Chapter 1: Chemical Equilibrium Part 11 SABIS Grade 11 (Level M) Chemistry


1.8 The Law of Equilibrium from Rates of Reactions

The Law Of Equilibrium From Rates Of Reactions


2NO(g) + O2(g) → 2NO2(g)       (1)
The reaction to the right, (R), proceeds with a rate that is found experimentally to depend upon the concentrations of the reactants as follows:
(rate)R = kR[NO]2[O2]                 (2)
The reverse reaction to the left, (l), has been studied as well. The rate of this reaction
        2NO(g)             +              O2(g)              ←               2NO2(g)             (3)

depends upon the concentrations as follows:
(rate)L = kL[NO2]2                        (4)
Expressions (2) and (4) show how the rates of reaction (1) and its reverse, reaction (3), depend upon the concentrations. Now we can apply our microscopic view of the equilibrium state. Chemical changes will cease (on the macroscopic scale) when the rate of reaction (1) is exactly equal to that of reaction (3). In this case, we can equate expressions (2) and (4):
(rate)R = (rate)L
or
kR [NO]2. [O2] = kL [NO2]2
By algebraic rearrangement, we write:
kRkL=[NO2]2[NO]2[O2]            (5)

Since both kR and kL are constants at a given temperature, their ratio is constant. Hence (5) is the equilibrium law expression for the equilibrium:
2NO(g) + O2(g) ⇌ 2NO2(g)
and the equilibrium constant is K=kRkL 

Thus we see that the experimental rate laws for this reaction and its reverse lead to the equilibrium law. In every reaction that has been sufficiently studied, this same result is obtained. We are led to have confidence in the molecular view of equilibrium as a dynamic balance between opposing reactions.




Next part