Recap

What is the order of reaction with respect to NO2(g)
and F2(g) given the following rate data at a certain
temperature?
[NO2(g)] / mol
dm–3
0.1
[F2(g)] / mol
dm–3
0.2
Rate / mol dm–3
min–1
0.1
0.2
0.2
0.4
0.1
0.4
0.2
Order with respect to NO2(g)
Order with respect to F2(g)
A.
first
first
B.
first
second
C.
second
first
D.
second
second
Lesson 7
HL Only
The Reaction Mechanism
Reaction Mechanisms Explained…with ping
pong balls

The task:

As a class seated in a big circle, you need to see how quickly
you can pass 10 ping pong balls around the circle.



Everyone starts with a spoon and then…
One of you must use THE SPATULA OF DOOM
The rules:


If the person you are trying to pass to already has a ball, hold
on to it until they have given it away
Each person can only have one ball at a time
Reflecting on the silly ping-pong balls thing

What were the main factors that influenced how fast the balls
could be passed around?

Was each step in the reaction the same speed?

This is a polite way of asking if someone found this activity
‘challenging’

Picture a million ping-pong balls being passed around the circle,
which person would have the biggest impact on how long it
took to pass them all around.

How much can the people present after the slow one
influence the rate?
Eating Elephants…

How do you eat an
elephant?

One mouthful at a time!

Chemical reactions are
similar.
Reaction Mechanisms
Example 1

Most reactions happen by a series of small steps, and this series is called the
‘reaction mechanism’

The individual steps can take place at very different rates.

For example:
NO2(g) + CO(g)  NO(g) + CO2(g)


The steps in the mechanism are:

Step 1: 2NO2(g)  NO3(g) + NO(g)
(slow)

Step 2: NO3(g) + CO(g)  NO2(g) + CO2(g)
(fast)
Note:


NO3 is an intermediate something that is made in one step and used up in another
The left at right sides of the mechanism cancel out to give you the overall equation.
A note on molecularity…

Molecularity describes the number of particles involved
in a single step:

Unimolecular: one molecule involved:
A  B +C

Bimolecular: two molecules involved
A+A  C
A+B  C

Termolecular: three molecules involved (extremely
rare, you will not encounter these at IB level).
Mechanisms and Rates
Example 1

Looking at our previous example


NO2(g) + CO(g)  NO(g) + CO2(g)

Step 1: 2NO2(g)  NO3(g) + NO(g)
(slow)

Step 2: NO3(g) + CO(g)  NO2(g) + CO2(g) (fast)
If you think about it….

Changing the concentration of CO will not affect the rate


Because it is involved in a fast step after the RDS, and so the only thing
relevant to this step is how quickly the NO3 can be made, and this is
made by a step which is very slow.
Changing the concentration of NO2 will affect the rate


Due to it being involved in the slow step
Since it appears twice in the slow step, changing it’s concentration will
have double the impact
Mechanisms and Rates
Example 2


2NO(g) + O2(g)  2NO2(g)

Step 1: 2NO(g)  N2O2(g)
(fast)

Step 2: N2O2(g) +O2(g)  2NO2(g)
(slow)
In this reaction:

Changing the concentration of O2 will affect the rate


Changing the concentration of NO will affect the rate


Because it is involved in the slow step.
Because it will affect the rate at which N2O2 is made, and this is needed in the slow
step.
In summary:



Rate is affected by everything up to and including the slow step
Rate is not affect by anything after the slow step
For this reason, the slow step is called the rate determining step.
How can we determine the reaction mechanism?
Example 1: NO2(g) + CO(g)  NO(g) + CO2(g)


First we identify a series of possible reaction mechanisms:
Possibility 1:
Possibility 2:
2NO2(g)  NO3(g) + NO(g)
NO2  NO + O
NO3(g) + CO(g)  NO(g) + CO2(g)
CO + O  CO2
Depending on which is RDS either:
•Rate = k[NO2]2
•Rate = k[NO2]2[CO]
Depending on RDS either:
•Rate = k[NO2]
•Rate = k[NO2][CO]
RDS most likely to be step 1.
RDS most likely to be step 1.
From experimental data we find that the rate law is Rate = k[NO2]2


This tells us that the reactions at/before the RDS involve two NO2 or things that can
be made from it.
The only possibility that fits this is the first, with the first step as the RDS.
Key Point

The order of reaction with respect to a reactant tells you
the number of molecules of the reactant involved up to
and including the rate determining step.
How can we determine the reaction mechanism?
Example 2: 2NO(g) + O2(g)  2NO2(g)

First we identify a series of possible reaction mechanisms:
Possibility 1:
Possibility 2:
Possibility 3:
NO + O2  NO2 + O
NO + O  NO2
2NO  N2O2
N2O2 + O2  2NO2
NO  N + O
NO + O  NO2
N + O2  NO2
Step 1 would be slow so: Depending on RDS:
•Rate = k[NO][O2]
•Rate = k[NO]2
•Rate = k[NO]2[O2]

Step 1 likely to be
slowest so:
•Rate = k[NO]
From experimental data we find that the rate law is Rate = k[NO]2[O2]


This tells us that the reactions at/before the RDS only at least two NO and one
O2.
The only possibility that fits this is the second, with the second step as the slow
one
Review

Rate equations are determined by the mechanism of a
reaction

We can use the rate equation to help us choose from
possible candidate mechanisms

The order of the reaction with respect to each reactant
tells you the number of times they are involved in the
mechanism up to and including the RDS.