How to find equilibrium constant of a reaction from rate constant?
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Answer:
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Explanation:
The Mechanisms of Chemical Reactions
What happens when the first step in a multi-step reaction is not the rate-limiting step? Consider the reaction between NO and O2 to form NO2, for example.
2 NO(g) + O2(g) <----> 2 NO2(g)
This reaction involves a two-step mechanism. The first step is a relatively fast reaction in which a pair of NO molecules combine to form a dimer, N2O2. The product of this step then undergoes a much slower reaction in which it combines with O2 to form a pair of NO2 molecules.
Step 1: 2 NO <----> N2O2 (fast step)
Step 2: N2O2 + O2 <----> 2 NO2 (slow step)
The net effect of these reactions is the transformation of two NO molecules and one O2 molecule into a pair of NO2 molecules.
2 NO <----> N2O2
N2O2 + O2 <----> 2 NO2
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2 NO + O2 <----> 2 NO2
In this reaction, the second step is the rate-limiting step. No matter how fast the first step takes place, the overall reaction cannot proceed any faster than the second step in the reaction. As we have seen, the rate of any step in a reaction is directly proportional to the concentrations of the reactants consumed in that step. The rate law for the second step in this reaction is therefore proportional to the concentrations of both N2O2 and O2.
Step 2: Rate2nd = k(N2O2)(O2)
Because the first step in the reaction is much faster, the overall rate of reaction is more or less equal to the rate of this rate-limiting step.
Rate k(N2O2)(O2)
This rate law is not very useful because it is difficult to measure the concentrations of intermediates, such as N2O2, that are simultaneously formed and consumed in the reaction. It would be better to have an equation that related the overall rate of reaction to the concentrations of the original reactants.
Let's take advantage of the fact that the first step in this reaction is reversible.
Step 1: 2 NO <----> N2O2
The rate of the forward reaction in this step depends on the concentration of NO raised to the second power.
Step 1: Rateforward = kf(NO)2
The rate of the reverse reaction depends only on the concentrations of N2O2.
Step 1: Ratereverse = kr(N2O2)
Because the first step in this reaction is very much faster than the second, the first step should come to equilibrium. When that happens, the rate of the forward and reverse reactions for the first step are the same.
kf(NO)2 = kr(N2O2)
Let's rearrange this equation to solve for one of the terms that appears in the rate law for the second step in the reaction.
(N2O2) = (kf /kr) (NO)2
Substituting this equation into the rate law for the second step gives the following result.
Rate2nd = k (kf /kr) (NO)2(O2)
Since k, kf, and kr are all constants, they can be replaced by a single constant, k', to give the experimental rate law for this reaction described in Exercise 22.6.
Rateoverall Rate2nd = k'(NO)2(O2)
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The Relationship Between the Rate Constants and the Equilibrium Constant for a Reaction
There is a simple relationship between the equilibrium constant for a reversible reaction and the rate constants for the forward and reverse reactions if the mechanism for the reaction involves only a single step. To understand this relationship, let's turn once more to a reversible reaction that we know occurs by a one-step mechanism.
ClNO2(g) + NO(g) <----> NO2(g) + ClNO(g)
The rate of the forward reaction is equal to a rate constant for this reaction, kf, times the concentrations of the reactants, ClNO2 and NO.
Rateforward = kf(ClNO2)(NO)
The rate of the reverse reaction is equal to a second rate constant, kr, times the concentrations of the products, NO2 and ClNO.
Ratereverse = kr(NO2)(ClNO)
This system will reach equilibrium when the rate of the forward reaction is equal to the rate of the reverse reaction.
Rateforward = Ratereverse
Substituting the rate laws for the forward and reverse reactions when the system is at equilibrium into this equation gives the following result.
kf[NO][ClNO2] = kr[ClNO][NO2]
This equation can be rearranged to give the equilibrium constant expression for the reaction.
equation
Thus, the equilibrium constant for a one-step reaction is equal to the forward rate constant divided by the reverse rate constant.