HL - Rate-determining step
Do now:
What is the ratedetermining step? What
E = Activation energy Jmol
R = gasfactors
constant (8.3 JKare
)
involved in
this?
How can we tell/find this
out?
-1
-1
Rate-determining step
Most reactions don’t take place in one step, the separate
steps which lead from the reactants to the products is
called a reaction mechanism.
In a reaction mechanism, the steps
nearly always follow on from each
E = Activation energy Jmol
other, where the products of one
R = gas constant (8.3 JK )
step is the starting material for the
next step.
-1
-1
What do you think is meant by the
term rate determining step?
Rate-determining step
The rate determining step is the slowest step in the
reaction mechanism. It governs the rate of the overall
reaction.
Knowing which is the rate determining step, we are able
to predict something about the mechanism of a reaction
E =from
Activation the
energy Jmol
rate equation.
-1
R = gas constant (8.3 JK-1)
Species found in the rate equation must be involved in the
rate determining step.
Rate-determining step
Look at the reaction below:
This reaction actually takes place in a 2 step mechanism:
E = Activation energy Jmol-1
R = gas constant (8.3 JK-1)
Which is the rate determining step?
What is X?
Rate-determining step
The slowest step is the rate determining step, which produces ‘X’,
this doesn’t appear in the overall equation as it is an
intermediate.
E = Activation energy Jmol-1
A and C react slowly to produce X, as soon as X is produced D
and E are formed.
Only species involved in the rate determining step, or steps
before it are involved in the rate expression.
R = gas constant (8.3 JK-1)
What species are found in the rate expression?
[A] and [C]
With your
partner, discuss
and write down
a solution for
this question.
3 mins.
With your partner,
discuss and write
down a solution
for this question.
2 mins.
With your
partner, discuss
and write down
a solution for
this question.
3 mins.
Rate-determining step
NO2(g) + CO(g)
NO(g) + CO2(g)
The overall balanced equation shows us that:
• 1 mol NO2(g) reacts with 1 mol CO(g) to produce 1 mol NO(g) and 1 mol
CO2(g)
The overall equation does not tell you anything about the mechanism.
To do this we need to carry out some rate experiments.
E = Activation energy Jmol-1
R = gas constant (8.3 JK-1)
If these experiments are carried out the rate equation is found to be:
rate = k[NO2]2
What are the orders of reaction with respect to the two reactants?
[NO2] = second order
[CO] = zero order
Rate-determining step
NO2(g) + CO(g)
•
NO(g) + CO2(g)
If a reactant appears in the rate equation then it must participate in the
rate determining step.
• The order with respect to that reactant tells you how many particles of
that reactant are involved in the rate determining step.
E = Activation energy Jmol-1
R = gas constant (8.3 JK-1)
NO2 + NO2 → slow, rate determining step.
Rate-determining step
• The rate determining step must be followed by further, faster, steps.
• Together, the sum of all the steps adds up to give the overall equation.
NO2(g) + CO(g)
NO(g) + CO2(g)
We can propose a two step mechanism for this reaction. First, summarise what
we know so far.
E = Activation energy Jmol-1
R =stgas constant (8.3 JK-1)
1 Step
NO2 + NO2
2nd Step
Overall equation NO2 + CO
NO + CO2
• CO and CO2 must be in the second step as they are in the overall equation
• NO2 must be a product of the second step as only one molecule of it is in
the overall reaction
Worked example
1st Step
NO2 + NO2
Slow (RDS)
2nd Step
+CO
Overall equation NO2 + CO
NO2 + CO2
NO + CO2
Fast
• The other reactant in the second step must be NO3
• So the products of the first step must be NO and NO3
So the combined mechanism is:
E = Activation energy Jmol-1
R = gas constant (8.3 JK-1)
1st Step
2nd Step
NO2 + NO2
NO3 + CO
NO + NO3 Slow (RDS)
NO2 + CO2
Fast
Overall equation NO2 + CO
NO + CO2
NO3 is an intermediate is it does not appear in the overall equation.
It is generated and consumed during the mechanism.
Your turn
Answers
With your partner, discuss
and write down a solution
for this question.
3 mins.
Question
4
1. For each scheme, identify the
species that are involved in the
rate expression?
2. Which species can you
identify as intermediates for
Scheme 3?
3. Form a rate equation for the
following rxn for Scheme 3.
Assess the following answer:
1. Scheme 1 = A and B,
Scheme 2 = A and E,
Scheme 3 = A and B
1. C, D, F
2. Rate = k [A] [B]
Mark scheme for the extension question
HL - SN1 and SN2 Reactions
Do now:
What is a
nucleophile?
What happens in
nucleophilic
substitution?
E = Activation energy Jmol-1
R = gas constant (8.3 JK-1)
SN1 + SN2 Reactions
S = Substitution
N = Nucleophilic
1 = Unimolecular
SN2
S = Substitution
N = Nucleophilic
2 = Bimolecular
SN1 + SN2 Reactions
1. The Slow Step:
First step of the SN1 reaction:
The leaving group leaves, and the substrate carbon now only
has three substituents, taking on a positive charge. This is called
a carbocation.
● Carbocations are most stable when there are more atoms
to distribute this positive charge.
● Carbocation stability:
3º > 2º >> 1º
● Tip: study the difference between reaction intermediates
and transition state
SN1 + SN2 Reactions
2. The Fast Step:
Second step of the SN1 reaction:
The nucleophile attacks the carbocation intermediate,
bringing its electron pair to resolve the positive charge.
The substrate loses any stereospecificity during the
carbocation inter
Comparison between SN1 + SN2
Reactions
Reaction Parameter
SN2
SN1
alkyl halide structure
methyl > primary >
secondary >>>> tertiary
tertiary > secondary >>>> primary
> methyl
nucleophile
high concentration of a
strong nucleophile
poor nucleophile (often the
solvent)
mechanism
1-step
2-step
rate limiting step
bimolecular transition state
carbocation formation
rate law
rate = k[R-X][Nu]
rate = k[R-X]
stereochemistry
inversion of configuration
mixed configuration
solvent
polar aprotic
polar protic