Revision notes on thermodynamics - entropy data and how to calculating entropy changes Advanced Level Theoretical-Physical Chemistry:

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Part 3: ΔS Entropy Changes and ΔG Free Energy Changes

3.2 Table of entropy data & comments and 3.3 Entropy values, entropy changes and feasibility of a physical change

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Page introduction

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.

3.2 Table of entropy data and comments

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.

 Examples – and brief explanation of entropy trend Entropy values below are the Sθ298K,1atm in J K–1 mol–1 Group 0 Noble Gases: A series of monatomic molecules – single atoms – an increasing complexity of electron arrangement – more electrons and energy levels used to distribute the electronic energy. He(g) Ne(g) Ar(g) Kr(g) Xe(g) 126 146 155 164 170 Linear alkanes – increasingly complex molecules, more electrons, more bond vibrations, rotations of groups of atoms. Sθpentane quoted for the gas only, even though it is a liquid at 25oC and 1atm pressure. CH4(g) C2H6(g) C3H8(g) C4H10(g) C5H12(g) 186 230 270 310 348 Water and benzene – illustrating the difference between the entropy of a gas, liquid and solid. The more condensed liquid form offers less ways of arranging the molecules. The entropy decrease for highly polar water from gas ==> liquid is particularly marked because even in the liquid there are clumps of molecules held together by hydrogen bonds – a bit of 'pseudo solid' structure decreasing the entropy more! This 'clumping' is far less likely in benzene in which only Van der Waals forces are operating so the entropy is greater. The entropy for a solid is even less than for that of the liquid – more ordered arrangement and less variation of possible states, though the entropy decrease from liquid to solid is not as marked as the entropy decrease from a liquid to gas. In an interactive sense i.e. intermolecular forces AND the possible ways of 'molecular arrangement', a liquid is closer to a solid than to a gas. H2O(g) H2O(l) H2O(s) C6H6(g) C6H6(l) 189 70 44.8 269 173

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

3.3a The Entropy Change of a state changes

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.

ΔSθsys = ΣSθfinal state – ΣSθinitial state

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

ΔSθsurr = –ΔHθsys/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.

ΔSθtot = ΔSθsys + ΔSθsurr

For a change to be feasible the total entropy change must be at least greater or equal to zero.

i.e. ΔSθtot >= 0

but note that when ΔSθtot = 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.

3.3a1 The melting of a pure substance – water

ΔSH2O melting = +22.0 J mol–1 K–1, ΔHmelt/fusion(solid H2O) = +6.01 kJ mol–1, T must be in K

For the change: H2O(s) ==> H2O(l)

ΔSθsys = Sθfinal state – Sθinitial state

ΔSθsys = SθH2O(l) – SθH2O(s) = +22.0 J mol–1 K–1

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

ΔSθsurr = –ΔHθfusion ice/T = –(+6010/T)

therefore for ice melting (endothermic) ...

ΔSθtot = ΔSθsys + ΔSθsurr = 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 (–5oC), (ii) 273 K(0oC) and (iii) 278 K(5oC) and see what the result is i.e. what is predicted ...

For (i) ΔSθtot =  22.0 – 6010/268 = –0.425 J mol–1 K–1

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

For (ii) ΔSθtot =  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 (0oC), 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) ΔSθtot =  22.0 – 6010/278 = +0.38 J mol–1 K–1

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

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

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3.3a2 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)

ΔSθsys = Sθfinal state – Sθinitial state

ΔSθsys = SθH2O(s) – SθNaCl(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.

ΔSθtot = ΔSθsys + ΔSθsurr = –(22.0 + x) + (+6010/T)

Note that ΔSθsurr = –ΔHθfusion ice/T = –(–6010/T) = +6010/T

Its reasonable here to assume ΔHmelt/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.

ΔSθtot = –(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 ΔSθtot. 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 ΔSθtot = 0 = ΔSθsys + ΔSθsurr = ΔSθsys + (–ΔHθsurr/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 ΔSθsys = ΔSθsurr 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!

A simple experiment to determine the enthalpy of vaporistion of water and the accompanying entropy change

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 100oC (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 ΔHvap(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

ΔSvap(H2O(l)) = ΔHvap(H2O(l))/Tbpt = 43563/373 kJmol-1 = +117 Jmol-1K-1

I couldn't find the entropy data for 373 K, but at 298 K, the entropy change can be calculated as follows

ΔSθsys = ΣSθfinal state – ΣSθinitial state

ΔSθsys = SθH2O(g) – SθH2O(l) = 189 - 70 = +119 J mol–1 K–1

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

Energetics-Thermochemistry-Thermodynamics Notes INDEX

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 QUICK INDEX for Energetics: GCSE Notes on the basics of chemical energy changes – important to study and know before tackling any of the three Advanced Level Chemistry pages Parts 1–3 here * Part 1a–b ΔH Enthalpy Changes 1.1 Advanced Introduction to enthalpy changes of reaction, formation, combustion etc. : 1.2a & 1.2b(i)–(iii) : 1.2b(iv) : 1.3a–b 1.4 : 1.4a : 1.4b : 1.4c : 1.4d : Extra Q page ** Part 2 : 2.1a–c 2.1d–e : 2.2a–c *** Part 3 : 3.1a–g : 3.2 * 3.3a : 3.3b : 3.4a–d : 3.5 Doc Brown's Chemistry Website content © Dr Phil Brown 2000+.  All copyrights reserved on revision notes, images, quizzes, worksheets. Copying of website material is NOT permitted. Exam revision summaries and references to science course specifications are all unofficial but all based on the official specifications.

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