Revision notes on thermodynamics: calculating an equilibrium constant from free energy change Advanced Level Theoretical-Physical Chemistry:

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Doc Brown's A Level Chemistry - Advanced Level Theoretical Physical Chemistry – GCE AS A2 IB Level Revision Notes – Basic Thermodynamics

Part 3: ΔS Entropy Changes and ΔG Free Energy Changes

3.5 Calculating Equilibrium Constants from Gibbs free energy equation data and 3.6 Kinetic stability versus thermodynamic instability

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Page introduction

This page describes and explains how equilibrium constants can be calculated from the Gibbs free energy data for a reaction and the equation relating the delta G (reaction) and the equilibrium expression.

Examples are given for a gaseous reaction (Kp expression), a homogeneous solution reaction (Kc expression) and  If the calculated free energy change is negative, will the reaction occur spontaneously on mixing the reactants?

i.e. just because a reaction is thermodynamically feasible according to the Gibbs free energy equation, will actually happen in practice?

i.e. the reaction is not just feasible BUT spontaneously happens too!

3.5 Calculating Equilibrium Constants from Gibbs free energy data

3.5a The relationship between free energy and equilibrium constants

To calculate the concentration equilibrium constant Kc or Kp

for a reaction aA + bB ... reversible dD + eE ...

Kc = ([A]a [B]b...)/([D]d [E]e...)  where [] are the concentration terms

Kp =(pAa pBb...)/(pDd pEe...)  where px are the partial pressures

The standard Gibbs free energy change ∆Gθsys = ∆Hθsys – T∆Sθsys  and

∆Gθsys = –RT ln(Kc) = –RT ln{([A]a [B]b ..)/([D]d [E]e...)}

or for a gaseous equilibrium

∆Gθsys = –RT ln(Kp) = –RT ln{(pAa pBb...)/(pDd pEe...)}

Two examples of Kp and a Kc equilibrium constant calculations are given below and both start from fundamental thermodynamic data, but in an exam, you might well start the calculation given a G value. A Kc equilibrium constant calculation for a redox reaction is described in section 3.5b.

Ex 3.5a1 Calculate the equilibrium constant at 298K for the dissociation of dinitrogen tetroxide

This colourless gas readily dissociates into nitrogen(IV) oxide (nitrogen dioxide).

equilibrium equation

N2O4(g) reversible 2NO2(g)
ΔHθf,298K/kJ mol–1 +9.7   2 x +33.9
Sθ298K/J mol–1K–1 304   2 x 240
ΔHθsys,298K = ∑ΔHf,products – ∑ΔHf,reactants = (2 x +33.9) – (+9.7) = +58.1 kJ mol–1
ΔSθsys,298K = ∑Sproducts – ∑Sreactants = (2 x 240) – (304) = 176 J mol–1K–1
ΔGθsys,298K = ΔHθsys – TΔSθsys = (+58.1) – (298 x 176/1000) = +5.652 kJ mol–1
∆Gθsys = –RT ln(Kp) = –RT ln{(pNO22)/(pNO2)}
substituting: 5.652 x 1000 = –8.314 x 298 ln(Kp)

ln(Kp) = –5652/(8.314 x 298) = 2.281

Kp = pNO22/pNO2 = exp(–2.281) = e(–2.281) = 0.102 atm

One data book value quoted is 0.115 atm, so not too bad!


Ex 3.5a2 Calculate the equilibrium constant for the formation of ammonia



+     3H2(g) reversible 2NH3(g)
ΔHθf,298K/kJ mol–1 0 0   2 x –46.2
Sθ298K/J mol–1K–1 192 3 x 131   2 x 193
ΔHθsys,298K = ∑ΔHf,products – ∑ΔHf,reactants = (0 + 0) – (2 x 46.2) = –92.4 kJ mol–1
ΔSθsys,298K = ∑Sproducts – ∑Sreactants = (2 x 193) – (192 + 3x131) = –199 J mol–1K–1
ΔGθsys = ΔHθsys – TΔSθsys = (–92.4) – (298 x –199/1000) = –33.1 kJ mol–1
∆Gθsys = –RT ln(Kp) = –RT ln{(pNH32)/(pN2 x pH23)}
substituting: –33.1 x 1000 = –8.314 x 298 ln(Kp)

ln(Kp) = 33100/2477.6 = 13.35, so Kp = exp(13.36) = 6.33 x 105 atm–2

Kp = pNH32/pN2 pH23 = 6.28 x 105 atm–2

You can also calculate Kp via Kp = exp(–ΔG/RT) = exp{33100/(8.314 x 298} = exp(13.36) = 6.33 x 105 atm–2

One data book value quoted is 6.76 x 105 atm–2, so again, quite good agreement.


Ex 3.5a3 Calculating the equilibrium constant for an esterification

equation, all (l) CH3COOH +   CH3CH2OH reversible CH3COOCH2CH3 + H2O
ΔHθf,298K/kJ mol–1 –487 –278   –481 –286
Sθ298K/J mol–1K–1 160 161   259 70
ΔHθsys = ∑ΔHf,products – ∑ΔHf,reactants = (–481 –286) – (–487 –278) = –2 kJ mol–1
ΔSθsys,273K = ∑Sproducts – ∑Sreactants = (259 + 70) – (160 + 161) = +8J mol–1K–1
ΔGθsys = ΔHθsys – TΔSθsys = –2 – (298 x +8/1000) = –4.38 kJ mol–1
∆Gθsys = –RT ln(Kc) = –RT ln{[CH3COOCH2CH3][H2O]/[CH3COOH][CH3CH2OH]}
substituting: –4.38 x 1000 = –8.314 x 298 ln(Kc)

ln(Kc) = –4380/–2478 = 1.77, so Kc = exp(1.77) = e(1.77) = 5.8

Kc = [CH3COOCH2CH3][H2O]/[CH3COOH][CH3CH2OH] = 5.8

However a figure of 4 is often quoted for this Kc esterification equilibrium constant at 298K (25oC), but in an early investigation in 1862 this value of 4 was quoted for 373 K (100oC).  Very confusing! though it should be noted that applying Le Chatelier's rule, Kc will increase at a lower temperature as the reaction is slightly exothermic which corresponds in principle to the value calculated here.


3.5b Free energy changes in redox reactions and calculation of the equilibrium constant Kc

The free energy change for a reversible electrode reaction is ∆Gθsys = –nEθF

n is the number of electrons transferred in the theoretical cell reaction

F is the Faraday constant (96500 C mol–1, i.e. equivalent to 1 mole of electrons)

and in general ∆Gθsys = –RT ln(Kc)

therefore for a redox equilibria –nEθF = –RT ln(Kc)

R = ideal gas constant, T absolute temperature in K, Kc = concentration equilibrium constant,

so nEθF = RT ln(Kc) from which the equilibrium constant Kc can be calculated for a redox reaction.


∆Gθ must be negative, and Eθ positive, for the theoretical overall redox reaction to be feasible.

Ex 3.5b1 Calculating the equilibrium constant for the redox equilibrium between silver and iron(III) ions at 298K

Ag+(aq) + e reversible Ag(s) Eθ298 = +0.799V  (silver(I) ion – silver half–cell reaction equilibrium)

Fe3+(aq) + e reversible Fe2+(aq) Eθ298 = +0.771V  (iron(III) – iron(II) ion half–cell reaction equilibrium)

For the 'cell' reaction: Ag+(aq) + Fe2+(aq) reversible Ag(s) + Fe3+(aq)

Kc = [Fe3+(aq)]/[Ag+(aq)] [Fe2+(aq)] mol–1 dm3

(remember solid concentrations are considered to be constant and treated as being unity)

As the cell reaction is set out,

silver ions are reduced to silver (Eθred)

and iron(II) ions oxidised to iron(III) ions (Eθox), therefore at 298K from

Eθcell = Eθred–half–cell – Eθox–half–cell

Eθcell = (+0.799) – (+0.771) = +0.028 V

This shows the reaction is feasible, but with Eθreaction close to zero, an equilibrium situation would be expected, i.e. the reaction will not go anywhere near completely to the right.

∆Gθsys = –nEθF = –1 x 0.028 x 96500 = –2702 J mol–1

∆Gθsys = –RT ln(Kc)

–2702 = –8.314 x 298 x ln(Kc)

ln(Kc) = 2702/(8.314 x 298) = 1.091

Kc = exp(1.091) = e(1.091) = 2.98 mol–1 dm3


3.6 Kinetic stability versus thermodynamic feasibility

If the calculated free energy change is negative i.e. the reaction is feasible and reactants theoretically suffer from instability! BUT will the reaction occur spontaneously on mixing the reactants?

See also "KINETICS" section 6.

A mixture of hydrogen and oxygen (e.g. in air at room temperature) is perfectly stable until a means of ignition, e.g. a lit splint, match or spark etc., is applied. Thermodynamically the mixture is highly unstable with a very negative  free energy ΔGθ and shouldn't exist! However, the activation energy to break the strong H–H or O=O bonds is so high, that they 'happily' co–exist without reacting, because the particle collisions are not energetic enough to cause a reaction. Therefore the mixture is kinetically stable. A high temperature from a match or spark etc., gives enough of the reactant molecules sufficient kinetic energy to overcome the activation energy on collision*.

2H2(g) + O2(g) ==> 2H2O(l), ΔGθ = –237 kJmol–1 or  ΔHθ = –286 kJmol–1 

Note: A very negative ΔHθ is usually, but not necessarily, indicative of a negative free energy change, which is the case here.

*The transition metal palladium can reduce the activation energy so much that it catalyses the spontaneous combustion/combination of hydrogen and oxygen at room temperature!

Like the methane–chlorine reaction below, the combustion of hydrogen is free radical chain reaction (a complex mechanism not dealt with at A level). The initiating energy produces the first free radicals.

A mixture of hydrogen/methane and chlorine is stable in the dark, but exposed to light (particularly ultra–violet), the mixture explodes to form hydrogen chloride/chloromethane gas. The strong H–H/C–H and Cl–Cl bonds ensure a high activation energy, but the absorption by chlorine (with the weakest bond) of a photon of light energy, to start the Cl–Cl bond breaking and so initiating the fast and exothermic free radical chain reaction (explosive!).

H2(g) + Cl2(g) ==> 2HCl(g), ΔGθ = –191 kJmol–1 or  ΔHθ = –185 kJmol–1 (again the free energy change is very negative)

mechanism: initiation: Cl2 ==> 2Cl

propagation: Cl + H2 ==> HCl + H and H + Cl2 ==> HCl + Cl

termination: 2Cl ==> Cl2 or 2H ==> H2 or H + Cl ==> HCl

CH4(g) + Cl2(g) ==> CH3Cl(g) + HCl(g), ΔGθ = –103 kJmol–1 or  ΔHθ = –99 kJmol–1

[full mechanistic details for methane]

Hydrogen peroxide is thermally unstable, its aqueous solution is kept in a brown bottle to avoid decomposition initiated by light. It has a decent shelf–life of a few weeks, i.e., reasonably kinetically stable in the short term, until manganese(IV) oxide powder is added and then the rapid exothermic reaction takes place!

2H2O2(aq)  ==> 2H2O(l) + O2(g)

See also "KINETICS" section 6.

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) Thermochemistry – Hess's Law and Enthalpy Calculations – reaction, combustion, formation etc. : 1.2b(iv) Enthalpy of reaction from bond enthalpy calculations  : 1.3a–b Experimental methods for determining enthalpy changes and treatment of results and calculations : 1.4 Some enthalpy data patterns : 1.4a The combustion of linear alkanes and linear aliphatic alcohols : 1.4b Some patterns in Bond Enthalpies and Bond Length : 1.4c Enthalpies of Neutralisation : 1.4d Enthalpies of Hydrogenation of unsaturated hydrocarbons and evidence of aromatic ring structure in benzene : Extra Q page A set of practice enthalpy calculations with worked out answers ** Part 2 ΔH Enthalpies of ion hydration, solution, atomisation, lattice energy, electron affinity and the Born–Haber cycle : 2.1a–c What happens when a salt dissolves in water and why? : 2.1d–e Enthalpy cycles involving a salt dissolving : 2.2a–c The Born–Haber Cycle *** Part 3 ΔS Entropy and ΔG Free Energy Changes : 3.1a–g Introduction to Entropy : 3.2 Examples of entropy values and comments * 3.3a ΔS, Entropy and change of state : 3.3b ΔS, Entropy changes and the feasibility of a chemical change : 3.4a–d More on ΔG, free energy changes, feasibility and applications : 3.5 Calculating Equilibrium Constants from ΔG the free energy change : 3.6 Kinetic stability versus thermodynamic feasibility - can a chemical reaction happen? and will it happen?

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