In last part-1 we seen how the rate of reaction is made with an example.Now here we will see rate laws and  its constant.

3.Rate Law :


The rate law is an expression relating the rate of a reaction to the concentrations of the chemical species present, which may include reactants, products, and catalysts. Many reactions follow a simple rate law, which takes the form

ν = k [A]a[B]b[C]c

i.e. the rate is proportional to the concentrations of the reactants each raised to some power. The constant of proportionality, k, is called the rate constant. The power a particular concentration is raised to is the order of the reaction with respect to that reactant. Note that the orders do not have to be integers. The sum of the powers is called the overall order. Even reactions that involve multiple elementary steps often obey rate laws of this kind, though in these cases the orders will not necessarily reflect the stoichiometry of the reaction equation. For example,

H2 + I2 → 2HI ν = k [H2][I2]

3ClO− → ClO3 + 2Cl− ν = k [ClO−]2

Other reactions follow complex rate laws. These often have a much more complicated dependence on the chemical species present, and may also contain more than one rate constant.Complex rate laws always imply a multi-step reaction mechanism. An example of a reaction with a complex rate law is 

H2 + Br2 → 2HBr ν =[H2][Br2]1/2 / [1 + k'[HBr]/[Br2] ]

In the above example, the reaction has order 1 with respect to [H2], but it is impossible to define orders with respect to Br2 and HBr since there is no direct proportionality between their concentrations and the reaction rate. Consequently, it is also impossible to define an overall order for this reaction.To give you some idea of the complexity that may underlie an overall reaction equation, a slightly simplified version of the sequence of elementary steps involved in the above reaction is shown below.

Br2 → Br + Br

Br + H2 → H + HBr

H + Br2 → Br + HBr

Br + Br → Br2

As well as having rate laws for overall reactions, we can of course also write down individual rate laws for elementary steps. Elementary processes always follow simple rate laws, in which the order with respect to each reactant reflects the molecularity of the process (how many molecules are involved).
For example,

Unimolecular decomposition A → B ν = k [A]

Bimolecular reaction A + B → P ν = k [A][B]

A + A → P ν = k [A][A] = k [A]2

Multi-step processes may follow simple or complex rate laws, and as the above examples have hopefully illustrated, the rate law generally does not follow from the overall reaction equation. This makes perfect sense, since the overall reaction equation for a multi-step process is simply the net result of all of the elementary reactions in the mechanism. The ‘reaction’ given in the overall reaction equation never actually takes place! However, even though the rate law for a multi-step reaction cannot immediately be written down from the reaction equation as it can in the case of an elementary reaction, the rate law is a direct result of the sequence of elementary steps that constitute the reaction mechanism. As such, it provides our best tool for determining an unknown mechanism. As we will find out later in the course, once we know the sequence of elementary steps that constitute the reaction mechanism, we can quite quickly deduce the rate law.
Conversely, if we do not know the reaction mechanism, we can carry out experiments to determine the orders with respect to each reactant (see Sections 7 and 8) and then try out various ‘trial’ reaction mechanisms to see which one fits best with the experimental data. At this point it should be emphasized again that for multi-step reactions, the rate law, rate constant, and order are determined by experiment, and the orders are not generally the same as the stoichiometric coefficients in the reaction equation.
A final important point about rate laws is that overall rate laws for a reaction may contain reactant,product and catalyst concentrations, but must not contain concentrations of reactive intermediates (these will of course appear in rate laws for individual elementary steps).

4. The units of the rate constant


A point which often seems to cause endless confusion is the fact that the units of the rate constant depend on the form of the rate law in which it appears i.e. a rate constant appearing in a first order rate law will have different units from a rate constant appearing in a second order or third order rate law. This follows immediately from the fact that the reaction rate always has the same units of concentration per unit time, which must match the overall units of a rate law in which concentrations raised to varying powers may appear. The good news is that it is very straightforward to determine the units of a rate constant in any given rate law.Below are a few examples.

(i) Consider the rate law

ν = k[H2][I2].

If we substitute units into the equation,we obtain

(mol dm-3 s-1) = [k] (mol dm-3) (mol dm-3)

where the notation [k] means ‘the units of k’. We can rearrange this expression to find the units of the rate constant, k.

[k] =(mol dm-3 s-1) / (mol dm-3) (mol dm-3) = mol-1 dm3 s-1

(ii) We can apply the same treatment to a first order rate law,

for example

ν = k [CH3N2CH3].

(mol dm-3 s-1) = [k] (mol dm-3)

[k] = (mol dm-3 s-1) / (mol dm-3) = s-1

(iii) As a final example, consider the rate law

ν = k [CH3CHO]3/2.

(mol dm-3 s-1) = [k] (mol dm-3)3/2

[k] = (mol dm-3 s-1) / (mol dm-3)3/2 = mol-1/2 dm3/2 s-1

An important point to note is that it is meaningless to try and compare two rate constants unless they have the same units.

                                                                                                              ... continued