# 2.2 The equilibrium constant

An expression for the equilibrium constant of a reaction can be put together from the concentrations of the reactants and products at equilibrium.

Chemists represent theconcentration of a reactant or product by enclosing its chemical formula in square brackets.

Thus, the concentration of NO(g) is written [NO(g)].

To write down the equilibrium constant of a reaction, you need to start with the concentrations of the products. Each one is raised to the power of the number that precedes it in the reaction equation, and the corresponding terms for each product are then multiplied together.

Do this now for the products of the equilibrium system (2)

The result is [N

_{2}(g)] × [CO_{2}(g)]^{2}, or, taking the multiplication sign as read, [N_{2}(g)][CO_{2}(g)]^{2}. In Equation (1), CO_{2}(g) is preceded by a two, so its concentration is squared.

So that’s the products dealt with, now for the reactants.

Repeat the operation for the reactants in Equation (2)

The result is [NO(g)]

^{2}[CO(g)]^{2}, both NO(g) and CO(g) are preceded by a two.

The equilibrium constant, *K*, is obtained by dividing the result for the products by the result for the reactants:

So returning to the earlier question, is the equilibrium position unfavourable?

In other words Reaction (1) does not happen because the equilibrium position for equilibrium system (2) lies well over to the left. As stated above, at equilibrium, the concentrations of NO(g) and CO(g) are very high, and those of N_{2} (g) and CO_{2}(g) are so small as to be undetectable.

If so, will

*K*be large or smallIt will be very small because the large quantities ([NO(g)] and [CO(g)]) occur on the bottom of Equation (1), and the small quantities ([N

_{2}(g)] and [CO_{2}(g)]) occur on the top.

Actally, the value of *K* can be determined experimentally. Assuming a typical temperature in a car exhaust system is 525 °C. At this temperature, *K* turns out to be 10^{40} mol^{−1} litre.

Given this information, does the equilibrium position lie to the left or right of Equation (2)?

*K*is__immense__, so at equilibrium, the concentrations of the products must be much greater than those of the reactants (which appear on the bottom). The equilibrium position for Reaction 7.2 at 525 °C therefore lies well over to the right.

Note: the units here are mol^{-1} litre for this equilibrium, mol is shorthand for moles, which as you know is a measure of an amount of substance**. **

So up to this point, the issue of why the polluting gases NO and CO predominate hasn’t been addressed, as the equilibrium is in fact favourable.

Another factor must be coming into play.

This is considered in the next section.