International A Level (IAL)ChemistryEdexcelRevision Notes4: Rates, Equilibria & Further Organic ChemistryChemical EquilibriaEquilibrium Constant & Entropy

Equilibrium Constant & Entropy (Edexcel International A Level (IAL) Chemistry): Revision Note

Exam code: YCH11

Written by: Stewart Hird

Reviewed by: Caroline Carroll

Updated on 24 October 2026

Equilibrium Constant & Entropy

The equation for calculating the total entropy change is:

ΔS^ꝋ_total = ΔS ^ꝋ_sys + ΔS^ꝋ_surr

(sys = system and surr = surroundings)

Remember that ΔS^ꝋ_total is positive for all spontaneous changes
There is very little change in ΔS^ꝋ_sys with a change in temperature unless there is a change in the state of one of the reactants or products
There will be a significant change in ΔS^ꝋ_surr however
The entropy change of the surroundings during a chemical reaction is ΔS_surroundings = -ΔH/T

Where ΔH is the enthalpy change and T is the absolute temperature (measured in kelvin)
We can use this information to determine whether a reaction is spontaneous at a given temperature

Worked Example

Is the decomposition of calcium carbonate into calcium oxide and carbon dioxide spontaneous at the given temperatures?

CaCO₃(s) → CaO(s) + CO₂(g) ΔH = +177.9 kJ mol^-1 ΔS_sys = +160.4 J K^-1 mol^-1

293K (20 °C)
1173K (900 °C)

Answer 1: at 293 K (20°C)

ΔS_surroundings= $- \frac{+ 177900 J {mol}^{- 1}}{293 K}$ = -607.2 J K^-1 mol^-1
ΔS_total(293K) = (+160.4 - 607.2) = -446.8 J K^-1 mol^-1
The decomposition of calcium carbonate is not spontaneous at 293K

Answer 2: at 1173K (900°C)

ΔS_surroundings= $- \frac{+ 177900 J {mol}^{- 1}}{1173}$ = -151.7 J K^-1 mol^-1
ΔS_total(1173) = (+160.4 - 151.7) = +8.7 J K^-1 mol^-1
The decomposition of calcium carbonate is spontaneous when heated to 1173K

Relationship between entropy change and equilibrium constant

For a reversible reaction that can reach equilibrium, the equilibrium position can be reached from either side of the reaction
This means that the change towards equilibrium is spontaneous from either side
- ΔS_total is positive for any mixture moving towards the equilibrium position, whether it starts from reactants or from products
For example, consider the following reaction:

N₂O₄ (g) ⇌ 2NO₂ (g)

Graph of entropy against percentage of NO2 in a mixture of N2O4 and NO2: entropy rises from pure N2O4 to a maximum at the equilibrium mixture and falls towards pure NO2, with the entropy change forward and backward to the equilibrium mixture both positive

A graph of entropy against the percentage of NO₂ in a mixture of N₂O₄ and NO₂

The entropy of N₂O₄ is less than the entropy of the equilibrium mixture
The change in entropy from pure N₂O₄ to the equilibrium mixture is positive
- The change is spontaneous
The entropy change for NO₂ to the equilibrium mixture is also positive
- This change is also spontaneous
The entropy change for a mixture of the gases in any proportions moving towards the equilibrium position is also positive
Neither the forward nor backward reaction can go to completion as the entropy change from the equilibrium mixture to either the reactants or products is negative
At equilibrium, the total entropy change for any further change is zero

ΔS_total [forward reaction] = ΔS_total [backward reaction]

The relationship between the total entropy of the reaction and the equilibrium constant (K_c or K_p) is

ΔS_total = R lnK

Using total entropy change to calculate an equilibrium constant

Rearranging the equation mentioned above we get:

lnK = $\frac{∆ S_{t o t a l}}{R}$

Hence:

K = $e^{\frac{∆ S_{t o t a l}}{R}}$

Relationship between equilibrium constant and equilibrium position

There is no hard rule for the relationship between equilibrium constant and the position of equilibrium
As a general rule, we can say a very large value of K suggests the equilibrium position is pushed towards the products (right-hand side)
Similarly, a very small value of K suggests the equilibrium position is pushed towards the reactants (left-hand side)

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

I would just like to say a massive thank you for putting together such a brilliant, easy to use website.I really think using this site helped me secure my top gradesin science and maths. You really did save my exams! Thank you.

Beth
IGCSE Student

This website is soooo useful and I can’t ever thank you enough for organising questions by topic like this. Furthermore, the name of the website could not have been more appropriate as it literally did SAVE MY EXAMS!

Fathima
A Level Student

Incredible! SO worth my money, the revision notes have everything I need to know and are so easy to understand. I actually enjoy revising! It makes me feel a lot more confident for my GCSEs in a few months.

Kate
GCSE Student

Absolutely brilliant, both my girls used it for A levels and GCSE. It's saves on paper copies, also beneficial exam questions ranked from easy to hard. It's removed a lot of stress from the exams.

Sameera
Parent

Just to say that your resources are the best I have seen and I have been teaching chemistry at different levels for about 40 years

Mark
Chemistry Teacher

Excellent