction is second order
for [C4H6].

Finding k
 Rate = k[C4H6]2
 The k is actually the slope of the
graph. To find slope, calculate y/ x
from your graph.
 Alternately, pick two values off your
data table to calculate the slope from.
 Visually, I used (0, 100) and
(5000, 400) and got k=.06
 Z did it with the data points and got 6.14 e -
2

 Calculate rate of consumption of a reactant or of creation of a product from
stoichiometry DONE
 Determine the rate law of a reaction from dataDONE
 Determine the order of a reaction from time/rate data DONE
 Solve for the rate constant from time/rate data DONE
 Determine the instantaneous rate of a reaction DONE
reaction DONE

The integrated rate law
 Take a look at page 562, equation 12.2
and at the AP “cheat sheet” page
ii, kinetics section.
 Note that the form of the integrated rate
law on the cheat sheet is somewhat
different than in Z. It is only marginally
different.
 This form of the integrated rate law is
only applied to first order reactions. If
you use it to work with zero or 2nd order
reactions, you will mess up!! The
lesson? Establish the order before you

Why is Zumdahl right this
time?
 Zumdahl has organized his version of
the integrated rate law into y=mx+b
format.
 Y= ln[A]
 M = -k
 X = time
 B = ln[A]o
 THAT is cool!

Integrated rate law for second
order reactions
 Page 567, eq’n 12.5
 Again, Z uses y=mx+b format, and the
AP cheat sheet does not.

 Calculate rate of consumption of a reactant or of creation of a product from
stoichiometry DONE
 Determine the rate law of a reaction from dataDONE
 Determine the order of a reaction from time/rate data DONE
 Solve for the rate constant from time/rate data DONE
 Determine the instantaneous rate of a reaction DONE
reaction DONE
concentration at time t DONE

Half life – not as simple as you
thought!
 The traditional concept of half life – the
amount of time it takes for half of a
sample to convert to some other form –
is based on first order reactions only
 In first order reactions, half life follows a
simple equation
 T1/2= .693/k
 This is NOT on the AP cheat sheet!
 This equation was derived on page 565
using the integrated rate law for first
order reactions. Follow the logic, and
you can reconstruct it if you need it.

Half life problem – first order
 See Z page 566
 Solve for k
 Solve for time to a particular
concentration
 Solve for a concentration at a given
time
 Plug and chug… not difficult, except
that you must use the right equations!

What you should notice about
that half life equation
 This is NOT dependent on
concentration! To calculate specific
concentrations at time t, you will need
the initial concentration, but you do not
need it to calculate k or t1/2

Half life of a second order
reaction
 t1/2 = 1/k[A]0
 This is specific to second order
reactions. It is not on the AP cheat
sheet, but can be derived from the
integrated rate law for second order
reactions.
 This IS dependent on initial
concentration.
 Unlike simple first order half life, second
order half lives do not mean half the
material changes in a set amount of
time. For these reactions, each half life
takes twice as long.

So to quote
Zumdahl, “Spontaneous does
not mean fast.”
 But time IS something that we can use
to learn more about a reaction, what
its mechanism is, and how it is likely to
proceed. We can calculate with it.
We can graph stuff.

Intro to kinetics

Recommandé

Recommandé

Contenu connexe

Tendances

Tendances (20)

En vedette

En vedette (7)

Similaire à Intro to kinetics

Similaire à Intro to kinetics (20)

Plus de lallen

Plus de lallen (12)

Dernier

Dernier (20)

Intro to kinetics