F325 Entropy

Entropy and Free Energy

All the energy changes we have considered so far have been in terms of enthalpy, and we have tried to predict whether a reaction is likely to occur on the basis of the enthalpy change associated with it. A reaction in which the overall enthalpy change is negative (exothermic) results in more stable products with stronger bonds, and hence is more likely to proceed. We have also considered that the activation energy (which we could relate to enthalpies of atomisation and ionisation) affects the conditions which might be needed to get the reaction started.

However, we also know that endothermic reactions do happen spontaneously even though they are energetically unfavourable in terms of enthalpy change. We don't yet have an explanation for this, and the reason is that we need to consider another kind of energy-related change as well – entropy.

It may be useful to think of entropy in terms of "amount of randomness" or "chaos". Consider that it takes energy to maintain order, but left to their own devices, natural tendencies are towards disorder.

A perfectly ordered crystal lattice at absolute zero (0K) would have an entropy of zero. In reality all substances possess some entropy – so what gives them entropy ?

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Candidates should be able to:(a) 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;(b) explain the difference in magnitude of entropy:

(i) of a solution and a gas,(ii) when a solid lattice dissolves,(iii) for a reaction in which there is a change in the number of gaseous molecules;

(c) calculate the entropy change for a reaction given the entropies of the reactants and products;(d) explain that the tendency of a process to take place depends on temperature, T, theentropy change in the system, ΔS, and the enthalpy change, ΔH, with the surroundings;(e) explain that the balance between entropy and enthalpy changes is the free energy change, ΔG, which determines the feasibility of a reaction;(f) state and use the relationship ΔG = ΔH – TΔS;(g) explain, in terms of enthalpy and entropy, how endothermic reactions are able to takeplace spontaneously.

Entropy is the quantitative measure of the degree of disorder in a system. The symbol used for entropy is S. Units are J K-1 mol-1

The more disordered a system is, the more energetically stable it is.

F325 Entropy

1) The entropy of a pure substance increases with increasing temperature. This is because as we increase temperature the particles vibrate more, making their relative positions a little more disordered and random.It follows that larger molecules, having more bonds to vibrate, have higher entropies than smaller molecules at the same temperature.

2) Entropy also increases during changes of state from solid to liquid (melting) or as a solid lattice dissolves (solution) as the particles are now able to move about within the liquid, becoming even more disordered than they were in the regular solid structure

3) Entropy increases again when a liquid turns into a gas as the particles become able to travel outside the volume originally occupied by the liquid, becoming even more disordered

4) Entropy increases when a reaction produces more moles of gaseous products than there were moles of gaseous reactants.

Thus- entropy is always a positive number- the more localised or concentrated the energy (vibrations, translations) in a system is, the lower the entropy- the more dispersed the energy in a system is, the higher its entropy

Entropy changesThe favoured direction for entropy changes (ΔS) is for entropy to increase. An increase in disorder corresponds to a more energetically stable situation.

Examples When a liquid perfume is spilt, the molecules spread throughout the room Ionic substances such as NaCl dissolve in water even though the enthalpy

change of solution is endothermic (ΔHs = +25 kJ mol-1) Ice melts in a warm room, even though it takes energy to overcome the

hydrogen bonding that holds the molecular lattice together

The standard entropy change when a reaction takes place can be calculated using the standard entropies of the reactants and the products:

ΔSө = ΣSөproducts - ΣSө

reactants

Example: Substance Sө (J K-1mol-1)O3(g) + 237.7O2(g) + 204.9

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Standard EntropiesThe standard entropy of a substance, Sө, is the entropy content of one mole of that substance under standard conditions of 298K and 101kPa.

F325 Entropy

For the reaction: 2O3 3O2

ΔSө = (3 x 204.9) – (2 x 237.7) = +139.3 J K-1mol-1

This sign of ΔS here is expected, as three moles of gas are formed from two. The increase in entropy favours this reaction taking place, but we'd also have to look at the enthalpy change to be sure if the reaction was feasible.

The standard entropies of oxygen and ozone are also as expected – ozone has more bonds than oxygen, so more vibrations, and a larger standard entropy.

Example: Substance Sө (J K-1mol-1)N2(g) + 192H2(g) + 131NH3(g) + 193

For the reaction: N2(s) + 3H2(g) 2NH3(g)

ΔSө = (2 x 193) – (192 + (3 x 131)) = -199 J K-1mol-1

So this reaction results in a decrease in entropy, which does not favour the reaction taking place. We'd expect this because we've fewer moles of gas as a result of the reaction, so a more ordered system. To be sure whether the reaction is feasible, we'd have to consider the enthalpy change as well, however…

Practice 1) Calculate the standard enthalpy changes of the following reactions, given these standard entropy values:

substance Sө (J K-1mol-1)NO(g) + 211O2(g) +205 N2O4(g) +304C6H6(l) +173CO2(g) +214H2O(l) +70

a) 2NO(g) + O2(g) N2O4(g)

b) C6H6(l) + 7½O2(g) 6CO2(g) + 3H2O(l)

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A reaction is spontaneous if a chemical system becomes more stable and its overall energy decreases as a result of the reaction. The overall energy decrease results from contributions from both enthalpy and entropy.

F325 Entropy

Using ΔH and ΔS togetherWe want to know whether a reaction is feasible, in other words whether it will happen spontaneously.

We know that when enthalpy decreases (ΔH is negative) a reaction is favoured because the products become more stable than the reactants were.

We know that when the entropy increases (ΔS is positive) a reaction is favoured because the amount of disorder in the system is increased.

We can conclude that:1) If ΔH is negative AND ΔS is positive, the reaction will always be feasible because both result in the products being more stable than the reactants.

2) If ΔH is positive and ΔS is negative, the reaction will never be feasible because both result in the products being less stable than the reactants.

But what if the entropy change favours the reaction but the enthalpy change doesn't ? (or vice versa ?)

In these latter two cases we need a way of comparing the effect of entropy and enthalpy, and the way to do this was developed by J.W. Gibbs. Gibbs recognised that the effect of entropy is small at low temperatures, but much larger at high temperatures

The value ΔG is the (Gibbs) free energy. A process can take place spontaneously when its ΔG is less than 0, i.e. when ΔG is negative.

3) If ΔH is positive and ΔS is positive as well, this is an endothermic reaction, but one in which the system becomes more disordered. For the reaction to be feasible ΔG must be negative, so the value of TΔS must be larger than the value of ΔH. The reaction will be feasible ABOVE a certain temperature.

4) If ΔH is negative and ΔS is negative as well, the reaction may be feasible. This is an exothermic reaction, but resulting in an ordered system. For ΔG to be negative, the value of TΔS must be smaller than the value of ΔH. This will be the case at low temperatures. The reaction will be feasible BELOW a certain temperature.

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He defined a quantity ΔG = ΔH – TΔS Where T = temp. in KΔS = entropy changeΔH = enthalpy change

N.B. ΔS and ΔH must both be in J or kJ not one of each!

F325 Entropy

ExamplesWe can use these ideas to explain why some processes occur as they do.

When ice melts, the enthalpy change associated with breaking the hydrogen bonds in the molecular lattice, H2O(s) H2O(l) ΔH = +6.01 kJ mol-1

= 6010 J mol-1

The standard entropy of ice Sө = +48 J K-1mol-1

The standard entropy of water Sө = +70 J K-1mol-1

So the entropy change on melting ice ΔSө = 70.0 – 48.0 = +22 J K-1mol-1

Lets consider what happens at -5°C (= 268K):

ΔG = ΔH – TΔS = 6010 – (268 X 22) =6010-5896 = +114 J mol-1

melting of ice is not a process which happens spontaneously at -5°C

Now consider what happens at +5°C (= 278K)

ΔG = ΔH – TΔS = 6010 – (278 X 22)=6010-6116 = -106 J mol-1

melting of ice is spontaneous at +5°C.

We can go further and say that the process of melting ice becomes feasible when ΔG stops being positive and becomes negative, i.e. at ΔG = 0

ΔG = 0 when ΔH = TΔS

6010 = T x 22 so T = 6010/22 T = 273K (to 3 sf) i.e. 0°C

While this was a change of state, the same thinking can be applied to reactions. Thermal decompositions are endothermic reactions – heating is required.

Consider the thermal decomposition of zinc carbonate

ZnCO3(s) ZnO(s) + CO2(g) ΔH = +71kJmol-1

( = 71000 J mol-1- )

We can see that the entropy change is going to be positive – a gas is being formed, so the products will be more disordered than the solid lattice we started with. We can therefore conclude that both ΔH and ΔS will be positive – the reaction will be feasible ABOVE a certain temperature.

If we look up the standard entropies, we can calculate ΔS, and we should then be able to use this to work out the temperature above which the zinc carbonate will thermally decompose.

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F325 Entropy

substance Sө J K-1mol-1

zinc carbonate +82zinc oxide +44carbon dioxide +214

ΔS = (214 + 44) – 82 = 176 J K-1mol-1

At room temperature, 298K, the value of ΔG = 71000 - (298 x 176) = +18552 J mol-1

Clearly ZnCO3 will not decompose at room temperature because ΔG is positive.

The reaction becomes feasible when ΔG = 0 so ΔH = TΔS

71000 = T x 176 so T = 71000/176 = 403.4K

Converting to °C T = 403.4 – 273 = 130.4°C

Reversing a reactionIf the sign of ΔG is positive for the forward reaction, then it is negative (with the same numerical value) for the reverse reaction.

ZnO(s) + CO2(g) ZnCO3(s) ΔG = -18552 J mol-1 at room temperature.

So this reaction is feasible. If we leave zinc oxide lying around it will absorb carbon dioxide from the air, reacting with it to form zinc carbonate.

Summary: If a reaction is not feasible below a certain temperature, the reverse reaction will be feasible below that temperature.

Practice:2) At high temperature, it is possible to react carbon with steam to produce a mixture of carbon monoxide and hydrogen known as water gas. This is a useful fuel. The equation for the reaction is:

C(s) + H2O(g) CO(g) + H2(g)

Use the enthalpy and entropy values below to calculate the temperature in °C at which the reaction becomes feasible.

ΔHf ө(H2O(g)) = -241.8 kJmol-1

ΔHf ө(CO) = -110.5 kJmol-1

Sө(C) = 5.7 JK-1mol-1

Sө(H2O(g)) = 188.7 JK-1mol-1

Sө(CO) = 197.9 JK-1mol-1

Sө(H2) = 130.6 JK-1mol-1

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F325 Entropy

Answers to practice questions:

1.a) 2NO(g) + O2(g) N2O4(g)

ΔS = 304 – (2x211 + 205) = -323 J K-1mol-1

b) C6H6(l) + 7½O2(g) 6CO2(g) + 3H2O(l)

ΔS = ((6x214) + (3x70)) - (173 + (7.5x205)) = 1494 – 1710.5 = -216.5 J K-1mol-1

2. Working out enthalpy change (using Hess Cycle):ΔHr = +218 – 110.5 = 107.5 kJmol-1 = 107500 Jmol-1

Working out entropy change: ΔS = 197.9 + 130.6 – 118.7 – 5.7 = 204.1 JKmol-1

Reaction becomes feasible when ΔG = 0 i.e. when ΔH = TΔSrearranging, T = ΔH/ΔS = 107500/204.1 = 526.7K = 253.7°C

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