 ### The equilibrium constant, Kc

A reversible reaction is one that does not go to completion, this means that we end up with an mixtur of reactant and products. Given enough time a reversible reaction in a closed system will reach equilibrium. We used Le Chatelier's principle to predict what would happen to the amounts of reactant and products in an equilibrium mixture if we changed the temperature, pressure or concentration of one of the reactants or products. However Le Chatelier's principle cannot give exact values for example on how the concentration of a particular reactant or product will change, that is where equilibrium constants come into play.

### Expressions for the equilibrium constant, Kc

Consider a reaction that we are all familar with, the Haber process for the synthesis of ammonia gas:

##### N2(g) + 3H2(g) ⇌ 2NH3(g)

If we filled lots of flasks with different amounts of nitrogen, hydrogen and ammonia and waited until all the flask were at equilibrium we could measure the concentration of each of the reactants and products and look for any relationship between them. Well, the relationship which was found can be summarised as follows.
For any reversible reaction at equilibrium, such as:

##### aA(l) + bB(l) ⇌ cC(l) + dD(l)
Where A,B,C and D are the reactants and product, while a, b, c and d are the coefficients in the balanced equation for the reaction, then we can say that the equilibrium constant is written as: Kc is the equilibrium constant and the subscript c refers to concentrations in mol dm-3. The [] are used to indicate that we are using concentrations in mol dm-3 for all of the reactants and products. The equilibrium constant is a ratio of the concentration of the products over the reactants, with each reactant and product raised to the power of the co-efficients in the balanced equation for the particular reaction in question. So for the Haber process we could write an expression for the equilibrium contant as:

##### Equation for Haber Process: N2(g) + 3H2(g) ⇌ 2NH3(g) The equilibrium constant is easily written from the balanced symbolic equation, for example consider the following reactions:
##### Equation for formation of sulfur trioxide from sulfur dioxide: 2SO2(g) + O2(g) ⇌ 2SO3(g) ##### Equation for formation of hydrogen iodide: H2(g) + I2(g) ⇌ 2HI(g) From the examples above you should note that the concentration of the products in mol dm-3 appears on the top line or numerator of the expression for the equilibrium constant. The denominator or base line of the equilibrium constant expression is where the concentration of the reactants in mol dm-3 are written. Each concentration term for the reactants and product is raised to power of the number in front of its formula in the balanced equation, that is the number of moles present.

### The units of the equilibrium constant

The units of the equilibrium constant depend upon the stoichiometry (number of reacting moles) of the particular equation. For example consider the following reaction:

##### A(g) + B(g) ⇌ C(g)
An expression for the equilibrium constant with units is shown below. Rememeber that for Kc all the concentration must be in moles per cubic decimetre (mol dm-3). To get the final units of the equilibrium constant simply cancel out similar units on the top and bottom lines of the equilibrium expression as shown below: Example 2: Consider the following reaction:

##### 2A(g) + B(g) ⇌ C(g)
This time in the equilibrium expression for A we have [A]2, all we do here is square the units of concentration so we get:
(mol dm-3)(mol dm-3) = (mol2 dm-6). As before we simply cancel out identical units from the top and bottom lines of the equilibrium expression. The expression for Kc is: Example 3: Consider the following reaction:
##### A(aq) + B(aq) ⇌ C(aq) + D(aq)
The expression for Kc is: ### The values of the equilibrium constants

Since the value of the equilibrium constant, Kc is the ratio of the concentration of the products over the reactants its value will given an excellent indication as to whether there are more reactants or products present at equilibrium. The values of kc vary tremendously, from very large numbers, for example 1 x 1050 to very small number such as 1 x 10-20. These numbers give us an excellent glimpse as to what is happening at equilibrium:

• If Kc is a large number such as 100 or a huge number such as 1 x 1010 then there are much more products than reactants in the equilibrium mixture. The larger the value of Kc then the more products and the less reactants are present in the equilibrium mixture. We would say that the position of equilibrium lies very much to the right (towards the products)
• If Kc is a small number, such as 0.0011 or 1x10-5 then there is much less product than reactant present in the equilibrium mixture. The smaller the value of Kc the more reactant and the less products is present in the equilibrium mixture. We would say that the position of equilibrium lies to the left (towards the reactants).
• If Kc has a value which is not particularly large, maybe even close to 1 then it is likely that there is approximately equal amounts of reactants and products in the equilibrium mixture.

This is summarised in the diagram below: ### Equilibrium constants and Le Chatelier's principle

You can use Le Chatelier's principle to predict what will will happen to the value of Kc when the reaction conditions are altered by changing say the temperature, concentration or pressure.

For example consider what happens to the value of Kc for a reaction when the temperature is changed. Remember that Kc is the ratio of [products]/[reactants], so when the reaction conditions are changed ask yourself what happens to the concentration of the reactants and products, from this you should be able to explain what happens to the value of Kc. As an example consider the reaction shown below:

##### A(aq) + B(aq) ⇌ C (aq)ΔH= -ve
Here the forward reaction is exothermic, this means that the back reaction will be an endothermic reaction. So what would happen if this reaction was at equilibrium and we changed the temperature?
• If the temperature is increased then according to Le Chatelier's principle the system will oppose any change made to it. This means that the position of equilibrium will shift in such a way as to reduce the temperature. That is turning some of the product C into reactants A and B, since this is an endothermic reaction. This means that the value of the equilibrium constant, Kc will decrease.
• Using the same argument if the temperature is reduced then the position of equilibrium will adjust in such a way as to oppose the temperature reduction, that is more reactants will be turned into products, since the forward reaction is exothermic. This means that the value of Kc will increase.

You should be able to make similar arguments to explain what will happen to the value of Kc when the concentration or pressure of the reactants or products are changed. This is discussed in detail on the page which deals with Le Chatelier's principle.

### Key Points

• The equilibrium constant Kc gives an indication of the amounts of the reactants and products in an equilibrium mixture.
• The values of Kc vary enormously, very large values for Kc indicate that the reaction maybe thought of as going to completion, that is almost all products present and next to no reactants. The oppsite would be true for very small values of Kc.
• The units of Kc vary and need to be worked out from the expression for the equilibrium constant.