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 (a sort of theoretical versus actual comparison)

3.5 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!

3.5a Examples of calculating Equilibrium Constants from Gibbs free energy data

3.5a The relationship between free energy and equilibrium constants

To calculate the concentration equilibrium constant K_c or K_p

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

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

K_p =(p_A^a p_B^b...)/(p_D^d p_E^e...)  where p_x are the partial pressures

The standard Gibbs free energy change ∆G^θ_sys = ∆H^θ_sys – T∆S^θ_sys  and

∆G^θ_sys = –RT ln(K_c) = –RT ln{([A]^a [B]^b ..)/([D]^d [E]^e...)}

or for a gaseous equilibrium

∆G^θ_sys = –RT ln(K_p) = –RT ln{(p_A^a p_B^b...)/(p_D^d p_E^e...)}

Two examples of K_p and a K_c 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 K_c 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)		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

equilibrium	N2(g)	+     3H2(g)		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		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   [CH3COOCH2CH3][H2O]   Kc =  -------------------------------------- = 5.8   [CH3COOH][CH3CH2OH]   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.

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

The free energy change for a reversible electrode reaction is ∆G^θ_sys = –nE^θF

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 ∆G^θ_sys = –RT ln(K_c)

therefore for a redox equilibria –nE^θF = –RT ln(K_c)

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

so nE^θF = RT ln(K_c) from which the equilibrium constant K_c can be calculated for a redox reaction.

REMEMBER ...

∆G^θ must be negative, and E^θ 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^– Ag_(s) E^θ₂₉₈ = +0.799V  (silver(I) ion – silver half–cell reaction equilibrium)

Fe³⁺_(aq) + e^– Fe²⁺_(aq) E^θ₂₉₈ = +0.771V  (iron(III) – iron(II) ion half–cell reaction equilibrium)

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

K_c = [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 (E^θ_red)

and iron(II) ions oxidised to iron(III) ions (E^θ_ox), therefore at 298K from

E^θ_cell = E^θ_{red–half–cell} – E^θ_{ox–half–cell}

E^θ_cell = (+0.799) – (+0.771) = +0.028 V

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

∆G^θ_sys = –nE^θF = –1 x 0.028 x 96500 = –2702 J mol^–1

∆G^θ_sys = –RT ln(K_c)

–2702 = –8.314 x 298 x ln(K_c)

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

K_c = exp(1.091) = e^(1.091) = 2.98 mol^–1 dm³

Ex

3.6 Explaining 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.

Example 1. An explosive gaseous mixture of hydrogen and oxygen!

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 ΔG^θ 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^*.

2H_2(g) + O_2(g) ==> 2H₂O_(l)

 ΔG^θ = –237 kJmol^–1  and   ΔH^θ = –286 kJmol^–1 

Note: A very negative ΔH^θ 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!

 

Example 2. A light initiated reaction - hydrogen/chlorine mixture

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!).

H_2(g) + Cl_2(g) ==> 2HCl_(g)

ΔG^θ = –191 kJmol^–1 or  ΔH^θ = –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

CH_4(g) + Cl_2(g) ==> CH₃Cl_(g) + HCl_(g)

ΔG^θ = –103 kJmol^–1 or  ΔH^θ = –99 kJmol^–1

[full mechanistic details for methane]

 

Example 3. The instability of hydrogen peroxide

Hydrogen peroxide is thermally unstable, its aqueous solution is kept in a brown bottle to avoid decomposition initiated by uv or visible light photons.

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!

2H₂O_2(aq)  ==> 2H₂O_(l) + O_2(g)

See also "KINETICS" section 6.

Energetics-Thermochemistry-Thermodynamics Notes INDEX

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