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Advanced A Level Chemistry

Kinetics-Rates Part 6

 

6.2 kinetic stability v thermodynamic stability

do all reactions happen spontaneously just because they are thermodynamically feasible?


Advanced A Level Kinetics Index


6.2 Kinetics versus thermodynamic stability and reaction feasibility

IF A REACTION IS THERMODYNAMICALLY FEASIBLE, WILL IT AUTOMATICALLY HAPPEN?

6.2 Introductory points

  • Even before rates factors are considered, the feasibility of a reaction is governed thermodynamically, that is in order for a reaction to be able to occur the energy changes must be favourable.

  • In order for a reaction to feasible the overall entropy change ...

    • ΔSθtot must be >=0 and the overall free energy change ΔGθ must be <=0.

  • For a reaction the overall entropy change is given by:

  • Total entropy change = entropy change of system + entropy change of surroundings

    • ΔSθtot = ΔSθsys +  ΔSθsurr

    • ΔSθsys= ΣSθproducts – ΣSθreactants

    • and ΔSθsurr = –ΔHθsys–reaction/T(K)

      • Note the connection between ΔS and ΔG.

      • From above and substituting ...

      • ΔSθtot = ΔSθsys –ΔHθsys/T

      • multiplying through by T gives

      • TΔSθtot = TΔSθsys –ΔHθsys

      • changing signs throughout gives

      • –TΔSθtot = –TΔSθsys + ΔHθsys

      • –TΔSθtot is called the Gibbs Free Energy change symbol G,

      • so giving the familiar free energy equation ...

      • ΔGθsys = ΔHθsys – TΔSθsys 

      • The free energy can be thought of as heat energy that is available to do work.

        • OR in the case of cells, where ΔGθsys =  –neEθF,

        • this gives the electrical energy available to do useful work.

        • PLEASE NOTE you do not have to derive the Gibbs Free Energy equation but some syllabuses require you to be able to use when supplied with enthalpy and entropy data.

    • The free energy change of the reaction, ΔGθ must be <=0 to be feasible.

    • If ΔSθtot or ΔGθsys is close to zero, then you are likely to be dealing with equilibrium situation.

  • BUT, however feasible a reaction might be thermodynamically, it does not necessarily mean it will happen spontaneously because of kinetic limitations.

  • The speed at which a reaction takes place depends on the many factors described here or on the GCSE rates notes page. However, a feasible reaction that you might expect to take place, may not happen because the activation energy is so high that virtually no molecules have enough kinetic energy to change on collision. This would be described as a kinetically stable mixture but thermodynamically unstable.

The thermochemistry & thermodynamics dealt with in section 6. have been re–written and extended via

Part 1 Thermochemistry – Enthalpy changes * Part 2 Born–Haber Cycle * Part 3 Entropy & Free Energy

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6b. Kinetic stability/instability examples

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

    • Note: A very negative ΔHθ is usually, but not necessarily, indicative of a negative free energy change.

    • *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, it is free radical chain reaction. 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   and   ΔHθ = –185 kJmol–1 

      • 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   and   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) 


The thermochemistry & thermodynamics dealt with in section 6. have been re–written and extended via

Part 1 Thermochemistry – Enthalpy changes * Part 2 Born–Haber Cycle * Part 3 Entropy & Free Energytop


6c. Thermodynamic 'instability' – reaction feasibility

  • Energy distributed not just in translational KE, but also in rotation, vibration and also distributed in electronic energy levels (if input great enough, bond breaks).

  • All four forms of energy are quantised and the quanta ‘gap’ differences increases from trans. KE ==> electronic.

  • Entropy (S) and energy distribution: The energy is distributed amongst the energy levels in the particles to maximise their entropy.

  • Entropy is a measure of both the way the particles are arranged AND the ways the quanta of energy can be arranged.

  • We can apply ΔSθsys/surr/tot ideas to chemical changes to test feasibility of a reaction:

    • ΔSθtot = ΔSθsys +  ΔSθsurr

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

    • For example: CaCO3(s) ==> CaO(s) + CO2(g) 

      • ΔSθsys = ΣSθproducts – ΣSθreactants 

      • ΔSθsys = SθCaO(s) + SθCO2(g) – SθCaCO3(s) 

      • ΔSθsurr is ΔHθ/T(K) and ΔH is very endothermic (very +ve),

    • Now ΔSθsys is approximately constant with temperature 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 800oC ΔSθtot becomes plus overall (and ΔGθ becomes negative), so the decomposition is now chemically, and 'commercially' feasible in a lime kiln.

The thermochemistry & thermodynamics dealt with in section 6. have been re–written and extended via

Part 1 Thermochemistry – Enthalpy changes * Part 2 Born–Haber Cycle * Part 3 Entropy & Free Energy


Advanced A Level Kinetics Index


Revision notes for GCE Advanced Subsidiary Level AS Advanced Level A2 IB Revise AQA GCE Chemistry OCR GCE Chemistry Edexcel GCE Chemistry Salters Chemistry CIE Chemistry, WJEC GCE AS A2 Chemistry, CCEA/CEA GCE AS A2 Chemistry revising courses for pre–university students (equal to US grade 11 and grade 12 and AP Honours/honors level courses) kinetic stability/instability examples, thermodynamic 'instability'

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Advanced A Level Kinetics Index

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