Elementary Steps
What is elementary step? Elementary step is a process or a reaction that occurs in a single event. Example is the reaction of NO(g) and O3(g):
NO(g) + O3(g) → NO2(g) + O2(g)
The reaction above is an example of single elementary step. There are other reactions that involves 2 or more elementary steps before the final products are formed and this are called multistep mechanisms. Multistep mechanism will be discussed below.
The number of molecules that participate as reactants in an elementary steps is what we call molecularity of reaction. Molecularity can be unimolecular, in which a single molecule is involved as reactant, for example A → products. Bimolecular, elementary steps that involves 2 molecules as the reactants; example A + A → Products and termolecular, elementary steps that involves 3 reactants molecules that collide simultaneously, like A + A + B → Products. Termolecular steps are far less probable than unimolecular and bimolecular processes and are rarely encountered.
Multistep Mechanism
The multistep mechanism is consist of a series of elementary steps before the final product is formed. An example is the reaction of NO2 and CO:
NO2(g) + CO(g) → NO(g) + CO2(g)
This reaction appears in two elementary steps, and each elementary step is bimolecular.
NO2(g) + NO2(g) → NO3(g) + NO(g)
NO3(g) + CO(g) → NO2(g) + CO2(g)
NO3(g) + CO(g) → NO2(g) + CO2(g)
The elementary steps in the multistep mechanism must always add to give the overall chemical reaction.
In the above example the sum of the the 2 elementary steps is:
2NO2(g) + NO3(g) + CO(g) → NO2(g) + NO3(g) + NO(g) + CO2(g)
Simplifying the equation by eliminating substances that appear both on the reactant side and on the product side will give:
You learned in my previous post that rate laws can be determined experimentally. Another method of determining the rate law is with the use of reaction mechanism. Reactions that only involved in single elementary step has a way to determine its rate law. The rate law of any elementary step is based directly on its molecularity. For example in a unimolecular process:
NO2(g) + CO(g) → NO(g) + CO2(g)
NO3 is called intermediate. Intermediate is a substance that is produced in the first step and consumed in the second step. Multistep mechanisms can have one or more intermediates.
Sample Problem:
It has been proposed that the conversion of ozone into O2 proceeds via two elementary steps
Rate Laws for Elementary StepsSample Problem:
It has been proposed that the conversion of ozone into O2 proceeds via two elementary steps
O3(g) → O2(g) + O(g)
O3(g) + O(g) → 2O2(g)
a) Describe the molecularity of each step in this mechanism.
b) Write the equation for the overall reaction.
c) Identify the intermediates.\
Solution:
a) Since the first elementary step only consists of a single reactant therefore it is unimolecular and the second elementary step is consist of 2 reactant molecules and so bimolecular.
b) The overall reaction is
2O3(g) + O(g) → 3O2(g) + O(g)
Canceling the intermediate the final overall reaction is
2O3(g) → 3O2(g)
c. The intermediate is O(g)
You learned in my previous post that rate laws can be determined experimentally. Another method of determining the rate law is with the use of reaction mechanism. Reactions that only involved in single elementary step has a way to determine its rate law. The rate law of any elementary step is based directly on its molecularity. For example in a unimolecular process:
A → Products
The rate law will be first order:
Rate = k [A]
Below is a table showing the rate law of some reactions consisting only of single elementary step.
It is important to remember that in writing the rate law, we should not follow the balanced chemical equation, it can be determined experimentally or with the use of reaction mechanism.
Sample Problem:
If the following reaction occurs in a single elementary step, predict the rate law:
Sample Problem:
If the following reaction occurs in a single elementary step, predict the rate law:
H2(g) + Br2(g) → 2HBr(g)
Solution:
Assuming the reaction occurs in single elementary step, the step is bimolecular involving two reactant molecules. The rate law would therefore
Rate = k[H2][Br2]
Experimental studies of this reaction show that it has a very different rate law:
Rate = k[H2][Br2]1/2
This indicates that the reaction do not only involve in single elementary step but more that one elementary steps.
Mechanism with an Initial Slow Step
In a multistep reaction mechanism the slowest step determines the overall rate of reaction. This is true for multistep mechanism with initial slow step. The rate of a faster step following the rate determining step does not affect the overall step. Consider the example below:
Step 1: NO2(g) + NO2(g) → NO3(g) + NO(g) (slow)
Step 2: NO3(g) + CO(g) → NO2(g) + CO2(g) (fast)
Overall : NO2(g) + CO(g) → NO(g) + CO2(g)
Since step 1 is a slow step, it is the rate determining step, and step 2 being fast will not affect the overall rate of the reaction. Thus the rate of overall reaction will be the rate of step 1. Step 1 is bimolecular and therefore will have the rate law:
There are also multistep reactions where step 2 is slow and the rate-determining. But it is not proper that the rate law contains an intermediate, a substance formed in the first step and consumed in the second step because this substance has an unknown concentration. If the second step is the rate-determining step it contains an intermediate. Let us have an example below:
However, the rate law contains the intermediate, NOBr2, which is generated in step 1 and consumed in step 2. Since intermediates are usually unstable and with unknown concentration and the rate law depends on the unknown concentration of the intermediate, we need to know how to express the concentration of this intermediate.
With the aid of some assumptions, the concentration of intermediate (NOBr2) can be expressed using the starting reactants in which it was formed. Since step 1 is very fast, there are two ways in which it can consume: first, it can either react immediately with NO to form NOBr or fall back apart forming the initial reactants NO and Br2. But because step 2 is very slow we assume that most of the NOBr2 falls apart according to the reaction below:
Using the equation above, we can solve for the concentration of NOBr2, [NOBr2], we have:
Substituting the relationship into the rate law for the rate-determining step,
Rate = k[NOBr2][NO]
Rate = k2(k1/k-1)[NO][Br2][NO]
Rate = k[NO]2[Br2]
In general, we have to remember that whenever a fast step precedes a slow one, we can solve for the concentration of the intermediate by assuming that an equilibrium is established in the fast step.
TRY THIS:
1. The following mechanism has been proposed for the gas phase reaction of H2 and ICl:
a) Write the balanced equation for the overall reaction.
b) Identify the intermediates in the mechanism.
c) Write the rate law for each elementary steps in the mechanism.
d) If the first step is slow and the second one is fast , what rate law do you expect to be observed for the overall reaction.
Rate Laws for Multistep Mechanism
There are reactions that involve more than one elementary step, wherein each step has its own rate constant and activation energy. Often steps have a different rate of reaction, sometimes the first step is slow and the second is fast and sometimes the first step is fast and the second step is slow. The overall rate of reaction cannot exceed the rate of the slowest elementary step of its mechanism, because the slow step limits the overall reaction rate. Therefore the slow step is called the rate-determining step.Mechanism with an Initial Slow Step
In a multistep reaction mechanism the slowest step determines the overall rate of reaction. This is true for multistep mechanism with initial slow step. The rate of a faster step following the rate determining step does not affect the overall step. Consider the example below:
Step 1: NO2(g) + NO2(g) → NO3(g) + NO(g) (slow)
Step 2: NO3(g) + CO(g) → NO2(g) + CO2(g) (fast)
Overall : NO2(g) + CO(g) → NO(g) + CO2(g)
Since step 1 is a slow step, it is the rate determining step, and step 2 being fast will not affect the overall rate of the reaction. Thus the rate of overall reaction will be the rate of step 1. Step 1 is bimolecular and therefore will have the rate law:
Rate = k[NO]2
The rate laws predicted by the reaction mechanism agree with the one observed experimentally.
Mechanism with an Initial Fast Step
Mechanism with an Initial Fast Step
There are also multistep reactions where step 2 is slow and the rate-determining. But it is not proper that the rate law contains an intermediate, a substance formed in the first step and consumed in the second step because this substance has an unknown concentration. If the second step is the rate-determining step it contains an intermediate. Let us have an example below:
Step 1 : NO(g) + Br2(g) ⇌ NOBr2(g) (fast)
Step 2: NOBr2(g) + NO(g) → 2NOBr(g) (slow)
Overall : 2NO(g) + Br2(g) → 2NOBr(g)
In this mechanism the step 2 is the slow step and therefore the rate-determining step, therefore the overall rate law will be
In this mechanism the step 2 is the slow step and therefore the rate-determining step, therefore the overall rate law will be
Rate = k[NOBr2][NO]
With the aid of some assumptions, the concentration of intermediate (NOBr2) can be expressed using the starting reactants in which it was formed. Since step 1 is very fast, there are two ways in which it can consume: first, it can either react immediately with NO to form NOBr or fall back apart forming the initial reactants NO and Br2. But because step 2 is very slow we assume that most of the NOBr2 falls apart according to the reaction below:
NOBr2 (g) → NO(g) + Br2(g)
Thus, we have both forward and backward reactions occurring much faster than in step 2. Because they occur rapidly, the forward and reverse processes in step 1 establish an equilibrium. In any dynamic equilibrium the rate of the forward reaction is equal to the rate of backward reaction and so:
k1[NO][Br2] = k-1[NOBr2]
rate of forward reaction rate of backward reaction
Using the equation above, we can solve for the concentration of NOBr2, [NOBr2], we have:
[NOBr2] = k1/k-1[NO][Br2]
Substituting the relationship into the rate law for the rate-determining step,
Rate = k[NOBr2][NO]
Rate = k2(k1/k-1)[NO][Br2][NO]
Rate = k[NO]2[Br2]
In general, we have to remember that whenever a fast step precedes a slow one, we can solve for the concentration of the intermediate by assuming that an equilibrium is established in the fast step.
TRY THIS:
1. The following mechanism has been proposed for the gas phase reaction of H2 and ICl:
H2(g) + ICl(g) → HI(g) + HCl(g)
HI(g) + ICl(g) → I2(g) + HCl(g)
b) Identify the intermediates in the mechanism.
c) Write the rate law for each elementary steps in the mechanism.
d) If the first step is slow and the second one is fast , what rate law do you expect to be observed for the overall reaction.
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