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Revision notes on thermodynamics-energetics - calculating an equilibrium constant from free energy change - for Advanced A/AS Level Theoretical-Physical Chemistry:
Doc Brown's A Level Chemistry Advanced Level Theoretical Physical Chemistry – AS A2 Level Revision Notes – Basic Thermodynamics
GCE Thermodynamics–thermochemistry sub–index links below
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
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. 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
The standard Gibbs free energy change ∆Gθsys = ∆Hθsys – T∆Sθsys and
or for a gaseous equilibrium
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).
Ex 3.5a2 Calculate the equilibrium constant for the formation of ammonia
Ex 3.5a3 Calculating the equilibrium constant for an esterification
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
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
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.
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