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[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. H2(g) + I2(g) ⇄ 2HI(g)

K= [HI]2 / [H2] [I2]

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= [HI]2 / [H2] [I2]

     = (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, Kc 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= [HI]2 / [H2] [I2]

     = x2/ x x

    =  x/ x2

    = 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 H2(g) + I2(g) ⇄ 2HI(g), equilibrium concentrations are:

H= 0.125 mol dm-3, I= 0.020 mol dm-3HI = 0.500 mol dm-3

K= [HI]2 / [H2] [I2]

     =  (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: H2(g) + I2(g) ⇄ 2HI(g) 

Starting amounts are: H2 = 0.35 mol, I= 0.25 mol and equilibrium amount of HI = 0.40 mol. Volume of 1 dm3.

Start by making an ICE table:

                    H2(g) + I2(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 H2 and Iamounts 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 H2 and I2 = -0.20 mol. To calculate equilibrium amount from here we simply add on this change to the initial amounts:

H2 = 0.35 + (-0.20) = 0.15 mol at equilibrium

I2 = 0.25 + (-0.20) = 0.05 mol at equilibrium

We can now plug these numbers into our table:

                             H2(g) + I2(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) ):

[H2] = 0.15 mol / 1dm= 0.15 mol dm-3

[I2] = 0.05 mol / 1dm= 0.05 mol dm-3

[HI] = 0.40 mol / 1dm= 0.40 mol dm-3


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

K= [HI]2 / [H2] [I2]

     =  (0.40)2 / (0.15) x (0.05)

     = 21.3 (no units)

Charlotte H. A Level Chemistry tutor, GCSE Chemistry tutor, A Level B...

3 years ago

Answered by Charlotte, an A Level Chemistry tutor with MyTutor

Still stuck? Get one-to-one help from a personally interviewed subject specialist


£30 /hr

Vicky H.

Degree: Dentistry (Masters) - Leeds University

Subjects offered:Chemistry, Science+ 12 more

Human Biology
English Literature
English Language
.BMAT (BioMedical Admissions)
-Personal Statements-
-Oxbridge Preparation-
-Medical School Preparation-

“Hi I'm Vicky, I'm a 2nd year dental student at the University of Leeds. I am friendly and patient and have a love for science, maths and teaching.”

£24 /hr

Eleanor H.

Degree: Biochemistry (Bachelors) - Bristol University

Subjects offered:Chemistry, Maths+ 1 more


“ABOUT ME: I am a biochemistry student, currently studying at the University of Bristol, and have had a passion for the natural sciences from a very young age. I am patient, hardworking and experienced, having tutored students in Maths...”

£26 /hr

Helen M.

Degree: Medicine (Bachelors) - Queen's, Belfast University

Subjects offered:Chemistry, Maths+ 6 more

Business Studies
-Personal Statements-
-Medical School Preparation-

“Durham law graduate and current first year medical student with four years tutoring experience.”

About the author

Charlotte H.

Currently unavailable: for new students

Degree: Chemistry (Masters) - Oxford, Hertford College University

Subjects offered:Chemistry, Science+ 2 more


“First year Chemistry student at Oxford University, looking to help students with GCSE and A level Biology, Chemistry and Maths.”

You may also like...

Other A Level Chemistry questions

How does the reactivity of group 2 elements change down the group, and what is the cause of this trend?

Why can endothermic reactions occur spontaneously if the entropy change is negative for a cooling process?

Dehydration reaction involving ethanol will produce ....

You have 3.51g of hydrated zinc sulphate. You heat up the zinc sulphate until all the water has evaporated from it. The weight after heating is 1.97g. Find how many H2O molecules per zinc sulphate molecule there are in the hydrated form of it.

View A Level Chemistry tutors

We use cookies to improve your site experience. By continuing to use this website, we'll assume that you're OK with this. Dismiss