# How do you calculate the equilibrium constant, Kc, of a reaction?

Given a reaction in the form:

aA + bB ⇄ cC + dD

where A and B are reactants, C and D are products and a, b, c and d are stoichiometric coefficients (relative number of moles of each molecule).

The equilibrium coefficient is given by:

K_{c }= [C]^{c}[D]^{d} / [A]^{a}[B]^{b}

i.e. The concentration of each product raised to the power of its stoichiometric coefficient, divided by the concentration of each reactant raised to the power of its stoichiometric coefficient. **Note: it's the concentration of the products over reactants, not the reactants over products.**

e.g. H_{2}(g) + I_{2}(g) ⇄ 2HI(g)

K_{c }= [HI]^{2} / [H_{2}] [I_{2}]

__Equilibrium constant units:__

The method for calculating units is similar to very simple algebra (though don't be put off if you never got on with this in school, it really is very simple). You can think of the units of concentration, mol dm^{-3}, as 'x'.

The first step is to replace the molecules in the square brackets with the units of concentration. Using our previous example:

K_{c }= [HI]^{2} / [H_{2}] [I_{2}]

= (mol dm^{-3})^{2} / (mol dm^{-3})(mol dm^{-3})

You can see that the numerator, (mol dm^{-3})^{2}, is the same as the denominator, (mol dm^{-3})(mol dm^{-3}) and so all the units cancel. Therefore in this case, K_{c} has no units.

It may be easier to understand this if you think of mol dm^{-3} as 'x' and so instead of replacing the molecules in the square brackets with mol dm^{-3}, you replace them with x:

K_{c }= [HI]^{2} / [H_{2}] [I_{2}]

= x^{2}/ x x

= x^{2 }/ x^{2}

=^{ }no units

__Calculating the equilibrium constant:__

If you the information in the question gives you the equilibrium concentration then this step is very easy. **Note: the information must specify equilibrium concentration and not starting concentration.**

To do the calculation you simply plug in the equilibrium concentrations into your expression for Kc. For example for H_{2}(g) + I_{2}(g) ⇄ 2HI(g), equilibrium concentrations are:

H_{2 }= 0.125 mol dm^{-3}, I_{2 }= 0.020 mol dm^{-3}, HI = 0.500 mol dm^{-3}

K_{c }= [HI]^{2} / [H_{2}] [I_{2}]

= (0.500)^{2} / (0.125) x (0.020)

= 100 (no units)

If the question gives you starting concentrations and equilibrium concentrations of one component rather than equilibrium concentrations of all components, you can calculate equilibrium concentrations using an ICE table (initial amount/mol, change in amount/mol, equilibrium amount/mol. For each component of reaction). **Note: you must convert starting concentration to amounts (in moles) using container volume before putting them into the table. n = (concentration) x (volume)**

e.g. For: H_{2}(g) + I_{2}(g) ⇄ 2HI(g)

Starting amounts are: H_{2} = 0.35 mol, I_{2 }= 0.25 mol and equilibrium amount of HI = 0.40 mol. Volume of 1 dm^{3}.

Start by making an ICE table:

H_{2}(g) + I_{2}(g) ⇄ 2HI(g)

Initial/mol 0.35 0.25 0

Change/mol 0.40 (we know that we have gained 0.40mol HI)

Equilibrium/mol 0.40 (at this point we only know equilibrium amount of HI)

From here we know that HI has changed by +0.40 mol and also that there are two moles of HI in the overall equation. This means that H_{2} and I_{2 }amounts must change by half that. Also because they are our reactants, we must be losing amounts of them, so their changes in amount will be negative. So changes in H_{2} and I_{2} = -0.20 mol. To calculate equilibrium amount from here we simply add on this change to the initial amounts:

H_{2} = 0.35 + (-0.20) = 0.15 mol at equilibrium

I_{2} = 0.25 + (-0.20) = 0.05 mol at equilibrium

We can now plug these numbers into our table:

H_{2}(g) + I_{2}(g) ⇄ 2HI(g)

Initial/mol 0.35 0.25 0

Change/mol -0.20 -0.20 +0.40

Equilibrium/mol 0.15 0.05 0.40

Now we convert our equilibrium amounts into concentrations (using

n = (concentration) x (volume) --> (concentration) = n / (volume) ):

[H_{2}] = 0.15 mol / 1dm^{3 }= 0.15 mol dm^{-3}

[I_{2}] = 0.05 mol / 1dm^{3 }= 0.05 mol dm^{-3}

[HI] = 0.40 mol / 1dm^{3 }= 0.40 mol dm^{-3}

Finally all we need to do is put these equilibrium concentrations into our expression for equilibrium constant:

K_{c }= [HI]^{2} / [H_{2}] [I_{2}]

= (0.40)^{2} / (0.15) x (0.05)

= 21.3 (no units)