Chapter 17 Chemical Kinetics
Last semester we studied different kinds of reactions in chapter 10, and then how much
heat is given off or absorbed in chapter 14. In this chapter we will study how fast a
reaction occurs. Some reactions are fast, the precipitation reactions you have observed
in the lab that seem almost instantaneous, and some reactions are S-L-O-W, like rust
forming on a piece of iron.
In this chapter we will emphasize the basics of reaction kinetics. How do we come up
with a number that describes a reaction rate, What factors make a reaction go faster or
slower, and find equations that will allow us to determine how much reactant or product
is left after a given time. Then in the next chapter (18) I will show you what these
numbers mean on the molecular level, and we will use the parameters with uncover in
this chapter to pick apart a chemical reaction to propose a step by step model of how a
chemical reaction actually works
17-1 Reaction Rates
Before we get to chemistry, let’s start with the everyday. We want to measure
how fast a reaction goes.
How do you measure how fast something goes, say a car?
You measure a speed, or more correctly a velocity.
How do you measure the speed of a car? Mph miles/hour or distance/time
Have a car here, start the stopwatch, peel rubber, and measure how far it got in
so many seconds so the velocity = Ädistance/Ätime
How would you measure the speed of a chemical reaction?
Put two solutions together of known concentration, start a stop watch, and some
time later measure how much any of the concentrations changed with time
velocity =Ä concentration /Ä time
Let’s start with the reaction
CH3-N=N-CH3 (g) 6 CH3-CH3 (g) + N2(g)
We could measure the appearance of either products, or the disappearance of
the reactant several different ways that I won’t get into here. The result is that
we could come up with data like that shown in Figure 17.1 or Table 17.1
Let’s focus on the appearance of product
1st notice in figure 17.1 the curves for [C2H6] = [N2]
Can anybody tell me why? (1:1 stoichiometry so have to be equal)
Next notice the line is curved. If we want the rate of the reaction at 90 minutes
we get a different rate if we use a different Ä around 90, as shown in figure 17.2
and your book calculates using data in table 17.1
2
As the Ä decreases the line that Ä[]/Ät describes, get closer and closer to the
slope of the line that is tangent to the curve. Now if you have had calculus you
already know that we are talking about the slope of the line, d[]/dt. But that is
calculus, and I don’t want you to worry about calculus
Bottom line for the rest of this chapter I will refer to the rate of a reaction as
Ä[]/Ät, but in reality what I am really thinking about in my head is the slope of the
line( d[]/dt for the math purists)
Now if you look back at figure 1 you should see that the curve fo the appearance
of product is + and the curve for the disappearance of reactant is -. This will
never do. A simple minded chemist cannot deal with both positive and negative
reaction rates so:
Key convention:
rate of reaction = + Ä[reactant]/Ät = - Äproduct/Ät, so you always have + rates,
whether you measure product or reactants
In this reaction we had nice 1:1:1 stochiometery. What happens if we mix it up
with a reaction like: 2NO2(g) 6 O2(g) + 2NO(g)
Here you can see we would get two different rates, and the rate of O2 production
would be 1/2 the rate of NO production because you make 2 moles of NO for
every mole of O2
Mathematically
Ä[O2]/Ät = ½ (Ä[NO]/Ät) = -½ (Ä[NO2]/Ät
So sign makes sure you get the same + rate for both products and reactants,
and the numerical factor used the stoichiometry to make sure the rates are the
same no matter which reactant or product is used
Let’s generalize this:
Key convention:
For the reaction aA + bB 6cC + dD
3
Practice problems
For the reaction 2N2O5(g) 6 4NO(g) + O2(g)
If the rate of the reaction is .45 moles/second, what is the rate of N2O5
consumption, NO production and O2 production?
Rate = .45 mole/second = -1/2 (ÄN2O5/Ät) ; (ÄN2O5/Ät) = -2(.45) = -.9 mol/sec
so N2O5 will disappear at a rate of 0.9 mol/sec
Rate = .45 mole/second = 1/4 (ÄNO/Ät) ; (ÄNO/Ät) = 4(.45) = 1.8 mol/sec
so NO will appear at a rate of 1.8 mol/sec
Rate = .45 mole/second = 1/1 (ÄO2 /Ät) ; (ÄO2/Ät) = (.45) = .45 mol/sec
so O2 will appear at a rate of .45 mol/sec
17-2 Rate and time
The rate of a reaction will vary with time. Look at Figure 17.3
If the rate of a reaction is the slope of the line, see how the slope varies from
point to point?
Now why do they vary? [concentration of reactant gets lower or concentration of
product get larger. Will study the effect of concentration in the next section]
Even though the rate changes at different time points, if we know the
concentration change of any reactant or product in a srt time period, we can
calculate the concentration of any other reactant or product over the same time
period
Example Problem 1
Given the reaction: 2N2O5(g) 6 4NO(g) + O2(g)
If the [N2O5]0 is 1.24x10-2 M and .23x10-2 M at t=55.0 minutes, what is the
concentration of [NO] at 55 minutes, given that it was 0 at t=0
Amount of N2O5 that reacted = 1.24x10-2 - .23x10-2 = 1.01x10-2M
Using stoichiometery
1.01x10-2 M X 4 NO = 2.02x10-2 M NO at 55 minutes
2N2O5
4
Example Problem 2
Given the reaction: 2N2O5(g) 6 4NO(g) + O2(g)
If [O2] is initially 0, and it .58x10-2 at 90 minutes, calculate the [N2O5] at 90
minutes, If the initial concentration was 1.24x10-2
The O2 concentration at 90 minutes is .58x10-2
Using stoichiometery
0.58x10-2 M X 2 N2O5 =1.16M N2O5
1O2
Since O2 is a product and N2O5 is a reactant, the [N2O5] must have decreased by
1.16x10-2 M at 90 minutes
If N2O5 was initially 1.24x10-2, it must now be 1.24x10-2-1.16x10-2 or .80x10-2 M
17-3 Initial Rates
In the last section I emphasized how the rate of a reaction changes with time,
and I tried to link that change to the change in the concentration of the reactants
and products. In this section we will nail down the relationship between reaction
rates and concentration
Many experiments have shown that the rate of a reaction is proportional to the
concentrations of the reactants, raised to small integer powers. For instance, in
the reaction 2N2O5(g) 6 4NO(g) + O2(g), many different kinetic experiments
would all indicate that the rate of the reaction %[N2O5]x
making this proportionality an equation
rate of reaction = k[N2O5]x
Key definitions/concepts:
The mathematical equation that relates the rate of a reaction to the
concentrations of the reactants is called the rate law of the reaction
In the rate low expression; rate = k[ ]x, k the proportionality constant is call the
rate constant, and x, the exponent to which the concentration is raised, is call
the order parameter
Both the rate constant and the order parameters must de derived from
experimental data. There is no relationship between the order parameter and the
stoichiometric coefficients in a balanced equation. (But you may see a cases that
make you think there is, so be careful!)
5
If the rate of a reaction is going to vary with time due to the changes that occur in
reactant and product concentrations, what point do we use to get an
experimental rate?
Key Concept
In the method of initial rates, we will study the rate of the reaction as a reaction
starts. At this time, the concentration of the product will be negligible, so no back
reaction can occur, and we will use a small time interval, so the concentrations of
the reactants do not change significantly.
Sounds great, but what are the nuts and bolts of the procedure?
Step 1. Write the rate law for the reaction
Say we want to determine the kinetics of the reaction 2N2O5(g) 6 4NO(g) + O2(g)
We have seen that the rate of the reaction is proportional to the reactants
concentrations raised to some power we write:
rate = k[N2O5]x
Since we only have one reactant, we move on to step 2
Step 2. Determine the initial rate of the reaction using different starting conditions
where the concentration of one reactant is changed
So lets’ do three experiments where we change the initial concentration of N2O5
Run
1
2
3
[N2O5] (M)
.01
.02
.035
Rate (mol/L@h)
.018
.036
.063
If you are clever, you can see that as we doubled the concentration the rate
doubled, so the order exponent mut be 1 thus
rate = k[N2O5]1
Key Concept:
When a reactant has an order parameter of 1, the reaction is said to be first
order with respect to that reactant.
In the above example, we found the order parameter using simple inspection.
That doesn’t always work, especially on a test, where I try to make things a little
less obvious. So what do you do when simple inspection doesn’t work?
You determine the order parameter by dividing one rate equation by another
6
Let’s try that with the data we have for runs 1 and 3
Run [N2O5] (M)
Rate (mol/L@h)
Rate equation
1
.01
.018
= k[N2O5]x
2
.02
.036
= k[N2O5]x
3
.035
.063
= k[N2O5]x
When we divide equations I always divide the equation with the larger rate by the
equation with the smaller rate:
In this case again you can see that x=1, but how would you prove that
mathematically? You would take the log (or ln) of both sides of the equation
log(3.5) = x log(3.5)
.544 = x(.544)
.544/.544 = x = 1
But wait, we aren’t done. We don’t have k. How do you get k? You choose any
of your sets of conditions into the rate equation and solve for k
Using data set and EQN 3
.063 M/hr = k(.035 M)1
.063 M/hr/.035M = k = 1.8 hr-1
So our final rate equation for this reaction is:
rate = 1.8 hr-1[N2O5]1
What happens when you have 2 reactants?
You do the same thing, but you need to have 3 sets of data, 2 in which only one
concentration is varied, and a third where only the other concentration is varied
7
Example:
2NO(g) + Cl2(g) 6 2NOCl(g)
[NO](mol/L)
.1
.1
.2
[Cl2](mol/l)
.1
.2
.2
Rate(mol/l@min)
.18
.36
2.02
rate = k [NO]x [Cl2]y
Using the first set of 2 rates
.18 = k(.1)x (.1)y
.36 = k(.1)x (.2)y
Divide larger by smaller
Here we have found an order parameter of 2
Key Concept:
When a reactant has an order parameter of 2, the reaction is said to be second
order with respect to that reactant.
Let’s continue with this example
Using the next set
.36 = k(.1)x (.2)y
2.02 = k(.2)x (.2)y
Divide the larger by the smaller
8
Note our order parameter does not always have to be an integer!! Especially with
real data!
So here we have rate = k[NO]2.5[Cl2]1
Two final notes before we leave this section
Note 1. Sometimes (Example 17-5 from text) you will find that the order
parameter for a reactant is 0
Key Concept:
When a reactant has an order parameter of 0, the reaction is said to be zero
order with respect to that reactant. A 0 order parameter means that the reaction
rate does not depend on the concentration of this particular reactant under the
set of condition you have run your experiment. I will explain what that means in
the next chapter.
Note 2. We have talked about a reaction being 0,1st or 2nd order with respect to a
particular reactant. We also talk about the overall order of a reaction, which is
the sum of all the individual order parameters. For instance
if rate = k[A]2[B]1 we would say that this reaction is 2nd order with respect to A, 1st
order with respect to B and third order overall
Key Concept:
The overall order of a reaction is the sum of the individual order parameters for
each reactant.
Clicker Question:
Given that the rate law for a chemical reaction is rate = k[A]o[B]1
What is the overall order of this reaction, and what is the order with respect to A
and B?
9
17-4 First order reactions
In the method of initial rates we set up an experiment like Figure 17.3, but we
only look at the initial slope. We then change one of the concentrations and try
again. This seems a bit wasteful, setting up a big experiment then only looking
at the first few seconds of data. There is a second way to do kinetics that looks
at the data over the entire time curve rather than just looking at the initial slope
If we have a reaction
A6B
and it is first order with respect to A then
rate = -Ä[A]/Ät = k[A]
Using elementary calculus you can integrate the above expression to derive the
following key equation:
Integrated equation for a 1st order reaction
ln[A] = ln[A]0 -kt
Where [A] is the concentration of A at time t
And [A]0 is the initial concentration of A
You can use the basic properties of logs to come up with the following equations
ln([A]/A]0 = -kt ; ([A]/[A]0) = e-kt ;
[A] = [A]o e-kt
All of these equation are equivalent and can be used in various situations. But I
will concentrate on the first equation I gave you
Practice problem:
If the reaction
2N2O5(g) 64NO2(g) + O2(g) is first order with respect to N2O5, and it has a rate
constant of 1.8 hr-1 at 45oC. What is the concentration of N2O5 at 50 minutes, if
the initial concentration is 1.24x10-2 M?
ln[A] = ln[A]0 -kt
Since k is in hr-1 we need to convert 50 minutes to hours
50 min x (1 hr /60 minutes) = .833 hr
Plug and chug
ln(X) = ln(1.24x10-2) - (1.8 hr-1 x .833 hr)
Ln(X) = -4.39 - 1.50
ln(X) = -5.89
X = e-5.89
X = 2.77x10-3 M
10
Graphic analysis
The equation
ln[A] = ln[A]0 -kt
Also lends itself to graphic analysis
Let’s rearange it a bit first ln[A] = -kt + ln[A]0
Remember the equation for a line? Y=mX +b?
If ln[A] = Y, and ln[A]0 = b, then you will have a straight line with slope = -k
Figure 17.4 and 17.5 on same figure?
Key Concept
Using the first order integrated rate expression - Instead of doing several
experiments and determining the initial slope of each experiment, we do 1
experiment, and let it run for a while. Then plot the ln[] vs t. If it is linear, then we
had a first order reaction and k = -slope
17-5 Half-life of First order reactions
The numerical value of the rate constant k is a measure of how fast a reaction
occurs, large k’s = fast rates
We have another way to measure how fast a reaction occurs, and that is the
time it takes for ½ of the reactant to disappear, called t1/2 or half-life
Key concept
half life = t1/2 = time it takes for ½ of the starting reactant to be consumed
The half-life of a first order reaction has a special property, it is independent of
the starting concentration
Figure 17.6
As you can see in this figure, even though the starting concentration changes
you get the same, constant t1/2
We can prove this is true with our integrated rate expression
ln[A] = ln[A]0 -kt
If t = t1/2 then [A] = [A]0 /2 so:
ln([A]0 /2) = ln[A]0 -kt1/2
ln[A]0 - ln(2) = ln[A]0 -kt1/2
-ln(2) = -kt1/2
ln(2) = kt1/2
.693 =k t1/2
11
Key equation
t1/2 for a first order reaction = .693/k
Practice problem 1
If a reaction has a t½ life of 3 minutes, what is the rate constant of the reaction?
3=0.693/k;
k = .693/3;
k=.231 min-1
Practice Problem 2
How long will it take for the reaction to reach 95% completion?
The integrated rate equation
K=.231 min-1 (from above)
[A]o is X
[A] = .05(X) (95 % complete means only 5% is left)
Ln(.05X) = -.231(t) + ln(X)
ln(.05X)-ln(X) = -.231(t)
ln(.05X)-ln(X) = ln(.05X/X) = ln(.05) = -3;
-3 = -.231t
t= -3/-.231 = 12.98 min
17-6 Radioactive decay & 17-7 C-14 Dating
While these two chapters are interesting, and I encourage you to read them and
ask me if you have any questions, I would like to push on with the chemistry and
will skip lecturing (and testing) over this material
17-8 Second order Kinetics
We can use integral calculus to find an equation for second order reactions as
well as first order
A6B
and it is second order with respect to A then
rate = -Ä[A]/Ät = k[A]2
Using elementary calculus you can integrate the above expression to derive
Key equation:
Integrated equation for a 2nd order reaction
1/[A] = 1/[A]0 + kt
And we plot 1/[A] vs t, if we get a liner plot then the reaction is second order, and
the slope = k
12
Example problem
Table 17.7, plots of [] vs time. ; ln[] vs time; 1/[] vs time
Using the data in table 17.7 determine the order of the reaction
NO2(g) + CO(g) 6 NO(g) + CO2(g)
Note plots of [] vs time and ln[] vs time are not linear
only plot of 1/[] vs time is linear, so is second order
The trendline (from excel) is Y = .4978X + 10.038
Slope = k = .4978 M-1s-1
17-9 Half Life of Second Order reactions
Figure 17.14
Notice in this figure that the t1/2 changes with concentration. The lower the
concentration the longer t1/2
using the integrated expression
1/[A] = 1/[A]0 + kt
1/([A]0/2) = 1/[A]0 + kt1/2
2/[A]0 = 1/[A]0 + kt1/2
2/[A]0 - 1/[A]0 = kt1/2
(2-1)/ [A]0 =kt1/2
1/[A]0 =kt1/2
1/([A]0 x k)=t1/2
Key Equation:
For a second order equation: t1/2 = 1/k[A]0
Example problem
if the reaction
NO2(g) + CO(g) 6 NO(g) + CO2(g)
is second order, and the concentration of NO2 is .08M how long will it take for the
concentration to drop to .04M .02 M?
From previous work the k for this reaction = k = .4978 M-1s-1
If [A]0 = .08 then
1st t1/2 = 1/(.08x.4978) = 25.1 sec
2nd t1/2= 1/(.04x.4978) = 50.2 sec
We should finish this chapter with Table 17.14, summarizing the 1st and second
order kinetics
I would like to add to this table an entry for zero order kinetics that was
introduces but no pursued in this chapter
13
Order Rate Law
0
rate =k
Units of k
M@s-1
Dependence on time
[A]=-kt + [A]0
t1/2
Test plot
[A]0/2k
[A] vs t