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# 2.1 Is the equilibrium position unfavourable?

The first possibility is that the reaction system has been able to reach chemical equilibrium, but the equilibrium position is not favourable.

How does this come about?

Look again at Reaction (1).

2NO(g) + 2CO(g) N2(g) + 2CO2(g) (1)

This indicates that both the forward reaction:

2NO(g) + 2CO(g) ➜ N2(g) + 2CO2(g) (3)

and the backward reaction:

N2(g) + 2CO2(g) ➜ 2NO(g) + 2CO(g) (4)

are taking place: at the microscopic, molecular level there is continuous change in both directions.

However, at equilibrium, the reaction system then seems static because, at the macroscopic level where measurements are made, there is no apparent change in the amounts or concentrations of any of the four gases involved.

Suppose that Reaction (1) appears not to occur because, although it has reached equilibrium, the equilibrium position is unfavourable. Then it must be that the rates of the forward and backward reactions become equal when the concentrations of the reactants (NO and CO) are very high, and those of the products (N2 and CO2) are very small, so small as to be undetectable.

This possibility can be tested by examining the equilibrium constant, K, for the reaction.

◄Contents►Next: 2.2 The equilibrium constant

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