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A Level Chemistry Chapter 23 Entropy and Gibbs free energy Detailed Summary Note £5.49
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A Level Chemistry Chapter 23 Entropy and Gibbs free energy Detailed Summary Note
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A Level Chemistry Chapter 23 Entropy and Gibbs free energy Detailed Summary Note including the topics - entropy changes, Hess cycle, calculating entropy change, and entropy equations

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  • September 17, 2024
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  • 2023/2024
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23 entropy and Gibbs free energy date


entropy (S) - the number of possible arrangements of the particles
and their energy in a given system
-system becomes more disordered entropy increases I




-increase in entropy means system becomes energetically more stable
entropy changes
-simpler substances with fewer atoms have lower entropy values
than complex substances with more atoms
-harder substances have lower entropy than softer substances of
the same type (ex. diamond graphite)
1) change in state 2) change in reactions
-entropy increases -gas liquid solid -CaCO (s) CaO(s) CO (g) N
Y t
2



more significantly CO gas is more disordered than 2



when liquid becomes solids so entropy increases
gas bcs particles -N (g) 3H (g) 2NH (g) 2
t
2
-
3



become very there are 4 gas molecules in
disordered (compared reactants and 2 gas molecules in
to solid liquid)
>
products so entropy decreases
-if total entropy is entropy changes in exothermic and endothermic reactions
positive feasible
-




exothermic reaction - energy released to the surroundings
increases the number of ways of arranging the energy, so likely
to be an increase in entropy and increased probability of the
chemical change occurring spontaneously (more feasible)
endothermic reaction - energy is absorbed from the surroundings
decreases the number of ways of arranging the energy, so likely
to be a decrease in entropy and reaction becomes less feasible

calculating entropy changes unit JK mol I I
- -




·




1
SSystem Sproducts ["reactants
=
S
1



ex) 2Ca(s) O (g) 2CaO(s)
1



S (Ca) 41.40 JK mol
⑧
-
-




I & -

2



SSystem Sproducts S
O




S (O ) 205.0 JK mol
O -
1 -
1


1 =I [ reactants 2
=




(2 39.70) - (2 41.40 205.0) S (CaO) 39.70 JK mol
⑧
-
I
-

I
= X X I =




-208.4 JK mol
-
I -
1

-




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