Title: Lesson 5 Reaction Mechanisms

Learning Objectives:

– Understand what a reaction mechanism is

– Understand the relationship between rate equations and reaction mechanisms

• This will explain how things can be ‘zero’ order

– Learn to identify possible reaction mechanisms from suitable rate data.

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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

[F2(g)] / mol dm–3

Rate / mol dm–3 min–1

0.1 0.2 0.10.2 0.2 0.40.1 0.4 0.2

  Order with respect to NO2(g)

Order with respect to F2(g)

A. first firstB. first secondC. second firstD. second second

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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. Most of you can use your hands 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

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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?

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Eating Elephants… How do you eat an

elephant?

One mouthful at a time!

Chemical reactions are similar.

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Reaction Mechanisms - Example 1 Most reactions happen by a series of small steps, and this series is called the

‘reaction mechanism’

The individual steps (elementary steps) can take place at very different rates and are not usually observed directly.

For example:NO2(g) + CO(g) NO(g) + CO2(g)

The (theoretical) steps in the mechanism are:

Step 1: NO2 + NO2(g) NO(g) + NO3(g) (slow)

Step 2: NO3(g) + CO(g) NO2(g) + CO2(g) (fast)

Overall reaction: NO2(g) + CO(g) NO(g) + CO2(g)

Note: NO3 is an intermediate something that is made in one step and used up in another The left and right sides of the mechanism cancel out to give you the overall equation.

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The rate determining step is the slowest step in the reaction mechanism

When city planners are trying to improve the flow of traffic through a particular area, they will look at where the traffic is moving most slowly as this will be the place that limits the overall traffic flow.

For chemical reactions with several steps, the overall rate will be determined by the slowest step in the sequence.

Products can only appear as fast as the products of this elementary step.

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Potential energy level profilesThe two maxima represent the transition states.

Minimum between the maxima represents the intermediate species after the first step.The energy of the

transition state for the first step is higher compared to the second step, hence is will be the slowest step.

The activation energy for the overall reaction is equal to the activation energy of the RDS.

Catalysts usually provide an alternate pathway for the RDS that has a lower activation energy

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Rate expression for an overall reaction is determined by the reaction mechanism RDS is an elementary step, a single molecular event, its rate law

comes directly from its molecularity.

Molecularity describes the number of particles involved in a single step.

The rate law for the equation in the RDS must contain the concentration of the reactants raised to the power of it’s co-efficient.

NOTE: This relationship only exists for elementary step equations, not overall reactions! (See lesson 3 to recap!)

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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 involvedA+A CA+B C

Termolecular: three molecules involved (extremely rare, you will not encounter these at IB level).

Experiment shows that each of the following reactions has the rate equation shown.

1. Hydrolysing a Primary HaloalkaneRCH2Br + OH- RCH2OH + Br-

Rate = k[RCH2Br][OH-]

2. Hydrolysing a Tertiary HaloalkaneR3CBr + OH- R3COH + Br-

Rate = k[R3CBr]

3. Nitrating Benzene using a c. H2SO4 Catalyst

C6H6 + HNO3 C6H5NO2 + H2O

Rate = k[HNO3]

4. Iodination of Propanone in Acid SolutionCH3COCH3 + I2 CH3COCH2I + H+ + I-

Rate = k[CH3COCH3 ] [H+]

RCH2Br and OH- both take part in RDS of

mechanism

Only R3CBr (not OH-) takes part in RDS of

mechanism

Only HNO3 takes part in the RDS of mechanism,

not H2SO4 or C6H6

CH3COCH3 and H+ both take part in RDS of

mechanism but not I2

Note: In example 4, H+ influences rate but is not a reactant H+ is a catalyst

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What about if the RDS is not the first step in the mechanism?

These examples show why the order of the reaction w.r.t each reactant is not linked to their co-efficients in the overall reaction equation.If a reaction is zero order w.r.t to that reactant (concentration will not affect the rate) it means it does not take part in the RDS.If a zero order reactant does appear in the rate equation, then the reactant or something derived from it must take part in the RDS.

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Solutions

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Mechanisms and Rates - Example 1

Looking at our previous example NO2(g) + CO(g) NO(g) + CO2(g)

Step 1: NO2(g) + NO2(g) NO(g) + NO3(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 Rate Determining Step (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

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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

Because it is involved in the slow step. Changing the concentration of NO will affect the rate

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.

e.g. NO2 + CO NO + CO2

can be shown to follow the following kineticsRate = k[NO2]2

2 molecules of NO2 only (not CO) are involved in the RDS.2 mechanisms were proposed:(A) 2NO2 N2O4 (slow RDS)

N2O4 + CO NO + NO2 + CO2 (fast)(B) 2NO2 NO3 + NO (slow RDS)

NO3 + CO O2 + CO2 (fast)Both are consistent with the rate equation and both give the overall reaction when steps addedbut they cannot both be correct!Other experimental work shows NO3 can be detected. mechanism (B) is acceptable

but only until further evidence contradicts it!

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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:

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.

Possibility 1:

2NO2(g) NO3(g) + NO(g)

NO3(g) + CO(g) NO(g) + CO2(g)

Depending on which is RDS either:•Rate = k[NO2]2

•Rate = k[NO2]2[CO]

RDS most likely to be step 1.

Possibility 2:

NO2 NO + O

CO + O CO2

Depending on RDS either:•Rate = k[NO2]•Rate = k[NO2][CO]

RDS most likely to be step 1.

FURTHER EXAMPLES:Suggest a possible mechanism for each of the following reactions which is consistent with the experimentally determined rate equation shown.

1. A + B C + D where rate = k[A][B]

2. A + B C where rate = k[A]

A + B [A-B]* RDS forms intermediate[A-B]* C + D Fast step

A [A]* RDS forms intermediate[A]* + B C Fast step

Q3 The reaction 2N2O5 4NO2 + O2

is shown by experiment to follow the following rate equation:

Rate = k [N2O5]

The mechanism is proposed to be

Step 1 N2O5 NO2 + NO3

Step 2 NO2 + NO3 NO + NO2 + O2

Step 3 NO + NO3 2NO2

For this mechanism to be valid, which step must be the rate determining step?

Step 1 Step 2 Step 3

Q4 The reaction H2O2 + 2H3O+ + 2I- I2 + 4H2O

is considered to take place in 3 steps

Step 1 H2O2 + I- IO- + H2O SLOW

Step 2 IO- + H3O+ HIO + H2O FAST

Step 3 HIO + H3O+ + I- I2 + 2H2O FAST

For this mechanism to be acceptable, experiment must show the rate equation to be:

Rate = K[H2O2][H3O+]2[I-]2 Rate = K[H2O2][H3O+][I-]

Rate = K[H3O+][IO-] Rate = K[HIO][H3O+][I-]Rate = K[H2O2][I-]

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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.

https://www.tes.co.uk/teaching-resource/video-the-rate-determining-step-6354488

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How can we determine the reaction mechanism?Example 2: 2NO(g) + O2(g) 2NO2(g) First we identify a series of possible reaction mechanisms:

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

Possibility 1:

NO + O2 NO2 + O NO + O NO2

Step 1 would be slow so:•Rate = k[NO][O2]

Possibility 2:

2NO N2O2N2O2 + O2 2NO2

Depending on RDS:•Rate = k[NO]2 •Rate = k[NO]2[O2]

Possibility 3:

NO N + ONO + O NO2N + O2 NO2

Step 1 likely to be slowest so:•Rate = k[NO]

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Complete the reactions mechanisms worksheet... Complete only questions: 1(a)+(b) 2(a)+(b)+(c)+(d)+(e) 3(a)+(b)+(c)+(d)

Leave the rest as it involves aspects of equilibrium which we haven’t covered yet...

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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.