Introduction to Entropy (College Board AP® Chemistry)
Study Guide
Written by: Alexandra Brennan
Reviewed by: Stewart Hird
Entropy & Phase Changes
Entropy (S) in a system reflects the variety of ways particles and their energy can be arranged
It is a measure of the system's disorder or randomness
When a system becomes more disordered, its entropy goes up
This rise in entropy signals increased energetic stability
Take the thermal decomposition of calcium carbonate (CaCO3) as an example:
CaCO3(s) → CaO(s) + CO2(g)
The creation of a gas molecule (CO2) adds disorder to the system
The production of two product molecules compared to one reactant molecule adds disorder to teh system
The constant movement of CO2 molecules makes them more disordered than the original solid (CaCO3)
This results in a higher overall entropy
Consider another example: when a solid transitions to a liquid, like the melting of ice:
H2O(s) → H2O(l)
In the solid state, water molecules in ice are fixed in position with limited movement
In the liquid state, these particles are more randomly arranged, allowing for freer motion
As a result, the liquid state is more disordered, leading to an increase in entropy compared to the solid state
In both cases, systems with higher entropy are energetically preferred because a disordered state allows for a more evenly spread distribution of energy
Increasing entropy in a system
Melting a solid will cause the particles to become more disordered resulting in a higher entropy state
For reactions involving reactants and products that are both in the gas phase, the entropy generally increases when the total number of moles of gas-phase products is greater than the number of moles of gas-phase reactants
Worked Example
In which reaction would you expect the largest decrease in entropy?
CH4 (l) + 2O2 (g) → CO2 (g) + 2H2O (g)
CH4 (g) + 2O2 (g) → CO2 (s) + 2H2O (g)
CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (g)
CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (l)
Answer:
All four reactions have the same number of molecules as reactants and products, which means that the number of molecules will not be a consideration for this question
Reaction 1 will show an increase in entropy as one of the reactants is a liquid but all products are gases
Therefore, reaction 1 cannot be the correct answer
Reaction 2 will show a decrease in entropy as the three gaseous reactants form one solid and two gaseous products
Reaction 3 will show no change in entropy as all of the reactants and products are gases
Therefore, reaction 3 cannot be the correct answer
Reaction 4 will show a decrease in entropy as the three gaseous reactants form one gaseous and two liquid products
Between reactions 2 and 4:
Reaction 4 will have the biggest decrease in entropy as there is a bigger change in the disorder of the particles for this reaction
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