2HI (g)
HI (g)
Since the equilibrium constant expression Kc includes the coefficients, it should be obvious that it doesn't make a lot of sense to write K without also including the balanced equation to which it applies. For example each of the following is a perfectly legitimate way to balance the reaction between hydrogen, iodine and hydrogen iodide gases. Notice that the Kc expression for each is different, as is the mathematical value of K (you'll see how to calculate this in a later section).
| Balanced Equation | Kc Expression | Value of Kc at 400 ºC |
| H2 (g) + I2 (g) 2HI (g) |
|
30.0 |
| 1/2H2 (g) + 1/2I2 (g) HI (g) |
|
5.48 |
Fractional coefficients are handled just like regular ones – they become the power of the concentration in the Keq expression. A fractional coefficient means to take the root (1/2 is the square root).
Make sure to always include the balanced equation with every equilibrium constant you write.
What do we do with the energy term in a chemical reaction? To this point we've
just ignored it, but every chemical reaction has a 
H, and is either exo- or endothermic. For example, the
equation for the Haber process
is exothermic:
N2 (g) + 3H2 (g) 2NH3(g)
H = - 91.8 kJ
or
N2 (g) + 3H2 (g) 2NH3(g) + 91.8 kJ
From le Châtelier's principle we know that an increase in temperature will cause this reaction to shift to the left (and a decrease in temperature will shift to the right). How is this reflected in K expression?
Since energy is not a substance it is not included in the Keq expression. Thus for:
N2 (g) + 3H2 (g) 2NH3(g)+ 91.8 kJ 
just as it would be if there were no energy term in the equation.
However, energy does affect the numerical value of K. This can be seen in the following table, which lists the value of Kc for this reaction at various temperatures.
| Temperature (ºC) | Kc |
|
| 300 | 8.85 | |
| 400 | 0.61 | |
| 500 | 0.081 | |
| 600 | 0.015 |
There is an obvious decrease in K as the temperature increases.
In order to completely define the K expression you must state: