Difference between equillibream limiting reaction and rate limiting reaction
Answers
Understanding how the different steps in a mechanism contribute to the overall rate of a reaction can be complicated if we demanded to know the rate with exquisite precision. However, if we simply want to get an understanding of the rate that is nearly (but not perfectly) correct, we can make an important simplification: the rate of the overall reaction is determined by the rate of the slowest step in the mechanism. The slowest step in the mechanism is called the rate determining step or rate limiting step.
Generally with any process, there is one key bottle neck (slow step) that is controlling the rate of the entire process. The same is true in a chemical mechanism. The overall rate of a reaction will depend on this rate limiting step as well as the steps that precede this step.
For example, look at the reaction
2NO(g)+O2(g)→2NO2(g)
The mechanism for this reaction is
NO(g)+NO(g)⇌N2O2(g)fast,equilibrium
N2O2(g)+O2→2NO2(g)slow
The second step in this reaction is the rate determining step as this is the "slow" step. This is a bimolecular step involving the collision of a reactant O2 and the intermediate N2O2. The rate is given by
rate=k2[O2][N2O2]
But, the empirical rate law should not contain any intermediates. Since the concentration of N2O2 depends on the concentrations of the reactants and the equilibrium constant for this step, the intermediate can be replaced in the rate law. For the fast equilibrium step, the concentrations of the products and reactants are related by the equilibrium constant
K1=[N2O2][NO]2
where K1 is the equilibrium constant for the reaction in the first step (Note: it is critical to keep track of the notation. Uppercase K for equilibrium constant. Lowercase k for rate constant). From this the concentration of N2O2 can be written in terms of the concentrations of the reactants and the equilbrium constant: [N2O2] = K1[NO]2. Plugging this in for the concentration of N2O2 in rate law for the slow step yields
rate=k[O2][NO]2
where is the rate constant "k" is a constant that contains the rate constant for the second step as well as the equilibrium constant for the first step: k = k2K1.
We can also write the rate law expression only in terms of rate constants. This is because the equilibrium constant can be expressed in terms of the rate constants for this elementary step. This is a very important relationship that utilizes the idea of dynamic equilibrium. At equilibrium, the rates of the forward reaction and the backward reaction are equal. For the example, reaction the forward rate is a bimolecular elementary step.
rateforward=k1[NO]2
The backward rate for this step is a unimolecular reaction with a rate
ratebackward=k−1[N2O2]
Where the rate constant for the backward step is given the notation k-1 to show that it is the reverse of step 1. Since the forward rate and the backward rate are equal
rate=k1[NO]2=k−1[N2O2]
Combining this with the relationship for the concentrations at equilibrium yields.
K=[N2O2][NO]2=k1k−1
The equilibrium constant for this fast equilibrium step is the ratio of the forward rate constant to the backward rate constant. This is a very important idea in chemistry and it unifies our ideas about kinetics and equilibria.