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Part 3: ΔS Entropy Changes and ΔG Free Energy Changes

3.3b Entropy Changes & Feasibility of a Chemical Reaction

 In Part 3.3a the total entropy change for the 'system' and its' surroundings' was considered and now applied to whether a chemical reaction is feasible i.e the criteria is a positive entropy is required. The entropy changes for thermal decomposition of limestone (a very endothermic reaction) and the burning of hydrogen in oxygen (a very exothermic reaction) are discussed in detail and in each case the feasibility of the reaction at different temperatures is discussed.

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3.3b Entropy Changes and feasibility of a Chemical Change

ΔSθtot must be >0 for a chemical change to be feasible.

Example 3.3b1 Thermal decomposition of calcium carbonate (limestone)

CaCO3(s) ==> CaO(s) + CO2(g)  ΔHθ = +179 kJ mol–1  (very endothermic)

This important industrial reaction for converting limestone (calcium carbonate) to lime (calcium oxide) has to be performed at high temperatures in a specially designed limekiln – which these days, basically consists of a huge rotating angled ceramic lined steel tube in which a mixture of limestone plus coal/coke/oil/gas? is fed in at one end and lime collected at the lower end. The mixture is ignited and excess air blasted through to burn the coal/coke and maintain a high operating temperature.

ΔSθsys = ΣSθproducts – ΣSθreactants

ΔSθsys = SθCaO(s) + SθCO2(g) – SθCaCO3(s) = (40.0) + (214.0) – (92.9) = +161.0 J mol–1 K–1

ΔSθsurr is –ΔHθ/T = –(179000/T)

ΔSθtot = ΔSθsys +  ΔSθsurr

ΔSθtot = (+161) + (–179000/T) = 161 – 179000/T

If we then substitute various values of T (in Kelvin) you can calculate when the reaction becomes feasible.

For T = 298K (room temperature)

ΔSθtot = 161 – 179000/298 = –439.7 J mol–1 K–1, no good, negative entropy change

For T = 500K (fairly high temperature for an industrial process)

ΔSθtot = 161 – 179000/500 = –197.0, still no good

For T = 1200K (limekiln temperature)

ΔSθtot = 161 – 179000/1200 = +11.8 J mol–1 K–1, definitely feasible, overall positive entropy change

Now assuming ΔSθsys is approximately constant with temperature change and at room temperature the ΔSθsurr term is too negative for ΔSθtot to be plus overall. But, as the temperature is raised, the ΔSθsurr term becomes less negative and eventually at about 800–900oC ΔSθtot becomes plus overall, so the decomposition is now chemically, and 'commercially' feasible in a lime kiln.

You can approach the problem in another more efficient way by solving the total entropy expression for T at the point when the total entropy change is zero. At this point calcium carbonate, calcium oxide and carbon dioxide are at equilibrium.

ΔSθtot–equilib = 0 = 161 – 179000/T, 179000/T = 161, T = 179000/161 = 1112 K

This means that 1112 K is the minimum temperature to get an economic yield. Well at first sight anyway. In fact because the carbon dioxide is swept away in the flue gases so an equilibrium is never truly attained so limestone continues to decompose even at lower temperatures.

Lime is actually formed at temperatures above 900oC (1173K) and a typical modern limekiln operating temperature range is 950–980oC (from web). These calculations are approximate above 298K because it assumed that enthalpy and entropy values do not change with temperature. This is not true BUT the above calculations exemplify the sort of calculation you can do to calculate at what temperature becomes feasible.

In the next few examples I haven't bothered listing the entropy values, I've just slotted them into the entropy equations and the reaction enthalpy values are by the relevant chemical equation.

Example 3.3b2 The entropy change in forming water from hydrogen and oxygen

(i) Calculate the entropy changes ΔS for the combustion of hydrogen to form water vapour.

H2(g) + 1/2O2(g) ==> H2O(g)  ΔHsys = –242 kJ mol–1

ΔSθsys–298K = Sθ298K(products) – Sθ298K(reactants) J mol–1 K–1

Watch the signs, all absolute S values are +ve, but ΔS can be +ve or –ve.

ΔSθsys–298K = Sθ298K(H2O(g)) – Sθ298K(H2(g)) – Sθ298K(O2(g))/2

ΔSθsys = 188.7 –130.6 –205/2 = –44.4 J mol–1 K–1

This shows what you might expect in terms of an entropy change, you have gone from 1.5 moles of gas particles down to just 1.0 moles of gas particles – therefore less gas particles means less microstates or ways of arranging the particles. BUT we know this reaction is very feasible with the help of a lit splint! POP! Therefore to complete the 'feasibility calculation' we must calculate the total entropy change.

ΔSθtot = ΔSθsys +  ΔSθsurr

ΔSθtot = ΔSθsys +  (–ΔHθsys/T)

ΔSθtot = –44.4 + (––242 x 1000/298) don't forget ΔH values come along in kJ

ΔSθtot = –44.4 +812.1  = +767 J mol–1 K–1

So, despite the loss of gas particles the total entropy change is very positive! Why?

The reason for this is that the reaction is very exothermic, and the heat released is absorbed by the surroundings whose particles absorb the energy and distribute it in many available ways in the various energy levels e.g. kinetic energy – translational levels or rotational/vibrational quantum levels.

(ii) Calculate the entropy changes ΔS for the combustion of hydrogen to form liquid water.

H2(g) + 1/2O2(g) ==> H2O(l)  ΔHsys = –286 kJ mol–1

ΔSθsys–298K = Sθ298K(H2O(l)) – Sθ298K(H2(g)) – Sθ298K(O2(g))/2

ΔSθsys = 69.9 –130.6 –205/2 = –163.2 J mol–1 K–1 notice the much smaller entropy for liquid water

The entropy decrease in this case is much larger in forming liquid water compared to forming gaseous water. The liquid state offers less ways of distributing the particles and there energy.

ΔSθtot = ΔSθsys +  ΔSθsurr

ΔSθtot = ΔSθsys +  (–ΔHθsys/T)

ΔSθtot = –163.2 + (––286 x 1000/298) don't forget ΔH values come along in kJ

ΔSθtot = –163.2 +959.7  = +796.5 J mol–1 K–1

So despite, the smaller entropy of the product liquid water, the overall entropy change is even more positive! because reaction (ii) involves condensation which is an exothermic process and more energy is transferred to increase the entropy of the surroundings.

Example 3.3b2 The entropy change in forming water from hydrogen ions and hydroxide ions – neutralisation

Calculate the entropy change for the neutralisation reaction

H+(aq) + OH–(aq) ==> H2O(l)  ΔHsys = –57.1 kJ mol–1

Sθ298(H+(aq)) = 0 by definition, Sθ298(OH–(aq)) = –10.7, Sθ298(H2O(l)) = 69.9

Its difficult to estimate absolute entropy values for aqueous ions, so the aqueous hydrogen ion is given an arbitrary value of zero and all the entropies of other ions are measured against it so that you can get negative entropy values here – but don't worry about it – I don't know if its required for UK A2 level courses.

ΔSθsys–298K = Sθ298K(products) – Sθ298K(reactants) J mol–1 K–1

ΔSθsys–298K = Sθ298K(H2O(l)) – Sθ298K(H+(aq)) – Sθ298K(OH–(aq))

ΔSθsys = 69.9 –0  ––10.7 = +80.6 J mol–1 K–1

Despite going from two particles to one particle, an apparent considerable decrease in the possible ways of arranging the particles there is still a large increase in in entropy of the system! Why?

Both of the ions are hydrated and the hydrogen ion is strongly associated with water molecules,

e.g. it forms H3O+, H5O2+, H7O3+ ions etc. (H+ and 1, 2, 3 etc. associated water molecules) because of its strong polarising power and electrostatic field effect on the negative lone pair electron oxygens of water molecules. So, when water molecules are formed, all the water molecules associated with the hydrated ions are freed! and thereby raising the entropy of the system. Just to complete the calculation, though its feasibility hardly needs any introduction ...

ΔSθtot = ΔSθsys +  ΔSθsurr

ΔSθtot = ΔSθsys +  (–ΔHθsys/T)

ΔSθtot = +80.6 + (––57.1 x 1000/298) don't forget ΔH values come along in kJ

ΔSθtot = +80.6 +191.6  = +272.2 J mol–1 K–1

A very healthy increase in entropy!


summary to do

These theoretical calculations can be used for any reaction BUT there are limitations:

You can’t say the reaction will definitely spontaneously happen (go without help!) because there may be rate limits especially if the reaction has a high activation energy or a very low concentration of an essential reactant.

However you can employ a catalyst, raise reactant concentrations or raise the temperature to get the reaction going! There is usually a way of getting most, but not all, feasible reactions to actually occur.

For more details on this last point see 3.6 Kinetic stability versus thermodynamic instability for a detailed discussion

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