Law of Mass action VS Rate law

[Raghav Dua]

Law of mass action have 2 parts-equilibrium part & reaction rate part

The reaction rate part states that rate of a reaction is directly proportional to the product of active masses of the reactants with each reactant term raised to it's stoichiometric coefficient. The rate law states that rate of a reaction is directly proportional to the product of active masses of the reactants with each reactant term raised to some power which may or may not be equal to stoichiometric coefficient of the reactant.

 

Why the two laws are different when we talk about power to which active masses of reactants are raised ? Which law is correct ?

Edited November 6, 2013 by Raghav Dua

[studiot]

The rate law is determined experimentally and so is the 'correct' one in the sense that it is what actually happens.

 

However that does not mean that the simple expression obtained from the stochiometric equation is useless. In fact comparison with the experimentally determined rate law tells us when something more complicated, such as intermediate reations, is going on. It also depends whether you insist on whole numbers in your stoichiometry.

 

Take, for instance, a reaction that requires two different molecules A and B to meet and react and form products C and D. Then the rate of reaction will depend upon the presence of both A and B so the assumption is that

 

A + B = C + D (stochiometry)

 

v = k [A]

 

ie is second order.

 

But suppose that the reaction is actually a decomposition

 

2N₂O₅ = 4NO₂ + O₂

 

experimentally rate equation is

 

v = k[N₂O₅]

 

not v = k[N₂O₅]²

 

This is only first order. But of course the dinitrogen pentoxide molecule does not need to meet another to decompose. The stochiometric equation could in fact be written in terms of a single reactant molecule and fractions for the products.

[Raghav Dua]

The rate law is determined experimentally and so is the 'correct' one in the sense that it is what actually happens.

However that does not mean that the simple expression obtained from the stochiometric equation is useless. In fact comparison with the experimentally determined rate law tells us when something more complicated, such as intermediate reations, is going on. It also depends whether you insist on whole numbers in your stoichiometry.

Take, for instance, a reaction that requires two different molecules A and B to meet and react and form products C and D. Then the rate of reaction will depend upon the presence of both A and B so the assumption is that

A + B = C + D (stochiometry)

v = k [A]

ie is second order.

But suppose that the reaction is actually a decomposition

2N₂O₅ = 4NO₂ + O₂

experimentally rate equation is

v = k[N₂O₅]

not v = k[N₂O₅]²

This is only first order. But of course the dinitrogen pentoxide molecule does not need to meet another to decompose. The stochiometric equation could in fact be written in terms of a single reactant molecule and fractions for the products.

Hey thank u for making my concepts more clear and sorry for late reply.

 

Edited November 11, 2013 by Raghav Dua