Reaction Mechanism

Do you know how reaction occurs?  Can a balanced chemical equation tells us how the reaction occurs? The answer is no.   Reaction mechanism is a process by which a reaction occurs.  It is a sequence of elementary steps that leads to product formation.

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)

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:

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

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)

Rate Laws for Elementary Steps

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:

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.

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

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

Rate = k[NOBr₂][NO]

    

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:

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)

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 Law

Rate law is an equation that relates the reaction rate to the concentration of the reactants.  Meaning it is dependent on the concentration of the reactants not the coefficient in the balanced chemical equation.  It can be determined experimentally and with the use of elementary steps.  For a general equation,

aA  +  bB  →  cC  +  dD

the rate law has the general form of

Rate = k[A]^m[B]ⁿ

where  

           k = is the rate constant

           [A] = concentration of reactant A

           [B] = concentration of reactant B

           m and n = typically small whole number (0,1, or 2)

The constant k in the rate law is called the rate constant.  The magnitude of k changes with temperature and therefore determines how temperature affects rate.    If we know the rate law and the rate of the reaction including the concentration of the reactants, rate constant can be calculated.

Sample Exercise:

1.  The initial rate of a reaction A +  B  → C, was measured for several different starting concentrations of A and B , and the results are as follows:

Using the data above, determine a) the rate law for the reaction; b) the magnitude of the rate constant; c) the rate of the reaction when [A] = 0.050 M and [B] = 0.100 M.

Solution:

a.  To determine the rate law, we need to analyze the table given, experiment 1 and 2 has constant concentration of A, therefore using this two  experiment number can give us the relationship between the rate and the concentration of B. While experiment 1 and 3 having constant value of [B] will give the relation between rate and [A].

This is how to get the value of n,

Using the relationship of rate above,

Therefore the value of n = 0, because any number raised to 0 equals 1.

To get the value of m we will analyze experiment 3 and 1:

Using the relationship of rate 3 and rate 1 above,

The value of m = 2, since the square of 2 equals 4 (2^m = 4).

The rate law now will be

Rate= k[A]²[B]⁰ = k[A]²

b.  For the magnitude of rate constant:

c.  Using the rate law in a:

Rate =k[A]² = (4.0 x 10^-3 M^-1 s^-1)(0.050 M)² = 1.0 x 10^-5 M/s

Exponents in the Rate Law

The exponent m and n in the rate law is called reaction orders.  Reaction orders can be first order,  second order with respect to reactants. It depends on the exponent of the reactants.  First order when the exponent is raised to 1 and second order if raised to 2. In the example above, reactant A is  second order and the overall reaction order is also second order overall.  The overall reaction order is the sum of the reaction orders with respect to each reactant in the rate law.


Units of Rate Constants

The units of rate constant is dependent on the overall reaction order of the rate law.  Third order overall will have different unit with that of second order overall.  In the above sample problem, the units of rate constant is M^-1 s^-1, that is for second order overall rate law.  For the third order overall will have the unit as shown below:

TRY THIS:

The following data were measured for the reaction of nitric oxide with hydrogen:

2NO(g)  +  2H₂(g)  →  N₂(g)  +  2H₂O(g)

a) Determine the rate law for this reaction.
b) Calculate the rate constant.
c) Calculate the rate when the [NO] = 0.050 M, [H₂]= 0.150 M

Reaction Rates

In my previous post I posted the factors affecting reaction rates, it means you know already how to increase or decrease the rate of chemical reaction if in case you are the one manipulating the reaction.  In this post you will learn how to write the expression for the rate of disappearance of reactants and rate of appearance of products and to calculate the rate of reaction using both the disappearance of reactants and appearance of products.

Reaction rates is defined as the change in the concentration of  reactants or products per unit time. It units is molarity per second (M/s).  molarity is for the change in the concentration and seconds is for the change in time.

The rate of reaction can be expressed either in the rate of disappearance of the reactants or the rate of appearance of the product.  Let us have an example the simple reaction A →  B, where A is the reactant and B the product.  Rate can be calculated either

Let us see have an example:

a. Calculate the rate of appearance of B over the time interval from 0 to 20s.
b. Calculate the rate of disappearance of A overt the time interval from 20 to 40s.

Solution:
'

a.

 

b.  

What if the stoichiometric relationship of the reaction is not one is to one ratio?  Like for example

2HI(g)  →  H₂(g)  +   I₂(g)

How do we calculate the rate of reaction?  As you can see there are two moles of HI in every 1 mole of H₂ and I₂, therefore the rate of disappearance of HI is twice the rate of appearance of either H₂ and I₂.  To equate  the rates, we must therefore divide the rate of disappearance of HI by 2 as shown below:

In general, for the reaction

aA  +  bB  →  cC  +  dD

the rate is given by

The coefficient of the reactants and products is the denominator in the fraction for each reactant and product.

Sample Exercise:

a) How is the rate of disappearance of ozone related to the rate of appearance of oxygen in the following equation:  2O₃(g)  → 3O₂(g) ?  b) If the rate of appearance of O₂, ∆[O₂]/∆t, is 6.0 x 10^-5 M/s at a particular instant, what is the value of the rate of disappearance of O₃, -∆[O₃]/∆t, at this same time?

Solution:

a) Using the balance equation the rate is 

b)  Solving the equation from a:

TRY THIS:

1.  The decomposition of N₂O₅ proceeds according to the following equation:

2N₂O₅(g)  →  4NO₂(g)  +  O₂(g)

If the rate of decomposition of N₂O₅ at a particular instant in a reaction vessel is 4.2 x 10^-7 M/s, what is the rate of appearance of a) NO₂;  b) O₂?

2. Consider the reaction,     4NO₂(g) +  O₂(g)   →  2N₂O₅(g)

Suppose that, at a particular moment during the reaction, molecular oxygen is reacting at the rate of 0.024 M/s. (a) At what rate is N₂O₅ being formed? (b) At what rate is NO₂ reacting?

Factors Affecting Reaction Rates

Reactions come everywhere in our environment, some reactions occurs very slow and some occurs very fast.  The area of chemistry that deals with the speed, or rate at which chemical reactions occur is called chemical kinetics.  The word "kinetic' means movement, motion or change.  Rates of reaction refers to the speed of chemical reaction, it is the change in the concentration of reactants or products per unit time.  Thus, the unit for reaction rate is molarity per seconds (M/s).  But before we discuss how to calculate the rate of reaction, let us have first the factors that affect rates of reaction.

Factors Affecting Reaction Rates

There are four factors that affects the rates of reaction, these are the physical state of the reactants, concentration of the reactants, temperature at which reaction occurs and the presence of catalyst.

1.  Physical state of the reactants.   Reactants need to collide in order to react.  The more the reactants collide with each other the faster is the reaction.  homogeneous reactions are more faster that heterogeneous reactions.  When reactants are in different phases, the reaction is limited to their area of contact that is why the reaction rate is slower compared to the same phases of reactants.  Reactions that involve solids tend to proceed faster if the surface area of solid is increased.  Like for example iodized salt dissolve faster in water than rock salt due to its large surface area.  The smaller the particle size of the substance the greater the surface area.

2.  The concentration of the reactants.  Increasing the concentration of the reactants increases the rate of reaction.  As concentration increases, the frequency of the molecules collision also increases, leading to increase rates.  For example, 70% alcohol solution is more concentrated than 40% alcohol solution, 70% alcohol solution is more effective in disinfecting your skin than 40% alcohol solution because of its high concentration.

3.  The temperature at which the reaction occurs.  The rate of chemical reaction increases when temperature is increased.   Just like for example sugar, sugar dissolve faster in hot water than in cold water.  Increasing the temperature increases the kinetic energy of molecules, thus molecules collide more frequently resulting faster rate of reaction.  This is also the reason why we refrigerate perishable foods like milk.

4.  The presence of catalyst.  Catalyst is a substance that is being added to the reactants in order that the reaction occurs faster and not being used up during the process. Enzymes for example act as our biological catalyst.  Enzymes increases the biological reaction in our body by factors ranging from  10⁶ to 10¹⁸.  Another example is adding papaya to our boiled chicken.  Papaya contains papain enzymes that increases the  rate of the tenderness of the chicken.

In general reactions can proceed faster or slower by controlling the different factors   If you want that reaction will go faster you can increase the reaction by increasing the surface area, increasing the concentration, increasing the temperature and adding catalyst to the reacting molecules. On the other hand if you want to decrease the rate of reaction you can do the opposite, decrease the surface area, decrease the concentration, decrease the temperature and add inhibitor instead of catalyst. Of course it will just depend on the given situation.  Avoiding for example the spoilage of food, food is placed inside the refrigerator not being left on the table or any container at room temperature.

Dimensional Analysis

In science we encounter problem solving where conversion of units are needed.  In conversion of units the easiest way is to use dimensional analysis or factor label method, a method of converting one unit to another unit provided the same quantities are involved.   It is a simple technique that needs a little of memorization of the conversion factors to use in calculation.  Dimensional analysis is based on the relationship between different units that express the same physical quantity either metric to metric conversion, metric to English and English to English conversion.

Let us have an example, dollar is different from that of peso but both are amount of money.  One US dollar is equivalent now to 49.72 in Philippine peso.  What if you have 200 US dollar, what is its value in peso?

First thing to do is to have a conversion factor between dollar and peso,  

1 dollar = 49.72 pesos

If you want to convert dollar to peso the conversion factor to use is

  49.72 pesos 
1 dollar

and  if you want to convert from peso to dollar, the conversion factor to use is

     1 dollar     

49.72 pesos

Since the problem above is asking the value of 200 dollars in peso, the first conversion factor will be used.  As shown in the calculation below:

How about if you are asked to convert 1.5 L to mL?  This problem can be expressed as

? mL = 1.5 L

The conversion factor to use is 

     1000 mL    

1L

since the problem is asking the value in mL, mL must be the numerator and L is the denominator, this is for us to cancel out L unit in the calculation as shown below:

                                                                                       =  1.5 x 10³ mL

To express the answer in proper number of significant figures the answer is converted to scientific notation, expressed in 2 significant figures.  1000 mL/ 1 L  will not matter since those are exact numbers.  

In general, to apply dimensional analysis we follow:

given quantity  x  conversion factor =  desired quantity

The calculation of units is shown below:

In dimensional analysis, we make it sure that the all undesired units are canceled out remaining only the desired unit.

If the desired unit is not obtained in the calculation, meaning something wrong with the calculation and it needs to be reviewed to locate the error.


PRACTICE EXERCISE:

Convert the following:
1.  25.5 mg to g
2.  4.0 x 10^-10 m to nm
3.  0.575 mm to µm
4.  1.48 x 10² kg to g
5.  7.25 x 10^-4 s to ms