Entropy and Free Energy

Why do reactions take place?

• Feasible reactions take place spontaneously, although the rate may be slow.

• Generally the more negative ΔH the more likely the reaction.

• So why are endothermic reactions possible?

Randomness

When a gas evaporates it spreads out.

This increases its degree of randomness.

The same is true on mixing gases;

+

And dissolving solids;

+

Entropy

• Endothermic reactions are feasible if they increase disorder.

• IE Product particles are more randomly arranged than reactant particles.

• Randomness is expressed mathematically as entropy (S).

• A feasible endothermic reaction will have a positive entropy change (ΔS).

• All entropies are +ve.• Entropies of elements in

their standard states are not zero.

• Entropies increase with temperature as the particles spread out.

• So entropies are quoted at 298K and 101kPa.

• NB Entropies of solids < liquids < gases

Substance S (JK-1mol-1)

Iron 27

Iron Oxide 88

Calcium carbonate

93

Ice 48

Water 70

Steam 189

Carbon dioxide

214

Calculating entropy changes

• 1) Add the entropies of the products.

• 2) Add the entropies of the reactants.

• 3) Subtract the entropies of the reactants from that of the products.

• 4) Then if ΔS is positive the reaction will be feasible.

• Eg; Calcium carbonate decomposes when heated to form calcium oxide.

• CaCO3 → CaO + CO2 ΔH = +178kjmol-1

CaCO3 → CaO + CO2

Entropies of products;

• CaO = 40• CO2 = 214• 40 + 214 = 254 • Entropy of reactants;

• CaCO3 = 93• Entropy change• ΔS = 254 - 93• = +161 JK-1mol-1

Gibbs Free Energy

• The feasibility of a reaction is determined by;

• 1) ΔH• 2) ΔS• These two factors are combined to

calculate Gibbs Free Energy (G).• ΔG = ΔH – TΔS• If ΔG is negative a reaction is feasible.

Influence of temperature on reactions

• ΔG is temperature dependent.

• This means that reactions can become feasible as temperature is raised.

• Eg; CaCO3 → CaO + CO2

• At 298K ΔH = 178 kjmol-1

• ΔS = 161 jK-1mol-1 = 0.161 kjK-1mol-1

• T ΔS = 298 x 0.161 = 47.98kjmol-1

• ΔG = 178 – 47.98 = 130 kjmol-1

• So the reaction is not feasible.

• Instead the reverse reaction occurs;

CaO + CO2 → CaCO3 ΔG = -130 kjmol-1

• But at 1500K • ΔH = 178 kjmol-1

• ΔS = 161 jK-1mol-1 = 0.161 kjK-1mol-1

• T ΔS = 1500 x 0.161 = 241.5kjmol-1

• ΔG = 178 – 241.5 = -63.5 kjmol-1

• The reaction now has a negative ΔG so it has become feasible.

Zero values of ΔG

• When ΔG is 0 the reaction is just feasible.

• The temperature at which this occurs can be calculated.

• Eg; CaCO3 → CaO + CO2

• ΔG = ΔH – TΔS• 0 = 178 – T0.161• T = 178 / 0.161 = 1106K

Kinetic factors

• Neither ΔH nor ΔS gives any indication of the rate of a reaction.

• Some reactions are predicted to be feasible on the basis of ΔH and ΔS but in practice are so slow that they unfeasible.

• This is because of a kinetic barrier.

• IE There is a high activation energy.

reactants

products

energy

exergonic reaction

Reaction profile

activation energy, Eaa

transition state

(or activated complex)

bonds breaking

bonds forming

Course of reaction Replay Close window

Eg; Stability of graphite

• C + O2 → CO2

• ΔH = -394kjmol-1

• S CO2 = 213.6 jK-1mol-1

• S O2 = 205 jK-1mol-1

• S C = 5.7 jK-1mol-1

• ΔS = +2.91 jK-1mol-1

• ΔG = -394 – (298 x 0.00291)

• = -394.86 kjmol-1

• So the reaction is feasible at 298K.

• But in practise it is too slow.