Colourful Solutions > The rate of chemical change > Mechanisms

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Higher-level only

Mechanism means the process that the reactant particles must go through in order to arrive at the product particles.

We have already established that collisions are an essential feature of a reaction procedure. The mechanism explains which particles actually collide and the order in which the collisions occur.

Syllabus ref: R2.2.6

Reactivity 2.2.6 - Many reactions occur in a series of elementary steps. The slowest step determines the rate of the reaction. (HL)

  • Evaluate proposed reaction mechanisms and recognize reaction intermediates.
  • Distinguish between intermediates and transition states, and recognize both in energy profiles of reactions.

Guidance

  • Include examples where the rate-determining step is not the first step.
  • Proposed reaction mechanisms must be consistent with kinetic and stoichiometric data.

Tools and links

  • Reactivity 3.4 - Which mechanism in the hydrolysis of halogenoalkanes involves an intermediate?

Mechanism

The term 'mechanism' refers to the actual processes taking place in a chemical reaction.

Normally a chemical reaction is expressed by a balanced equation. This tells us the relative amounts of reactants that react and products formed. It gives us no information about any intermediate species that may be formed, or any information about the actual collisions taking place.

In reality most reactions proceed via a series of steps involving intermediates, or high energy transition states. These steps, when put together, constitute the 'story' of the reaction - they are called the mechanism.

Example: The reaction between nitrogen(II) oxide and chlorine can be represented by the equation:

2NO(g) + Cl2(g) → 2NOCl(g)

A suggested mechanism for this reaction has two steps:

1. NO(g) + Cl2(g) → NOCl2(g)

2. NO(g) + NOCl2(g) → 2NOCl(g)

You should notice that if all of the steps of a mechanism are added up, cancelling out species that appear on each side, it gives the stoichiometric equation:

NO(g) + Cl2(g) → NOCl2(g)
NO(g) + NOCl2(g) → 2NOCl(g)
2NO(g) + Cl2(g) → 2NOCl(g)

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Why study the mechanism?

One of the aims of the chemist when studying kinetics is to come up with suggested mechanisms for reactions. By increasing the knowledge base of a specific reaction, scientists can then suggest ways of controlling or modifying the reaction.

This could lead to more efficient and cheaper processes for making novel and existing products in industry.

Industry, both on a large and small scale, is always on the lookout for ways of making things cheaper and more cost effective.

Understanding the mechanism of formation of a specific drug may help scientists to modify it in a beneficial way, or to come up with novel hitherto unconsidered methods of synthesis.

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The rate determining step

The steps of a mechanism may be fast or slow relative to one another. The slow step of a mechanism is the rate determing step.

It is called this because the slowest stage of any operation has a greater effect on the time taken for the operation to be completed.

Imagine a construction line with five different operational steps needed to make the final product. If one of the steps is very time consuming, then this part of the process will affect the time taken far more than the others.

It would be the rate determing step of the construction line.

In a chemical reaction exactly the same constraints apply. If one step of the mechanism is much slower than the others, then it will determine the overall rate of the reaction.

Example

1. NO(g) + Cl2(g) → NOCl2(g) - fast
2. NO(g) + NOCl2(g) → 2NOCl(g) - slow

When we conduct experiments to measure the rate of a reaction, we are really measuring the rate of the slowest step, the rate determining step.

Any fast steps in the mechanism of the reaction will have little influence on the overall rate.

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Molecularity

All of our rates investigations are used to follow the rate determing step (the slowest step) of a mechanism. The rate equation is solved for the reaction using the process described in section 6.21.

The final rate expression gives us the orders of the reaction with respect to the individual reactants.

The actual number of particles colliding in the rate determining step is called the molecularity.

The order of each component gives the molecularity of the rate determining step (or that of the rate determining step plus any steps that feed into the rate determining step).

Example: The reaction: 2NO(g) + Cl2(g) → 2NOCl(g)

1. NO(g) + Cl2(g) → NOCl2(g) - fast

2. NO(g) + NOCl2(g) → 2NOCl(g) - slow

The molecularity of the rate determining step (slowest) is 2, but the rate expression for this reaction is:

Rate = k[NO]2[Cl2]1

Althought the slow step of the mechanism has only one particle of NO and one particle of NOCl2, the particle of NOCl2 is not one of the reactants, it has to be made in a prior step, step 1.

In step 1, NOCl2 is made from the collision between one NO and one Cl2 particle.

Therefore overall we have needed two NO particles and one Cl2 particle to carry out the slowest step.

In the above example the mechanism was used to explain the rate expression.

In kinetics studies this is, of course, usually the reverse; the rate expression is first determined by experiment and then mechanisms are proposed that fit the rate expression.

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Determining the mechanism

A series of experiments are carried out to determine the rate expression for the reaction under study. This will reveal the order of reaction with respect to the concentration of each component.

The next stage is a little more complex in that the order wth respect to each component concentration gives us the molecularity of the rate determining step.

The problem is there may be several possible mechanisms that emerge from a consideration of this slowest step. At this point the investigator will use his knowledge of chemistry to suggest the most likely mechanism based on chemical stabilities and other factors.

It is not suggested that a student be able to do this, however it should be possible to suggest mechanisms based on the following three facts:

1. The order of reaction from the rate expression gives us the number of particles of that type involved in the rate determing step (or used in producing species that appear in the rate determing step, but that do not appear in the stoichiometric equation)

2. Three particle collisions are deemed improbable.

3. The sum of all of the steps must add up to the stoichiometric equation.

Mechanisms summary

Example: The following data were obtained for the reaction between nitrogen(II) oxide gas and hydrogen. Use this data to find the rate expression and suggest a mechanism for the reaction:

2H2(g) + 2NO(g) → 2H2O(g) + N2(g)

Initial [NO]/mol dm-3
Initial [H2]/mol dm-3
Initial rate/mol dm-3 s-1
6 x 10-3
1 x 10-3
3.19 x 10-3
6 x 10-3
2 x 10-3
6.36 x 10-3
6 x 10-3
3 x 10-3
9.56 x 10-3
1 x 10-3
6 x 10-3
0.48 x 10-3
2 x 10-3
6 x 10-3
1.92 x 10-3
3 x 10-3
6 x 10-3
4.30 x 10-3

By inspection order wrt [NO] = 2, order wrt [H2] = 1, therefore the rate expression is: Rate = k[NO]2[H2]1

This suggests that two particles of NO are involved and one particle of H2, but, as three particle collisions are highly improbable, there must be an equilibrium step (fast) leading into a rate determining step.

Step 1: NO + NO ⇋ N2O2 (fast)

Step 2: N2O2 + H2 → H2O + N2O (slow)

Step 3: N2O + H2 → H2O + N2 (fast)

There is no certainty that this is the actual mechanism of the reaction, but it does fulfill all of the requirements for a possible mechanism.

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Worked examples

Q633-01 A proposed mechanism for the reaction 2NO(g) + Br2(g) → 2NOBr(g) is as follows:

NO(g) + Br2(g) → NOBr2(g) slow
NO(g) + NOBr2(g) → 2NOBr(g) fast

If this mechanism is correct then the rate law could be expected to be which of the following:

  1. rate = k[NO][Br2]
  2. rate = k[NO]2[Br2]
  3. rate = k[NO][NOBr2]
  4. rate = k[NO]2
Answer

It can be seen that in the slow (rate determining) step of the reaction one molecule of NO collides with one molecule of Br2. We would expect that the order of each of these reactants is 1.
Rate = k [NO]1[Br2]1
Q633-02 The rate of the reaction 2A + B → Products, is consistent with the rate equation, rate = k[A][B].
Which reaction mechanism is consistent with this information?

A. A + B → AB (slow)
  AB + A → Products (fast)
B. A + A → A2 (slow)
  A2 + B → Products (fast)
C. A + B → AB (fast)
D. A + A → A2 (fast)
  A2 + B → Products (slow)
Answer

The order of the concentrations in the rate equation suggests the molecularity of these particles in the rate-determining step (in addition to equilibria or previous reactions that feed new particles into the rate determining step)
Only mechanism (A) has one particle of A and one of B in the rate determining step.
Q633-03 The following reaction, X2(g) + 2Y(g) → 2XY(g), has the proposed mechanism:

X2 → 2X slow
X + Y → XY fast

Based on this information the rate expression for this reaction is:

  1. rate = k[XY]
  2. rate = k[X2][Y]2
  3. rate = k[X2]
  4. rate = k[2X]
Answer

The first step is rate determining and involves only the decomposition of X2 molecules. The rate expression can only reflect this fact:

rate = k[X2]
Q633-04 The reaction of hydrogen and iodine monochloride is represented by this equation.

H2(g) + 2ICl(g) → 2HCl(g) + I2(g)

This reaction is first order in H2(g) and also first order in ICl(g). Which of these proposed mechanisms can be consistent with the given information about this reaction?

Mechanism 1 H2(g) + 2ICl(g) → 2HCl(g) + I2(g)  
Mechanism 2 H2(g) + ICl(g) → HCl(g) + HI(g) slow
  HI(g) + ICl(g) → HCl(g) + I2(g) fast
  1. 1 only
  2. 2 only
  3. Both 1 and 2
  4. Neither 1 nor 2
Answer

First order in H2(g) and also first order in ICl(g) suggests that one molecule of each of the species are involved in rate determining processes. In mechanism 1 there appears to be a three particle collision, making it extremely improbable. In mechanism II the collision is between one molecule of H2 and one of ICl in the slow (rate determining ) step. Thus only mechanism 2 is a possibility.
Q633-05 For the hypothetical reaction 2X + 2G → Q + 2M, the rate expression is Rate = k[X]2[G]. Which mechanism is most likely?

A. 2X + G ⇋ 2Q + R (fast equilibrium)
  Q + R + G → 2 M (slow, rate determining)
B. X + G ⇋ Q + R (fast equilibrium)
  R + X → 2M (slow, rate determining)
C. X + X ⇋ X2 (fast equilibrium)
  X2 + G → Q + T (slow, rate determining)
  T + G → 2M (fast)
D. G + G ⇋ G2 (fast equilibrium)
  G2 + X → Q + T (slow, rate determining)
  T + X → 2M (fast)
Answer

For a hypothetical mechanism to be valid it must fulfil three conditions:

  • Firstly the sum of all of the steps must give the stoichiometric equation.
  • Secondly the orders of the reactants must coincide with the particle collisions involved in the slowest step and steps leading into the slowest step.
  • Thirdly, there must be no three-particle collisions.

Mechanism A can be discarded as both steps involve three-particle collisions.

Mechanism B does not add up to the stoichiometric equation overall and so cannot be possible.

Mechanism C has no three-particle collisions, the sum of all of the equations add up to the stoichiometric reaction equation and the orders of the components in the rate expression suggest that there should be two particles of X and one particle of G in the slowest step, or in the showest step plus steps leading into the slowest step. This is the case, as step 1 has two particles of X, and this leads into the second (slow) step, which has one particle of G. The answer is response C.

Mechanism D would have a rate expression with orders of reaction of two with respect to G and one with respect to X.
Q633-06 The reaction between chloroform, CHCl3(g), and chlorine, Cl2(g), to form CCl4(g) and HCl(g) is believed to occur by this series of steps.
  1. Step 1: Cl2 → Cl(g) + Cl(g)
  2. Step 2: CHCl3(g) + Cl(g) → CCl3(g) + HCl(g)
  3. Step 3: CCl3(g) + Cl(g) → CCl4(g)

If this reaction is first order in CHCl3 and half order in Cl2, which statement about the relative rates of step 1, 2, and 3 is correct?

  1. Step 1 is the slowest.
  2. Steps 1 and 2 must both be slow.
  3. Step 2 must be slower than step 1.
  4. Step 3 must be the slowest
Answer

As the order of reaction with respect to CHCl3 is one, one molecule of this would be expected to figure in the slowest step. This is also backed up by a faster equilibrium leading into step 2 as chlorine also is involved in the rate expression, albeit with a fractional rate. Hence, step 2 must be slower than step 1.
Q633-07 The mechanism for the decomposition of nitrogen(V) oxide, has been identified as follows:
k1
N2O5 → NO2 + NO3 (slow)
k2
NO2 + NO3 → NO2 + O2 + NO (fast)
k3
NO + N2O5 → 3NO2 (fast)

The stoichiometric equation for this decomposition is:

  1. N2O5 → NO2 + NO2
  2. N2O5 → NO2 + O2 + NO
  3. 2N2O5 → 4NO2 + O2
  4. N2O5 + NO → 3NO2
Answer

The stoichiometry is obtained by adding all of the equations and cancelling out the same substances from each side.

N2O5 → NO2 + NO3 (slow)
NO2 + NO3 → NO2 + O2 + NO (fast)
NO + N2O5 → 3NO2 (fast)
N2O5 + NO2 + NO3 + NO + N2O5 → NO2 + NO3 + NO2 + O2 + NO + 3NO2
2N2O5 → 4NO2 + O2
Q633-08 The rate law consistent with the mechanism shown in question 07 above is:
  1. rate = k1 [N2O5]
  2. rate = K3 [NO][N2O5]
  3. rate = k1k2k3 [N2O5]2
  4. rate = k2 [NO2][NO3]
Answer

As only N2O5 is involved in the slowest step and nothing leads into this step, the only reactant that should appear in the rate expression is N2O5.

Therefore response A, rate = k1 [N2O5] seems most likely.
Q633-09 For the stoichiometric reaction:

CHCl3(g) + Cl2(g) → CCl4(g) + HCl(g)

the rate law, rate = k[CHCl3][Cl2]1/2 has been determined. The mechanism given below has been proposed.

  1. step 1: Cl2 ⇋ 2Cl
  2. step 2: CHCl3 + Cl → CCl3 + HCl
  3. step 3: CCl3 + Cl → CCl4

For this reaction, the rate determining step must be:

  1. step 1.
  2. step 2.
  3. step 3
  4. 1/2 times step 1
Answer

Chlorine appears in the rate expression and so does CHCl3.

If step 2 were the slowest step with step 1 feeding into it then this would give the particles in the rate expression.

Therefore step 2 is the slowest.
Q633-10 The ionic reaction:

3I- + S2O82- → 2SO42- + I3-

is found to be described by the experimental rate equation, rate = k[I-] [S2O82-]. This suggests that:

  1. the reaction occurs when three iodide ions collide simultaneous with one S2O82- ion
  2. there must be only one step in the overall reaction since the rate is first-order in both reactants
  3. the rate of reaction is inhibited by the presence of I3-
  4. a slow step in the overall reaction sequence may involve the collision of a single I- ion with a single S2O82- ion.
Answer

The rate expression suggests a collision between one iodide ion and one peroxodisulfate ion in the slowest step, as the order with respect to each of these components is 1. Response D.