doc bDoc Brown's Advanced A Level Chemistry

Advanced A Level Chemistry

Kinetics Part 2

Selected Case Studies, Arrhenius Equation, Kinetic v Thermodynamic Stability

Selected Case Studies of a variety of chemical reactions and their rate expressions – 4.1 Hydrolysis of halogenoalkanes : 4.2 Oxidation of iodide ions by peroxodisulphate : 4.3 Enzyme kinetics : 4.4 The hydrogen H2/iodine I2/ hydrogen iodide HI equilibrium : 4.5 Ester hydrolysis : 4.6 Acid catalyzed iodination of propanone : 4.7 Acid – thiosulphate/thiosulfate reaction  (use case studies index), the Arrhenius Equation – its implications and use to calculate activation energies and finally kinetic stability v thermodynamic stability – do all reactions happen spontaneously just because they are thermodynamically feasible?

Advanced Kinetics Part 1 index: 1. advanced particle theory: 1a. distribution of particle energies

1b. effect of increasing temperature : 1c. effect of a catalyst * 2. catalyst mechanisms : 2a. introduction

2b. heterogeneous examples–theory : 2c. homogeneous examples–theory

3a. Examples of obtaining rate data : 3b. rate expressions and orders of reactions

3c. deducing orders of reaction : 3d. Simple exemplar rates Q's

Advanced Kinetics Part 2 index: 4. case studies in kinetics * 5. Arrhenius equation for calculating activation energy

6. Kinetic versus thermodynamic stability : 6a. Introductory points

6b. Kinetic stability/instability examples : 6c. Thermodynamic 'instability'

Also Computer simulations of kinetic particle theory – Maxwell Boltzmann Distribution of particle speeds/KE's

You must know the basic GCSE stuff and most is NOT repeated here.

This page is directed at advanced level written papers but should be useful for kinetics study projects.

REMINDER before studying these 2 pages (c) doc b You should know everything in the GCSE/IGCSE NOTES

4. Selected case studies of kinetics


Case Studies 4.1 Hydrolysis of halogenoalkanes : 4.2 Oxidation of iodide ions by peroxodisulphate

4.3 Enzyme kinetics : 4.4 The H2/I2/HI equilibrium : 4.5 Ester hydrolysis

4.6 Acid catalyzed iodination of propanone : 4.7 Acid– thiosulphate reaction

Case study 4.1 The kinetics of the hydrolysis of halogenoalkanes

There is also a series of pages entirely devoted to organic reaction mechanisms!


  • Reaction kinetics: The possibility of two reaction mechanisms for the hydrolysis of halogenoalkanes (RX) with sodium hydroxide or water has consequences for the rate expressions.

    • (i) R3C–Cl + 2H2O ==> R3C–OH + Cl + H3O+ 

      • e.g. direct hydrolysis with water for tertiary halogenoalkanes, 3 R's = alkyl/aryl

    • (ii) R3C–Cl + OH ==> R3C–OH + Cl 

      • e.g. hydrolysis with sodium hydroxide for primary halogenoalkanes, 3 R's=2 H's and 1 alkyl/aryl.

  • The SN1 mechanism is sometimes described as 'unimolecular' because the rate only depends on the concentration of the halogenoalkane.

  • The mechanism for (i) has three steps of bimolecular collisions. Here the rate is only dependent on one reactant, the halogenoalkane, R3C–X, shown in step (1), (but it still has to collide with the solvent, which never seems to be shown at AS–A2 level and whose concentration is essentially constant!)

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  • Experimental results produce the overall 1st order rate expression: rate = k1[RX]

  • 1st order kinetics suggests there is a rate determining step involving one of the reactants, irrespective of the total number of steps, which in this case is three.

  • This is because the activation energy of the 1st step, forming the carbocation intermediate by heterolytic bond fission, is so high, that the speed is relatively low. Therefore step (1) alone determines the speed of the reaction. This is referred to as the rate determining step (or rds in shorthand!). Steps (2)/(3) have much lower activation energies and are much faster. You would register zero order for the order of reaction with respect to e.g. any hydroxide ion present or it might even hydrolyse just in water!

  • Note the simplicity of the rate expression, despite the complexity of the mechanism!

    • The 1st diagram (mechanism 10) shows the full reaction mechanism.

    • The 2nd diagram (mechanism 42) shows the reaction profile with step (1) having much the bigger activation energy and hence acting as the rate determining step. The two 'troughs represent the formation of the intermediates or transition states whatever their lifetime maybe!

  • topThe SN2 mechanism below for reaction (ii), is referred to as 'bimolecular' because the rate depends on the concentrations of both reactants.

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  • Experimental results produce the overall 2nd order rate expression: rate = k2[RX][OH],

    • and here the orders do indeed match the stoichiometry of the equation!

  • 2nd order kinetics suggests there is a rate determining step involving both of the reactants, irrespective of the total number of steps, though in this case it is just one step.

  • This is because it is a simple single step mechanism involving a bimolecular collision of the two reactant molecules/ions. The rate depends on both the halogenoalkane and hydroxide ion concentrations (1st order with respect to both reactants). 

    • The 1st diagram above (mechanism 2) shows the 'simple' mechanism.

    • The 2nd diagram (mechanism 33) shows the 'activated complex' or 'transition state'* which is the peak of the potential energy of the system (see diagram 41 below it) where the 'incoming' hydroxide ion is half–bonded to the carbon and the 'outgoing' chloride ion is still 'half–bonded' to the carbon. No intermediate is formed.

    • The 3rd diagram (mechanism 41) shows the energy changes as a reaction profile.

    • * Note that an 'activated complex' or 'transition state' is not the same as an intermediate like a carbocation which is a definite entity in its own right, however short its lifetime.

  • For full details and discussion of all halogenoalkane mechanisms see Mechanisms Part II



Case study 4.2 Oxidation of iodide ions by peroxodisulphate described in section on homogeneous catalysis


Case study 4.3 Enzyme kinetics – Biological Catalysts

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  • KEY in sequence: E = free enzyme, S = free substrate reactant molecule, ES = enzyme–reactant complex, EP = enzyme–product complex, E = free enzyme, P = free product



  • The mechanism of enzyme catalysed:

    • Step (1) E + S ==> ES

      • The 'docking in' of the substrate S of the onto the 'active site' of E to form the enzyme–substrate complex ES. This is often quoted as the rate determining step (rds), but it isn't (see step (2) below!).

    • Step (2) ES ==> EP

      • The conversion of the substrate–enzyme complex ES into the enzyme–product EP, is the rate determining step. The docking in of the substrate to be held e.g. by inter molecular forces is likely to have a low activation energy and a relatively high probability initially,  so E + S ==> ES cannot be the rds. However, ES ==> EP involves bond breaking and will have a much higher activation energy giving a much lower probability of ES  ==> EP transformation.

      • However, [ES] cannot be measured directly, so a simplified kinetics approach is to treat the formation and transformation of ES/EP through to P as a function of the concentrations of E and S (see the rate expressions below, which is a big bad fudge seen here and in most textbooks!). The situation should be dealt with via the Michaelis–Menten equation, but this goes way above AS–A2 chemistry demands.

    • Step (3) EP ==> E + P

      • The breakdown of the EP complex leading to the departure of the product P and leave the enzyme E, as a free 'active site' completing the catalytic cycle.

topA possible detailed reaction profile is shown below.

The uncatalysed profile would be a single, and much higher 'hump' (see 2nd simplified diagram below).

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  • Here Ea1 is small for the formation of the enzyme–substrate complex or intermediate ES. The energy of ES is lower than the reactants because it involves an energy releasing interaction whether it be via intermolecular forces or chemical bonding.

  • Ea2 is much larger for the transformation of the ES intermediate into the EP intermediate because it involves bond breaking in the substrate molecule.

    • Ea2 is still much smaller than Ea for the uncatalysed reaction, which is not shown here, but is shown in the simplified diagram below e.g. the uncatalysed decomposition of hydrogen peroxide solution has an Ea of 76kJmol–1, but with an enzyme like catalase/peroxidase, the Ea is reduced to 30kJmol–1 AND the rate of reaction is increased by a factor of 108, impressive biochemistry!!!

    • Ea3 is relatively small for the breakdown of the EP intermediate into the free product and free enzyme.

  • However, the simplified reaction profile diagram below should be good enough in an exam and shows the relevant activation energies.

    • Here, Ea1 is for E+S ==> ES, Ea2 is for ES ==> EP ==> E + P

      • (Ea for EP ==> E + P isn't shown separately)

    • and Ea3 for the uncatalysed reaction, which with no enzyme is significantly greater.

      • My thanks to Professor Martin Chaplin of London South Bank University for his most helpful emails concerning reaction profiles.

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  • Treating step (2) as the rds, gives the rate expression ...

    • rate = kES[ES],

    • that is the rate of reaction = rate of transformation of ES into EP .

    • However, since you cannot measure the concentration of the enzyme–substrate complex, the rate is treated as a function of the concentrations of the enzyme and substrate, [E] and [S], which form the ES intermediate, assuming a steady state situation, giving

      • rate = k1[E][S]

      • (Which is a considerably simplified approach if you care to research further in a thick undergraduate textbook of biochemistry and researching the Michaelis–Menten equation!)

    • So we are treating the reaction as 1st order with respect to both enzyme and substrate, 2nd order overall.

    • For a set of experiments, using a fixed amount of enzyme, the initial rate should be proportional to the concentration of substrate at relatively low concentrations, so the effective rate expression is pseudo 1st order ... 

      • rate = k2[S]

    • If you keep the substrate concentration constant and vary the quantity of enzyme, the effective rate expression is also pseudo 1st order ...

      • rate = k3[E]

  • topHowever, if the substrate concentration is very high, the maximum number of active centres on the enzymes are occupied and the rate is completely controlled by the transformation of the ES complex at its maximum possible concentration and independently of the high substrate concentration. This means you are now dealing with the maximum possible rate, which only depends on the amount of enzyme present,

    • so the rate expression becomes zero order with respect to the substrate,

      •  rate = k3[E]

    • and for a constant [E], and varying [S] at high concentrations of S the rate = k4,

    • The two situations are summed up by graph (1) below for a constant amount of enzyme. The graph moves from 1st order kinetics to zero order.

    • Graph (2) is a 'simple' comparison of the effects of no inhibitor, a constant amount of a competitive or a non–competitive inhibitor for a constant amount of enzyme.

      • Vo is the initial reaction rate for a constant amount of enzyme and varying the substrate concentration.

      • Vmax is the maximum possible rate when the maximum number of active centres are occupied with S or P.

      • Competitive inhibition: Although the inhibitor initially decreases the reaction rate, the substrate can increasingly compete with the inhibitor to occupy the 'active sites' as the substrate concentration increases (Le Chatelier's equilibrium concentration principle) so that the maximum reaction rate Vmax can eventually equal Vmax for the non inhibited enzyme.

      • Non–competitive inhibition: Since the inhibitor molecule does not bind to the 'active site' of the enzyme it cannot be 'replaced' by increasing the substrate concentration. However, because it constantly interferes with a fraction of the enzymes, the reaction rate will be reduced giving the lower the Vmax which can never reach Vmax however high the substrate concentration.

      • See also enzyme structure and function notes for more details on inhibition

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Case study 4.4 The H2/I2/HI equilibrium

The relationship between rate expressions and Kc equilibrium expressions



  • The gaseous phase equilibrium and kinetics involving hydrogen, iodine and hydrogen iodide has been very well studied quantitatively at temperatures of 250–500oC.

  • The reaction is: H2(g) + I2(g) doc b 2HI(g)  

  • The reaction mechanism, in either direction, is controlled by an initial bimolecular collision (rds) with a 'transition state' or 'activated complex' consisting of two hydrogen atoms and two iodine atoms.

  • The structure of the 'transition state' is not known and there are two possible mechanisms of either 2 or 4 steps.

  • However, the proposed mechanism of ...

    • (i) an initial rate determining step (rds) of I2 + H2 ==> intermediate state ==> products for the forward reaction,

    • and HI + HI ==> intermediate state ==> products for the backward reaction

    • ... is supported by the kinetics data which shows that ...

  • the rate expression for the forward reaction at equilibrium is:

    • ratef = kf[H2(g][I2(g)]

    • 1st order with respect to both reactants, hydrogen and iodine, 2nd order overall,

  • and the rate expression for the backward reaction at equilibrium is:

    • rateb = kb[HI(g]2 

    • 2nd order with respect to the only 'reactant', hydrogen iodide.

  • Now the equilibrium expression for the reaction is ...

    • Kc =  [HI(g]2/[H2(g][I2(g)], the equilibrium constant Kc has no units (dimensionless),

    • but since the rate expressions involve the same concentration expressions as the equilibrium expression and the rates of the forward and backward reaction are the same at equilibrium,

      • we can rearrange the rate expressions so that (where f = forward, b = backward)

    • Since for a dynamic equilibrium ratef = rateb

    • therefore ...

    • kf[H2(g][I2(g)] = kb[HI(g]2 

    • kf / kb = [HI(g]2 / [H2(g][I2(g)] = a constant at constant temperature

    • and this constant is the equilibrium constant Kc

    • Hence, an equilibrium can be derived from well proven rate expressions.

    • You can write the logic down in another way e.g.

    • [HI(g]2 = rateb/kb and [H2(g][I2(g)] = ratef/kf

    • therefore we can write: Kc = (rateb/kb)/(ratef/kf) = kf/kb (since the 'rates' cancel out)

    • so the equilibrium constant is the ratio of the two rate constants for the forward and backward reactions.

  • This a nice simple example to combine the concept areas of equilibrium and rates of reaction, but many other equilibrium reactions are not so simple to analyse in terms of rate expressions!

  • When a system is a dynamic equilibrium the rate of the forward reaction = rate of the backward reaction, so here the H2/I2/HI concentrations remain constant, but two reactions are simultaneously occurring.

  • Four points should be emphasised ...

    1. Rate expressions can only be obtained from experimental results.

    2. If both the rate expressions are known for a true dynamic equilibrium reaction, then it is possible to derive the correct Kc equilibrium expression and the Kc value at a given temperature.

    3. It is NOT possible to derive rate expressions from either (i) the stoichiometric (balanced) equation or (ii) the Kc equilibrium expression.

    4. It is of course possible, to derive the equilibrium expression from the stoichiometric equation, which can of course be verified by experiment, and more importantly, used to predict equilibrium concentrations for a given set of conditions.


Case study 4.5 Ester hydrolysis

  • Esters can be hydrolysed by (i) water alone (but very slow), but is (II) catalysed by acids (H+(aq) specifically) and (iii) alkalis (OH(aq) specifically). The latter hydrolysis (iii), is sometimes called a saponification reaction.

    • equation for (i)/(ii): RCOOR'(aq) + H2O(l) doc b RCOOH(aq) + R'OH(aq)

    • equation for (iii): RCOOR'(aq) + OH(aq) doc b RCOO(aq) + R'OH(aq)

  • Hydrolysis with water alone is usually too slow to obtain meaningful rate data.

  • However, with a fixed amount of acid catalyst e.g. HCl(aq), it is possible to follow the reaction by alkali titration and show that ...

    • rate = k1[RCOOR'(aq)], that is 1st order with respect to ester, however,

    • if the concentration of acid catalyst is varied, things are more complicated,

    • and the 2nd order expression, rate = k2[RCOOR'(aq)][H+(aq)] then applies.

  • With alkali, 2nd order kinetics are observed overall, and the reaction can be followed by titrating the remaining alkali with acid. Phenolphthalein (Pk(ind) = 9.3), would be a suitable indicator because one of the reaction products is the salt of a weak carboxylic acid (approx. pH 9). The product of the titration is neutral sodium chloride.

    • i.e. rate = k2[RCOOR'(aq)][OH(aq)], 1st order with respect to each reactant, 2nd order overall.


Case study 4.6 Acid catalyzed iodination of propanone


  • Propanone readily forms 1–iodopropanone on reaction with acidified iodine solution, as do all 2–ones ('methyl ketones') I assume?

    • CH3COCH3(aq) + I2(aq) ==> CH3COCH2I(aq) + H+(aq) + I(aq) 

  • However, the rate expression is: rate = k2[CH3COCH3(aq)][H+(aq)]

    • and iodine is not in the rate expression but one of the products is!

    • Therefore the reaction is zero order for iodine and it also zero order for bromine in the similar bromination reaction.

  • This suggests there is a slow rate determining step involving the ketone and the hydrogen ion in the mechanism and what ever happens next e.g. involving the iodine, is much faster. A proposed mechanism is shown below, where R = CH3 if it was propanone. Note that very little can be absolutely proved in mechanistic detail and you will find variations of this diagram on the web.

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  • The mechanism for propanone, with only one slow step suggested is ...

    • Step (1) (CH3)2C=O + H3O+ doc b (CH3)2C=O+H + H2O

      • The ketone is reversibly protonated on the oxygen (+) by the acid in an acid–base reaction (proton transfer).

    • Step (2) (CH3)2C=O+H doc b  (CH3)2C+–OH

      • The electrons 'between' the C–O partly shift to form a carbocation i.e. the positive charge is transferred from the oxygen to the carbon.

    • Step (3) (CH3)2C+–OH + H2O doc b CH3C(OH)=CH2 + H3O+ 

      • The carbocation, derived from the protonated ketone, loses a proton and slowly changes into the 'enol'* form.

      • This involves breaking a strong C–H bond, hence a high activation energy and slow speed. The positive charge on the adjacent carbon of the carbocation facilitates in 'pulling' the C–H bond pair to form the C=C bond and release the proton to a water molecule.

      • The rate of formation of the enol thus depends on the concentrations of the ketone and the acid, explaining the rate equation experimentally found. Also note that it is an example of autocatalysis because one of the reaction products is the oxonium ion!

      • * An 'enol' has both alkene and alcohol functional groups and is isomeric with the original ketone. This is an example of functional group isomerism involving a dynamic equilibrium of the two isomers (the original ketone and enol formed) and is sometimes called an example of tautomerism.

    • Step (4)  CH3C(OH)=CH2 + I2 doc b CH3C(=O+H)–CH2I + I 

      • Half of the iodine molecule (an electrophile) then quickly adds to the 'enol' (just like any reactive alkene), and the oxygen then carries the positive charge (not on the carbon as in electrophilic addition to alkenes), and the protonated iodoketone is formed.

    • Step (5) CH3C(=O+H)–CH2I + H2doc b  CH3COCH2I + H3O+ 

      • Then a water molecule rapidly, and reversibly, removes the proton in another acid–base reaction to leave the iodo–ketone.

    • For the slow step (3), the rate depends on the isomerisation of protonated ketone/carbocation, which in turn will depend on the concentration of the ketone AND the acid providing the H3O+ ion.

  • Note:

    • The same reaction is catalysed by bases and proceeds by a different mechanism and gives different products ultimately. Multiple substitution takes place, initially forming 1–iodopropane, then 1,1–diidopropane, and then 1,1,1–triiodopropane. Finally, the carbon chain splits to give triiodomethane, CHI3, i.e. its the 'iodoform' reaction given by ethanol, ethanal, and all 2–ones ('methyl ketones').

    • The individual products are almost impossible to isolate in the base catalysed reaction, but in the acid catalysed reaction, the rate of halogenation decreases with successive halogen atom substitution, so it is possible to isolate e.g. 1–iodopropanone, 1,1–diiodopropanone and 1,1,1–triiodopropanone and presumably? molecules such as 1,3–diodopropanone may be formed, but I'm not sure on this one?

    • This mechanism is often presented, and not unreasonably at UK A level, as a three step mechanism, with Step (1) as the rds and clearly showing the proton's role in this acid catalysed reaction.

      1. Step (1)  (CH3)2C=O + H+ reversible mechanism step (CH3)2C=O+H

        • SLOW protonation of the ketone by the hydrogen ion

      2. Step (2)  (CH3)2C=O+H reversible mechanism step CH3C(OH)=CH2 + H+

        • FAST deprotonation and rearrangement to give the enol form)

      3. Step (3)  CH3C(OH)=CH2 + I2 mechanism step CH3COCH2I + HI

        • FAST equivalent of adding I–I across the carbon–carbon double bond >C=C< and then elimination of HI.


Case study 4.7 The acid catalysed decomposition of sodium thiosulphate

  • The reaction between e.g. dilute hydrochloric acid and sodium thiosulphate is a redox reaction catalyzed by hydrogen ions. I do not know of any other catalysts?

  • Its a typical 'rates' reaction at GCSE level to illustrate temperature and concentration factors or used as a coursework investigation. Its followed by the time it takes to form enough sulphur to obscure a black X marked on white paper. The method described in more detail on the (c) doc b GCSE rates page.  It is possible to follow the reaction with a colorimeter due to the light scattering effect of the colloidal sulphur particles but the absorbance does not follow Beers Law and processing results is apparently difficult!

  • The reaction is ...

  • Na2S2O3(aq) + 2HCl(aq) ==> 2NaCl(aq) + SO2(aq) + S(s) + H2O(l) 

  • which for advanced level is much more appropriately written in the ionic form ...

  • S2O32–(aq) + 2H+(aq) ==>  SO2(aq) + S(s) + H2O(l) 

  • The thiosulphate ion on face value has an S=S bond, one S=O and two S–O bonds, but all four are 'merged' in the same delocalised pi bonding system.

    •  You could say that one sulphur is in the zero oxidation state and one in the +4 ox. state,

    • OR, perhaps 'safer' to argue, that on average each sulphur is in the +2 state (oxygen is –2, overall charge on ion 2–).

  • In the products, the oxidation states of sulphur are much clearer to define, +4 in SO2(aq) and 0 in S(s).

    • On the basis that two S's start off in the +2 state, you could argue that this is a disproportionation reaction, in which an element in an ion/compound/compound is simultaneously oxidised (+2 to +4) and reduced (+2 to 0).

  • You would expect that the rate might be controlled by the interaction of the negative thiosulphate ion and a positive hydrogen ion. You would expect the interaction of oppositely charged ions to have a relatively low activation energy, so in the rate expression:

  • rate = k[S2O32–(aq)]t[H+(aq]h, you might expect the order t and h to be 1, t=1 is quoted on the web.

  • but its likely to be at least a two step mechanism, so whether h is 1 or 2 or ?, I don't know? Whatever, the orders t and h can only be found by experiment.

    • S2O32–(aq) + H+(aq) ==> intermediate

    • intermediate ==> SO2(aq) + S(s) + H2O(l)

      • or intermediate + H+(aq) ==>  SO2(aq) + S(s) + H2O(l) 

      • or intermediate + H+(aq) ==>  intermediate 2 ==> SO2(aq) + S(s) + H2O(l) 

      • I don't know the details but you would expect the negative thiosulphate ion to combine with the 1–2? protons to form some intermediate that breaks down in one or more steps to give sulphur dioxide, sulphur and water.


5. Deriving the activation energy Ea from kinetics–rates data



  • The Arrhenius equation quantitatively describes the relationship between the rate constant k, temperature and the activation energy. The rate constant value increases with increase in temperature and nothing else varies it!

    • k = A exp(–Ea/RT) 

      • k = rate constant (from the rate expression)

      • A = a constant for a given reaction, sometimes called the 'frequency' or 'pre–exponential' factor and it seems to be linked to stereochemical factors i.e. the spatial aspects of reactant particle collision.

      • Ea = activation energy in Jmol–1 

      • R = ideal gas constant = 8.314 Jmol–1K–1 

      • T = temperature in K (Kelvin = oC + 273)

  • Rewriting the Arrhenius equation in logarithmic form gives:

    • ln(k) = ln(A) – Ea/(RT)

    • or log10(k) = log10(A) – Ea/(2.303RT)

  • From accurate rate data at different temperatures (e.g. 5 or 10o intervals and a minimum of four results) you can calculate the values of k OR more simply, for a fixed 'recipe', a set of 'rate' results.

  • You then plot  the value of ln(k or relative rate) versus the reciprocal temperature in Kelvin. 

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  • The negative gradient of the graph is equal to –Ea/R,

  • so, Ea = –R x gradient (in J, and /1000 => kJ).

    • In terms of y = mx + c, m = gradient = dy/dx = Ea/R, c = a constant = ln(A),

    • so, in terms of the Arrhenius equation, the algebra equates to

    • ln(k) = – Ea/(RT) + ln(A)

  • Example calculation 5.1

    • Some accurate rate constant data for various temperatures is tabulated for the reaction between hydrogen and iodine to form hydrogen iodide is presented on a sub–web page.

    • H2(g) + I2(g) ==> 2HI(g)

    • rate = k2 [H2(g)] [I2(g)]

    • The Arrhenius plot of ln(k) versus 1/T is also shown.

    • The Excel plot routine conveniently provides the best line fit.

    • y = –19867 + 25.651

    • therefore gradient = –19867 = –Ea/R

    • so, Ea = 19867 x 8.314 = 165174 J mol–1,

    • activation energy, Ea = 165 kJ mol–1 (3 sf, 165.2 4sf)

    • The constant A can be calculated as follows:

    • The constant 25.651 = ln(A), therefore A = e25.651 , so A = 1.325 x 1011,

    • so the full Arrhenius expression for the hydrogen iodine reaction is

    • k = Ae(Ea/RT) = 1.325 x 1011 x e(165174/8.314T)


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

6. Kinetic versus thermodynamic stability


6a. 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


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

REMINDER before studying these two connected advanced A level kinetic pages STUDY THE BASICS !


1. What do we mean by the rate/speed of reaction? An INTRODUCTION to REACTION RATES

Why is it important to know how fast reactions occur?

How can we measure the speed or rate of a chemical reaction?

Experimental methods are described in this introduction & subsequent sections

2. Collision theory of chemical reaction rate factors A general introduction, then applied to sections 3a to 3e

3. Factors affecting the speed/rate of a chemical reaction:

3a Effect on rate of changing reactant concentration in a solution/gas mixture

3b Effect on rate of changing pressure in reacting gases (gaseous equivalent of 3a)

3c Effect on rate of changing particle size/surface area & stirring of a solid reactant

3d Effect on rate of changing the temperature of reactants

3e Effect on rate of using a catalyst in a chemical reaction

Activation energy and reaction profiles

3f Light initiated reactions - effect of changing intensity

4. Examples of graphs of rate data, interpretation a summing up of ways your rate data can be analysed and interpreted

Some ideas to help you with RISK ASSESSMENT when designing and performing 'Rates of Reaction' experiments and many other experimental procedures. Some aspects listed are applicable to 'rates/kinetics' assignments, others are not applicable at all. Its up to you to decide!

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)

keywords–phrases: 4. case studies in kinetics * 5. Arrhenius equation for calculating activation energy * 6. Kinetic versus thermodynamic stability : 6a. Introductory points : 6b. Kinetic stability/instability examples : 6c. Thermodynamic 'instability' Also Computer simulations of kinetic particle theory – Maxwell Boltzmann Distribution of particle speeds/KE's 4.1 Hydrolysis of halogenoalkanes : 4.2 Oxidation of iodide ions by peroxodisulphate : 4.3 Enzyme kinetics : 4.4 The H2/I2/HI equilibrium : 4.5 Ester hydrolysis : 4.6 Acid catalyzed iodination of propanone : Case study 4.7 Acid– thiosulphate reaction : 

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