Part 3: ΔS Entropy Changes and ΔG Free Energy Changes

3.2 Table of entropy data for various substances and comments

3.3 Entropy values, entropy changes & feasibility of physical change

ALL my advanced A level theoretical chemistry revision study notes

3.2 Table of entropy data for a variety of substances and comments

Having introduced the idea of entropy this section looks at the entropy values (S) of variety of elements and compounds AND in some cases their entropy in different physical states and then discusses why they are different

i.e. why is one molecule gat a higher/lower entropy than another and then the entropy change for physical process like melting are described. The total entropy change for the 'system' and its' surroundings' must be considered.

The examples below illustrates some of the ideas described in section 3.1 above.

S denotes the entropy content of a substance and it always has a positive value. It increases with increase in temperature as more energy levels (rotational, vibrational, electronic etc.) become more available as energy is absorbed and redistributed as the temperature increases.

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 3.3 ΔS, Entropy changes and the feasibility of a change – physical or chemical

3.3a The Entropy Change of a state change and feasibility

The entropy change associated with a physical state change:

The direction of change of a system which is not at equilibrium (physical or chemical) is determined by the overall entropy change.

Up to now we haven't considered the whole situation where there is potential for change.

The whole situation meaning system and surroundings.

From data tables of entropy (e.g. as in table 3.2 above)  you can readily calculate the entropy change of the system.

ΔS^θ_sys = ΣS^θ_{final state} – ΣS^θ_{initial state}

The entropy change for the surroundings is minus the enthalpy change of the system change divided by the absolute temperature in Kelvin.

ΔS^θ_surr = –ΔH^θ_sys/T(K)

You do not need to know where this comes from.

The total entropy change is then given by the 'simple' equation i.e. the sum of all the entropy changes.

ΔS^θ_tot = ΔS^θ_sys + ΔS^θ_surr

For a physical or chemical change to be feasible the total entropy change for that direction must be at least greater or equal to zero.

i.e. ΔS^θ_tot >= 0

but note that when ΔS^θ_tot = 0, an equilibrium formed, in fact this is the criteria for equilibrium formation!

In this section 3.3a we will apply this rule to two physical state changes.

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3.3a1 The entropy change for melting of a pure substance – water

ΔS_{H2O melting} = +22.0 J mol^–1 K^–1, ΔH_{melt/fusion(solid H2O)} = +6.01 kJ mol^–1

(T must be in K)

For the change: H₂O_(s) ==> H₂O_(l)

ΔS^θ_sys = S^θ_{final state} – S^θ_{initial state}

ΔS^θ_sys = S^θ_H2O(l) – S^θ_H2O(s) = +22.0 J mol^–1 K^–1

(I couldn't find the individual S values for water at 0^oC, just the delta S value)

ΔS^θ_surr = –ΔH^θ_{fusion ice}/T = –(+6010/T)

therefore for ice melting (endothermic) ...

ΔS^θ_tot = ΔS^θ_sys + ΔS^θ_surr = 22.0 – 6010/T

watch the units! convert kJ to J here and T in Kelvin

So what we can do now is to try three temperatures in the above expression (i) 268 K (–5^oC), (ii) 273 K(0^oC) and (iii) 278 K(5^oC) and see what the result is i.e. what is predicted ...

For (i) ΔS^θ_tot =  22.0 – 6010/268 = –0.425 J mol^–1 K^–1

negative total entropy change, therefore melting not possible at 268 K (–5^oC)

For (ii) ΔS^θ_tot =  22.0 – 6010/273 = –0.014 J mol^–1 K^–1 (theoretically it is zero with sufficiently accurate data)

total entropy change close to zero at 273 K (0^oC), in fact at 273.15 K ice and water would coexist in equilibrium and delta S would zero.

Whether melting or freezing would take place at 273.15 K depends on the direction of heat transfer to or from the surroundings – remember there is no temperature change in this state change but melting occurs if heat is absorbed by the system or freezing happens if heat is lost to the surroundings in this case.

For (iii) ΔS^θ_tot =  22.0 – 6010/278 = +0.38 J mol^–1 K^–1

positive total entropy change, therefore melting possible at 278 K (5^oC)

This simple example of a physical state change exemplifies feasibility calculations i.e. what changes are possible theoretically?

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3.3a2 Entropy changes and the freezing of a mixture e.g. salt solution

Approach (i)

The spreading of salt (sodium chloride) on roads is a tried and tested method of de–icing roads or preventing ice formation. So, why does it work? why is the melting/freezing point of water lowered? – well its all down to the entropy change! Unfortunately I can't find accurate enough data to do an illustrative calculation. However, using the entropy equations, we can do a 'thought experiment' and reason out why adding salt lowers the freezing point of water, or to state it another way, why salt solution has a lower freezing point than pure water.

The most important point to grasp is that salt solution has a greater entropy than pure water. There are now three different particles in the liquid instead of one i.e. sodium and chloride ions as well as water molecules. When salt solution freezes pure water will crystallise out of the liquid.

For the change: H₂O_(l) ==> H₂O_(s)  (but from salt solution)

ΔS^θ_sys = S^θ_{final state} – S^θ_{initial state}

ΔS^θ_sys = S^θ_H2O(s) – S^θ_NaCl(aq) = –(22.0 + x) J mol^–1 K^–1

Note (i) The entropy change is negative as we are going from liquid to the more ordered solid state – decreased possible particle arrangements.

(ii) x represents the extra entropy of the salt solution compared to pure water and the more salt in the water, the higher the entropy of the mixture, so x will be even bigger.

ΔS^θ_tot = ΔS^θ_sys + ΔS^θ_surr = –(22.0 + x) + (+6010/T)

Note that ΔS^θ_surr = –ΔH^θ_{fusion ice}/T = –(–6010/T) = +6010/T

Its reasonable here to assume ΔH_melt/freeze is ~same for the salt solution i.e. the crystallisation of pure water to form ice, but the sign is changed, freezing is an exothermic process.

ΔS^θ_tot = –(22.0 + x) + 6010/T

Now think! the greater x, the more salt, the greater the entropy change, then the smaller T must be to give a positive total entropy change. When x is zero, T is 273, when x is over zero then T must be less than 273 to give a positive ΔS^θ_tot. So lowering of the freezing point of salt solution is all down to the higher entropy state of salt solution.

 

Two other approaches to the salt solution–ice situation

(ii) Alternative calculation approach but using the data from above:

When water and ice co–exist in equilibrium ΔS^θ_tot = 0 = ΔS^θ_sys + ΔS^θ_surr = ΔS^θ_sys + (–ΔH^θ_surr/T)

so if x represents the extra entropy change for ice <=> salt solution compared to just ice <=> pure water ...

for melting (endothermic ΔH +ve, increase ΔS +ve), so 0 = +(22 + x) – (+6010/T)

for freezing (exothermic ΔH –ve, decrease ΔS –ve), so 0 = –(22 + x) –(–6010/T)

therefore either way ΔS^θ_sys = ΔS^θ_surr at equilibrium and if x increases then T decreases numerically in order for the criteria for equilibrium to be satisfied.

 

(iii) A simple 'limited' conceptual approach:

In terms of entropy: ice + salt < ice + pure water < ice + salt solution (<== increasing order)

Therefore the highest entropy state will be attained at the lowest possible temperature.

However, unless you put data into the equations it doesn't exactly prove a lot. However using the total entropy equation plus a conceptual picture is a most satisfying situation – the ideal in chemistry!

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3.3a3 A simple experiment to determine the enthalpy of vaporisation of water and the accompanying entropy change plus the theoretical calculation

Procedure

You need an electric kettle of known power that can be accurately weighed to the nearest 0.1g.

Add about a litre of water to the kettle and get the water to boiling point.

Turn off the power and weigh the kettle and water.

Quickly, reconnect with the power and boil for two minutes, stop boiling and quickly reweigh the kettle.

Typical results

Mass of water evaporated/boiled off = 119.0 g.

Time of boiling = 60 x 2 = 120 s at 100^oC (373 K)

Calculation

(i) Enthalpy of vaporisation of water

Assuming the kettle's power rating is 2.4 kW = 2400 W = 2400 Js^-1

Energy transferred = power x time = 2400 x 120 = 288,000 J

Ar's: H = 1, O = 16, Mr(H₂O) = 18, Moles of water = 119/18 = 6.611

Enthalpy of vapourisation = 288,000/6.611 = 43563 Jmol^-1

Therefore ΔH_vap(H₂O(l)) = 43.6 kJmol^-1

The quoted book value of the enthalpy of vaporization of water is 41.1 kJmol^-1

(ii) Entropy change of the vaporisation of water

Since heat is transferred to liquid water converting it to gaseous water (steam!) there is a large positive increase in entropy.

ΔS = ΔH/T, ΔS = entropy change, ΔH = enthalpy change,

T = temperature K at which the state change takes place

ΔS_vap(H₂O(l)) = ΔH_vap(H₂O(l))/T_bpt = 43563/373 kJmol^-1

= +117 Jmol^-1K^-1

Theoretical calculation

I couldn't find the entropy data for 373 K, but for accurate entropy data at 298 K (and 1 atm, 101kPa), the entropy change can be very simply calculated as follows

ΔS^θ_sys = ΣS^θ_{final state} – ΣS^θ_{initial state}

ΔS^θ_sys = S^θ_H2O(g) – S^θ_H2O(l) = 189 - 70 = +119 J mol^–1 K^–1

Its a simple experiment to do, not that accurate, but very instructive!

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