Chem Complement
3 Reactants must collide with energy equal to or greater than the
activation energy, Ea.
What is the probability that two ethane molecules and five oxygen molecules will simultaneously collide in the correct orientation with sufficient energy for reaction? It is far more likely that your hard-working Chemistry teacher will win the lottery, or be struck by lightning (several times!) than that a seven-molecule successful collision will ever occur. The balanced reaction equation gives the overall picture, but it doesn’t explain how the reaction occurs. It’s a bit like a friend arriving at your house. You know where they started from, you know where they ended up; but you don’t know the exact route they took to get there, or how fast they travelled on each part of the journey.
Very few reactions occur in one step; most are multi-step processes, in which each step rarely involves more than two molecules. Often, intermediate species are formed, which are a fundamental part of the process, but do not
[CH3OCH3] (mol dm–3) 0 0.02 0 0.04 0.06 0.08 0.10 0.000025 0.000050 rate (mol dm –3 s –1) time (s) 0 0 2000 4000 6000 0.12 0.10 0.08 0.06 0.04 0.02 [CH 3 OCH 3 ] (mol dm –3)
5.2 ReaCtion meChanism
Figure 5.2.1 The chance of being struck by lightning twice in an 80-year lifetime is 1 in 9 000 000. It is even less likely that more than three molecules will be involved in a simultaneous collision resulting in a reaction.
16.2.1
Explain that reactions can occur by more than one step and that the slowest step determines the rate of reaction (rate- determining step). © IBO 2007
appear in the final reaction equation. The reactionmechanism is the actual
step by step process by which a reaction occurs.
Reactions may proceed via many steps, or only one or two. The photochemical conversion of ozone to oxygen in the stratosphere
2O3(g) → 3O2(g)
is believed to occur in two steps: Step 1: O3(g) → O2(g)+ O·(g)
Step 2: O3(g)+ O·(g) → 2O2(g)
Each step is called an elementary step or elementary process. In the first step, an ozone molecule decomposes to form diatomic oxygen and an oxygen atom (oxygen radical). In the second step, the reactive oxygen atom collides with another molecule of ozone, forming two diatomic oxygen molecules. Adding the two elementary steps gives:
2O3(g) + O·(g) → 3O2(g) + O·(g)
Cancelling species that are common to both sides gives the overall equation. 2O (g) + O (g)3 · →3O (g) + O (g)2 ·
Overall: 2O3(g) → 3O2(g)
The equations for the elementary steps in a reaction mechanism must always add to give the balanced, overall equation for the reaction.
In this reaction, O·(g) is an intermediate—it is formed in one step and consumed in the next.
The molecularity of an elementary step describes how many molecules participate in that step. In the example above, step 1 is unimolecular and step 2 is bimolecular. If a step involved three molecules, it would be called
termolecular. Termolecular reactions are very rare; it is 1000 times more
likely that a bimolecular reaction will occur than a termolecular one.
Another way to describe the molecularity of a reaction is to give it a number: • A unimolecular reaction has a molecularity of 1.
• A bimolecular reaction has a molecularity of 2. • A termolecular reaction has a molecularity of 3.
No examples of reactions with higher molecularity are known. This must be kept in mind when proposing possible reaction mechanisms. If you include a termolecular step, it makes your proposed mechanism unlikely; if you include a step in which more than three molecules simultaneously collide and react, it makes your proposed mechanism impossible and therefore incorrect.
Although rate expressions for overall reactions must be determined
experimentally, rate expressions for elementary steps can be derived from the equation for that step. For step 2 of the ozone reaction:
O3(g) + O·(g) → 2O2(g)
two molecules must collide for a reaction to occur. Doubling the concentration
bimolecular reaction Collision
Collision
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table 5.2.1 Rate expRessions foR elementaRy steps general equation
for elementary step example molecularity general rate expression
A → B + C O3(g) → O2(g) + O·(g) unimolecular Rate = k[A] A + B → C + D O3(g) + O·(g) → 2O2(g) bimolecular Rate = k[A][B] 2A → B + C 2NO2(g) → NO3(g) + NO(g) bimolecular Rate = k[A]2 A + B + C → D + E ClO·(g) + NO2(g) + N2(g) → ClONO2(g) + N2∗(g) termolecular Rate = k[A][B][C] 2A + B → C + D 2Cl·(g) + CH4(g) → Cl2(g) + CH4∗(g) termolecular Rate = k[A]2[B]
The symbol ∗ is used to indicate a high-energy state. Many termolecular reactions occur in the atmosphere and involve transitions from low-energy to high-energy states.
If the rate expression for an overall reaction is known, it can be used to test whether the process occurs in one step (rare) or in several steps (common). A reaction such as
4HBr(g) + O2(g) → 2H2O(g) + 2Br2(g)
will obviously take place in more than one step because there is such a low probability of five molecules participating in a one-step reaction. Remember that no instances of one-step reactions involving more than three molecules have ever been found.
With a reaction such as
2NO2(g) + F2(g) → 2NO2F(g)
it is possible, though unlikely, that the reaction occurs in only one step. If it occurred in one step, the rate expression would be
Rate = k[NO2]2[F2]
The experimentally determined rate expression for this reaction is Rate = k[NO2][F2]
The rate expressions don’t match, proving that the reaction must occur in two or more steps.
A certain take-away pizza shop has a production line involving four workers. The first person takes the pre-prepared base and covers it in tomato paste. The second person adds the toppings. The third person puts it in the oven, and takes it out again when cooked. The fourth person slices it up and puts it in the box. In this four-step process, the slowest step is the cooking of the pizza. This should take about 10 minutes, while the other steps should take a minute or less. The process can be summarized as:
Step 1: tomato paste on base Fast
Step 2: toppings on pizza Fast
Step 3: cooking Slow
Step 4: slicing and boxing Fast
The rate of pizza production depends almost entirely on step 3. The slowest step in a process is called the rate-determining step.
Rate-determining step
Figure 5.2.3 The rate-determining step in pizza production is the cooking of the pizza.
16.2.2
Describe the relationship between reaction mechanism, order of reaction and rate- etermining step. © IBO 2007
Multi-step reactions can be considered in the same way; there is always one elementary step that has a slower rate than the others, and this step will govern the rate of the overall reaction. It doesn’t matter if the slow step occurs at the beginning, middle or end of the reaction.
One example of a reaction in which the slow step occurs at the start of the reaction is the redox reaction between nitrogen(IV) oxide and carbon monoxide at 200°C:
NO2(g) + CO(g) → NO(g) + CO2(g)
If this reaction occurred in one step, the rate expression would be first order with respect to each reactant and second order overall:
Rate = k[NO2][CO]
In fact, the rate expression has been experimentally determined as being second order with respect to nitrogen(IV) oxide and zero order with respect to carbon monoxide, and thus second order overall:
Rate = k[NO2]2
Since the rate is governed by the slowest step in the process, the proposed reaction mechanism must contain a slow step with a rate expression that matches the rate expression for the overall reaction.
Note that knowing the rate expression and the value of k provides evidence for a proposed reaction mechanism, but does not prove it.
So what is one plausible reaction mechanism for the nitrogen(IV) oxide–carbon monoxide reaction? From the rate expression, we know that one step must involve two nitrogen(IV) oxide molecules colliding. One possible mechanism is shown in table 5.2.2.
table 5.2.2 pRoposed mechanism foR nitRogen(iV) oxide–caRbon monoxide Reaction
step equation Rate expression molecularity Rate
1 NO (g) + NO (g)2 2 →NO (g) + NO(g)3 Rate = k[NO2]2 Bimolecular Slow
2 NO (g) + CO(g)3 →NO (g) + CO (g)2 2 Rate = k[NO3][CO] Bimolecular Fast
Three important points indicate the validity of this mechanism:
• Adding the two equations for the elementary steps gives the correct overall equation.
• It contains a slow step (the rate-determining step) with a rate expression matching the experimentally determined rate expression.
• No step involves more than two molecules colliding.
The time taken for the fast step contributes negligibly to the overall rate. This indicates that k for step 2 is much larger than k for step 1. The NO3 intermediate is slowly produced in step 1, and then quickly consumed in step 2.
The importance of the rate-determining step is also reflected in the molecularity of the reaction.
DO ALL
ELEMENTARY STEPS INVOLVE ONLY
ONE OR TWO MOLECULES, OR THREE IF THE RATE EXPRESSION INDICATES A TERMOLECULAR STEP? IS THERE A STEP WITH A RATE EXPRESSION MATCHING THE EXPERIMENTALLY DETERMINED RATE EXPRESSION? DO THE EQUATIONS FOR THE ELEMENTARY STEPS ADD TO GIVE THE CORRECT OVERALL EQUATION?
Figure 5.2.4 If the answer to all these questions is ‘yes’, your proposed reaction mechanism is valid.
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In this case, it is bimolecular (molecularity = 2), making the reaction second order overall.
Figure 5.2.5 Kinetic relationships for the reaction NO2(g) + CO(g) → NO(g) + CO2(g). Order of reaction
= 2 (second order) Experimentally determined
rate expression:
Rate = k[NO2]2
Reaction mechanism must contain a slow step with a rate expression matching the one above:
NO2(g) + NO2(g) m NO3(g) + NO(g) NO3(g) + CO(g) m NO2(g) + CO2(g) Slow Fast Order of reaction = molecularity = 2
In this course, you will only be assessed on one-step or two-step mechanisms; however, it is useful to consider examples of reactions that occur in more than two steps. The reaction between hydrogen bromide and oxygen at 420°C has five molecules on the reactant side of the equation and so is likely to have a reaction mechanism involving more than two steps:
4HBr(g) + O2(g) → 2H2O(g) + 2Br2(g)
This is a second-order reaction, with the experimentally determined rate expression
Rate = k[HBr][O2]
This means that any proposed mechanism must contain a slow step in which one HBr molecule and one O2 molecule collide and react.
One proposed mechanism is shown in table 5.2.3.
table 5.2.3 pRoposed mechanism foR the hydRogen bRomide–oxygen Reaction
step equation Rate expression molecularity Rate
1 HBr(g) + O (g)2 →HBrOO(g) Rate = k[HBr][O2] Bimolecular Slow
2 HBrOO(g) + HBr(g)→2HBrO(g) Rate = k[HBrOO][HBr] Bimolecular Fast
3 HBrO(g) + HBr(g)→H O(g) +Br (g)2 2 Rate = k[HBrO][HBr] Bimolecular Fast
4 HBrO(g) + HBr(g)→H O(g) +Br (g)2 2 Rate = k[HBrO][HBr] Bimolecular Fast
In some cases, the rate expression may not be known, but the slow step may be determined via quantitative rate studies. Imagine that the reaction
2A + B + C → 2D + E
occurs via the mechanism shown in table 5.2.4, and that rate studies had determined the initial rates shown, at constant reactant concentration.
table 5.2.4 mechanism and Rates foR the a–b–c Reaction
step equation Rate expression molecularity Rate
(mol dm–3 s–1)
1 A +B→D + F Rate = k[A][B] Bimolecular 0.0810
2 F + C→E + A Rate = k[F][C] Bimolecular 0.1000
3 A + C→D + G Rate = k[A][C] Bimolecular 0.0001
4 A + G→C Rate = k[A][G] Bimolecular 0.0990
Step 3 is much slower than the other three steps; thus, it is the rate- determining step, and the rate expression for the overall reaction is
Rate = k[A][C]
There are two molecules involved in the rate-determining step, therefore the reaction is bimolecular (molecularity = 2) and second order overall.
Although we have focused on multi-step reactions, many examples of one-step reactions exist, such as SN2 reactions. One example is the dehalogenation of 2,3-dibromobutane using iodide ions:
CH3CHBrCHBrCH3 + I−→ CH
3CH = CHCH3 + IBr + Br−
Since the overall equation represents the only elementary step, this reaction will be second order, bimolecular, and have the rate expression:
Rate = k[CH3CHBrCHBrCH3][I−]
tHeORY OF KnOWLeDGe
The difficulty with relying on a rate expression to describe the relationship between the concentration of the reactants and the rate of the reaction is that nearly all reactions have more than one step. To understand how a chemical reaction occurs, chemists use deductive reasoning and intuition to predict a number of different possible reaction steps that might be consistent with the overall rate expression. Further experiments can be carried out, if needed, to eliminate the least likely mechanisms. The final proposed mechanism, however, is not proof that the reaction takes place this way, but it does provide chemists with a reaction pathway that can be compared with the experimentally determined rate expression to see if there is a match. If there is agreement, the proposed mechanism is more probable. If there is no agreement, this provides evidence to disprove the mechanism. Although chemists can never be entirely certain of the exact steps in a reaction mechanism, this does not prevent them from continuing the search for the most probable mechanism. Instead it reinforces the fact that scientific knowledge can be simultaneously certain and tentative, helping chemists remain open to new ideas, new theories and new knowledge.
• Why are we more likely to believe a proposed reaction mechanism when it is in agreement with the rate expression?
Worksheet 5.2 Reaction mechanisms
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1 Identify which of the following reactions is more likely to occur in a single