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:
Nitrogen(g) + hydrogen(g) ⇌ ammonia(g)
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.
Practice questions
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