Entropy Change, ∆S (Oxford AQA International A Level Chemistry)
Revision Note
Written by: Philippa Platt
Reviewed by: Stewart Hird
Entropy Change, ∆S
In endothermic reactions the products are in a less stable, higher energy state than the reactants
So, why do endothermic reactions take place?
The majority of chemical reactions we experience everyday are exothermic
But, ΔHꝋ alone is not enough to explain why endothermic reactions occur
Chaos in the universe
The entropy (S) of a given system is the number of possible arrangements of the particles and their energy in a given system
In other words, it is a measure of how disordered or chaotic a system is
When a system becomes more disordered, its entropy will increase
An increase in entropy means that the system becomes energetically more stable
Entropy of a chemical reaction
Using the thermal decomposition of calcium carbonate (CaCO3) as an example:
CaCO3 (s) → CaO (s) + CO2 (g)
In this reaction:
One reactant molecule forms two product molecules
Two molecules are more disordered than one
A solid molecule forms a solid molecule and a gas molecule
The gas molecule is more disordered than the solid reactant (CaCO3), as it is constantly moving around
As a result, the system has become more disordered and there is an increase in entropy
Entropy changes for different chemical reactions are different because they depend on the number of reactants and products
Entropy of a state change
Using the melting of ice to water as an example:
H2O (s) → H2O (l)
In this example:
The water molecules in ice are in fixed positions and can only vibrate about those positions
They are not very disordered
The water molecules in liquid water are still quite close together but are arranged randomly and can move around each other
Therefore, water molecules in the liquid state are more disordered
For any given substance, entropy increases when:
Its solid state melts into a liquid
Its liquid state boils into a gas
For any given substance, entropy decreases when:
Its liquid state freezes into a solid
Its gaseous state condenses into a liquid
For chemical reactions and state changes, the system with the higher entropy will be energetically favourable
This is because the energy of the system is more spread out when it is in a disordered state
Feasible or spontaneous reactions
Chemists talk about reactions being feasible or spontaneous
This means that reactions take place of their own accord
In other words, a reaction is energetically favourable
This is an outcome of the second law of thermodynamics which broadly states the the entropy of the universe is always increasing
We can see examples of this all around us:
Hot objects always cool and spread their heat into the surroundings, never the other way around
Feasibility does not consider the rate of reaction
Feasibility states what is possible, not what actually happens
A feasible reaction might be incredibly slow, such as the rusting of iron
Calculating entropy changes
Entropy changes are an order of magnitude smaller than enthalpy changes, so entropy is measured in joules rather than kilojoules.
The full unit for entropy is J K-1 mol-1
The standard entropy change (ΔS) for a given reaction can be calculated using the standard entropies (S) of the reactants and products
The equation to calculate the standard entropy change of a system is:
ΔSꝋ = ΣSproductsꝋ - ΣSreactantsꝋ
For example, the formation of ammonia from nitrogen and hydrogen:
N2 (g) + 3H2 (g) ⇋ 2NH3 (g)
The standard entropy change for this reaction is:
ΔSsystemꝋ = (2 x ΔSꝋ(NH3)) - (ΔSꝋ(N2) + 3 x ΔSꝋ(H2))
NOTE: The standard entropies for each chemical in the reaction must be multiplied by its stoichiometric coefficient
Unlike enthalpy of formation for elements, entropy for elements is not zero
Entropy values for elements and compounds are found in data books
Worked Example
Calculate the entropy change of the system for the following reaction:
2Mg (s) + O2 (g) → 2MgO (s)
Sꝋ[Mg(s)] = 32.60 J K-1 mol-1
Sꝋ[O2(g)] = 205.0 J K-1 mol-1
Sꝋ[MgO(s)] = 38.20 J K-1 mol-1
Answer:
ΔSsystemꝋ = ΣΔSproductsꝋ - ΣΔSreactantsꝋ
ΔSsystemꝋ = (2 x 38.20) - (2 x 32.60 + 205.0) = -193.8 J K-1 mol-1
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