1. Week 24
Explain that entropy is a measure of the disorder of a system, and that a system becomes energetically more stable when it becomes more disordered.
Explain: the difference in entropy of a solid and a gas; the change when a solid lattice dissolves; the change in a reaction in which there is a change in the number of gaseous molecules.
Calculate the entropy change for a reaction given the entropies of reactants and products.
© Pearson Education Ltd 2009
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2. Entropy S
Entropy is the quantitative measure of the degree of the disorder of a system.
An alternate description for entropy is the dispersal energy.
At absolute zero a perfectly ordered pure crystal will have zero entropy.
Otherwise since particles are in constant motion S is always a positive number.
The entropy of a ‘chemical system’ is the distribution of energy (disorder) within the chemicals present, either as reactants, products or different states of matter.
It is possible to calculate absolute standard entropy values SӨ and hence standard entropy changes Δ SӨ .
As disorder increases so does the entropy value.

3. Week 24 Increasing entropy 3 © Pearson Education Ltd 2009
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4. What can entropy tell us?
The increase in entropy relates to how many ways the particles can be arranged AND how many ways the energy of the system can be distributed between the particles.
In nature entropy tends to increase.
Disordered systems are more likely than ordered systems.
To impose order requires input of energy (think bedrooms!)
Generally speaking a reaction is favoured if entropy increases.
In an increasingly disordered system energy changes from being localised to spread out.

5. Increasing Entropy This leads to an increase in energetic stability.
Entropy increases whenever particles become more disordered.
What is the effect of increasing temperature on Entropy?
Increasing temperature increases the movement of particles and so increases entropy.
Change of state solid → liquid → gas increases randomness and S↑.
Dissolving increases disorder so S↑.
When a gas is evolved S↑.
Increasing the number of gas molecules S↑

6. Standard Entropy Change Δ SӨ
Standard Entropy Change Δ SӨ of a reaction is the entropy change that accompanies a reaction in the molar quantities expressed in the chemical equation under standard conditions, all reactants and products being in their standard states.
Standard entropy values are supplied in tables of data and have the units JK-1mol-1.
Note the value in J not kJ. Entropy values are small.
Δ SӨ =Σ SӨ(products)- ΣSӨ (reactants)
This is also written Δ SӨ = SӨ final- SӨ initial
If a change makes a system more random Δ S is positive.
If a change makes a system more ordered Δ S is negative.

7. Use this! S /JK-1mol-1 H2(g) 131 diamond 2.4 O2(g) 205 graphite 5.7
N2(g)
192
HCl(g)
187
Cl2(g)
223
HNO3(l)
156
H2O(g)
189
NO2(g)
240
H2O(l) 70
NaCl(s) 72

8. Calculation For the reaction: 2H2(g) + O2(g) → 2H2O(l)
Use the previous data to work out a value for the Δ SӨ
Δ SӨ =Σ SӨ(products)- ΣSӨ (reactants)
Σ SӨ(products) = 2x 70 = 140
ΣΔSӨ (reactants) = (2x 131) = 467
Δ S= 140 – 467 = -327 J K-1mol-1
What information does this value give you?
That the system has become more ordered as the gas molecules have reacted to give liquid molecules. Is this likely?
This reaction should not take place since entropy is supposed to increase when reactions take place.

9. Calculation For the reaction: 4HNO3(l) → 4NO2(g) + O2(g) + 2H2O(l)
Repeat the above calculation.
Ans = +681 J K-1 = + 170JK-1mol-1 (for 1 mole acid)
What does this tell you?
That there is a significant increase in disorder as the liquid decomposes to form gaseous products.
Ans q. on p. 179

10. Total entropy changes
Entropy changes in a chemical reaction are described as SYSTEM ENTROPY CHANGES.
The entropy changes in surroundings must also be considered.
If a chemical reaction is EXOTHERMIC the heat released from the system is used to increase the disorder of the surroundings so the entropy of the surroundings increases.
If the reaction is ENDOTHERMIC then the entropy of the surroundings decreases.
So we need to consider both when looking at chemical or physical changes.

11. Spontaneous Changes
A spontaneous change is one which proceeds on its own. (This is an energetic not kinetic statement – more later).
Spontaneous processes lead to lower energy and increased stability.
For a spontaneous change Δ Stotal must be positive ie the system and its environment considered together must increase in entropy.
Δ Stotal = Δ Ssystem + Δ Ssurroundings
The effect on the surroundings and their entropy changes can drive endothermic reactions to take place.

12. Spontaneous Changes
Energy derived from entropy is defined as T Δ S, where T is the temperature measured in Kelvin.
As temperature increases the energy derived from entropy becomes more significant.
Whether or not a process is spontaneous depends on 3 factors:
i) the temperature,. T in Kelvin
ii) the entropy change for the system Δ S
iii) the ENTHALPY change Δ H, with the surroundings

13. Free Energy
Free energy is released from a system and is available to do work e.g electrical energy to run a cell or battery or as heat to run an engine or mechanical device.
It is NOT the same as the energy given out in a reaction because some of this energy is used to change the vibrational energy of the particles present and to increase entropy.
The change in the free energy of a system determines whether or not a process is spontaneous – whether the change will happen.

14. Gibbs Free Energy
Free energy, enthalpy and entropy are related by this equation:
Δ GӨ = Δ HӨ - T Δ S Ө
Δ GӨ is the Gibbs free energy term.
For a reaction to be spontaneous (feasible) there must be an overall energy decrease.
So if Δ GӨ is <0 (-ve) the reaction is spontaneous.
If Δ GӨ is >0 (+ve) the reaction is non spontaneous and is spontaneous in the reverse reaction.
If Δ GӨ = 0 then the system is in equilibrium.

15. Will a reaction work?
In general since Δ HӨ is much larger than Δ S Ө :
If Δ HӨ is –ve and Δ S Ө is +ve then Δ GӨ must be negative and the reaction will work.
If Δ HӨ is +ve and Δ S Ө is -ve then Δ GӨ must be positive and the reaction will not work.
If both Δ HӨ and Δ S Ө are negative then Δ GӨ will be negative at low temperatures and the reaction will be feasible at low temps.
If both Δ HӨ and Δ S Ө are positive then Δ GӨ will be negative at high temperatures and the reaction will be feasible at high temps.

16. Consider these qualitatively:
H2(g) + F2(g) → 2HF(g)
What is the value for Δ HӨ ?
High and –ve because it’s a very exothermic reaction.
What is the value for Δ SӨ ?
Close to 0 because the number of gas molecules is the same but slightly negative because the system becomes slightly more ordered.
Overall Δ GӨ for this reaction must be negative so the reaction can go.

17. Consider these qualitatively:
Na+(g) + Cl-(g) → NaCl(s)
What is the value for Δ HӨ ?
High and –ve because it’s a very exothermic reaction (mostly lattice enthalpy)
What is the value for Δ SӨ ?
-ve because there is more order in the solid.
Overall Δ GӨ for this reaction is negative so the reaction can go.

18. Consider these qualitatively:
NH4NO3(s)) → NH4+(aq) + NO3-(aq)
What is the value for Δ HӨ ?
+ve because it’s an endothermic reaction.
What is the value for Δ SӨ ?
+ve because there is more disorder when the ions are aqueous.
Overall Δ GӨ for this reaction will be negative if T or the value for Δ SӨ is big enough – which it must be because dissolving takes place at room temperature.

19. Consider these:
Calculate Δ HӨ ; Δ SӨ ;Δ GӨ at 298K for the thermal decomposition of calcite.
At what temperature might the reaction go spontaneously?
CaCO3(s) → CO2(g) + CaO(s)
Data:
S/JK-1mol-1
Δ HfӨ/kJ mol-1
CaCO3(s)
92.9
CO2(g)
213.6
-393.5
CaO(s)
39.7
-635.1

20. a) Enthalpy change Δ HrӨ = Δ HprodӨ - Δ HreactӨ
= – ( )
= –( )
= kJmol-1

21. b) Entropy change CaCO3(s) → CO2(g) + CaO(s) 92.9 39.7 + 213.6
Entropy change = ( ) – 92.9
= J mol-1

22. c) Free energy change at 298K
Δ GӨ = Δ HӨ - T Δ S Ө
This is best taken in steps.
i) T Δ S Ө at 298
= 298 x J mol-1
= 298 x 160.4/1000 kJ mol-1
= 47.8 kJ mol-1
ii) Δ GӨ = Δ HӨ - T Δ S Ө =
= kJ mol-1
This positive value tells us that this reaction is not feasible at room temp.

23. At what temperature does the reaction become feasible?
When Δ GӨ = 0 then a reaction becomes just feasible and so the Δ GӨ equation can be rearranged to take this into account:
Δ GӨ = Δ HӨ - T Δ S Ө
If Δ GӨ is 0 then
T Δ S Ө = Δ HӨ
T = Δ HӨ / Δ S Ө
= kJ/160.4J
= 178.3/0.1604
= 1111K = = 838oC
Thus heating to at least 838oC may start this reaction. It gives no information about the RATE of the reaction.

24. Try this
Calculate the temperature at which the thermal decomposition of sodium hydrogencarbonate becomes feasible.
The balancing of the equation does NOT affect the feasibility temperature because halving the number of moles will affect both
Δ H and ΔS equally.
2NaHCO3(s) → Na2CO3(s) + H2O(g) + CO2(g)

25. Data 102 -951 135 -1131 189 -242 214 -394 S/JK-1mol-1 Δ HfӨ/kJ mol-1
NaHCO3(s)
102
-951
Na2CO3(s)
135
-1131
H2O(g
189
-242
CO2(g)
214
-394
Ans: Δ H = +135 kJ mol-1
Δ S =334 J mol-1
ΔG = 35.5kJ mol-1
T = 404K (131 oC)

26. Gibbs Free Energy and Equilibrium (NOS)
This equation relates free energy to the equilibrium constant.
For an equilibrium reaction when a reaction is just feasible and ΔG = 0 then T an K are effectively inversely proportional.
Increasing the temperature above the temperature at which the reaction becomes feasible for an exothermic reaction has the effect reducing the value of the equilibrium constant and of tipping an equilibrium back towards reactants.
At this point the reaction will no longer take place spontaneously.
For an endothermic change the reaction will take place spontaneously if the temperature is increased above the value when ΔG =0

27. Energetic vs Kinetic Stability
2H2(g) + O2(g) → 2H2O(l)
Δ S= 140 – 467 = -371 J K-1
Remember this? The decrease in entropy of the system should mean this reaction doesn’t go.
The total entropy of the system and its surroundings DOES increase so it is favoured but you can leave a 2:1 mixture of hydrogen and oxygen indefinitely and the reaction still doesn’t happen. Why not?
These gases are KINETICALLY stable. The activation energy is too high to be reached under normal conditions – a lighted splint is needed and THEN the reaction will go.
Work through box example on p.181 and then q. 1 and 2