Week 24

© Pearson Education Ltd 2009This document may have been altered from the original

• 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.

Entropy S• Entropy is the quantitative measure of the degree

of the disorder of a system.• 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.• 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.• 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.

Week 24

© Pearson Education Ltd 2009This document may have been altered from the original

Increasing entropy

What can entropy tell us?• In nature entropy tends to increase.• Disordered systems are more likely than

ordered systems.• Generally speaking a reaction is favoured if

entropy increases.• In an increasingly disordered system energy

changes from being localised to spread out.• This leads to an increase in energetic

stability.• Entropy increases whenever particles

become more disordered.

Increasing Entropy

• 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↑

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)- ΣΔSӨ (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.

Use this!

S /JK-1mol-1

H2(g) 131 diamond 2.4O2(g) 205 graphite 5.7N2(g) 192 HCl(g) 187Cl2(g) 223 HNO3(l) 156H2O(g) 189 NO2(g) 240H2O(l) 70 NaCl(s) 72

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) + 205 = 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.

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

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.

Spontaneous Changes• A spontaneous change is one which proceeds

on its own. (This is an energetic not kinetic statement – more later)

• 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.

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.

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.

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.

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.

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.

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.

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 Hf

Ө/kJ mol-1CaCO3(s) 92.9 -1206.9CO2(g) 213.6 -393.5CaO(s) 39.7 -635.1

a) Enthalpy change• Δ Hr

Ө = Δ HprodӨ - Δ Hreact

Ө

• = -635.1 -393.5 – (-1206.9)• = - 1028.6 –(-1206.9)• = +178.3 kJmol-1

b) Entropy change• CaCO3(s) → CO2(g) + CaO(s)

• 92.9 39.7 + 213.6• Entropy change = (39.7 + 213.6) – 92.9• = 160.4 J mol-1

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 160.4 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

• = 178.3 -47.8• = +130.5 kJ mol-1 • This positive value tells us that this reaction is not

feasible at room temp.

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

• = 178.3 kJ/160.4J• = 178.3/0.1604• = 1111K = 1111-273 = 838oC• Thus heating to at least 838oC may start this

reaction. It gives no information about the RATE of the reaction.

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)

Data

S/JK-1mol-1 Δ HfӨ/kJ mol-1

NaHCO3(s) 102 -951Na2CO3(s) 135 -1131H2O(g 189 -242CO2(g) 214 -394Ans: Δ H = +135 kJ mol-1

Δ S =334 J mol-1

ΔG = 35.5kJ mol-1

T = 404K (131 oC)

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

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