Advanced Level Chemistry – Kinetics Part 1 Temperature, Catalysis, Orders of Reaction, Rate Expressions

This pages the advanced particle collision theory with reference to the Maxwell–Boltzman distribution of particle kinetic energies and using the distribution curves–graphs to explain the effect of increasing temperature and the theory of catalysed reactions. The theory of catalytic mechanisms is discussed using heterogeneous examples and homogeneous examples. Experimental techniques i.e. examples of how to obtain rate data and deriving rate expressions and explaining and deducing orders of reactions with some exemplar rates questions involving deducing and using orders of reactants and rate expressions.

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 YOU SHOULD KNOW ALL THE RATES IDEAS IN THE GCSE/IGCSE NOTES

1. More advanced particle theory to help explain the kinetic effects of temperature change and catalysis

1a. The statistical distribution of particle kinetic energies

• THE PARTICLES IN A LIQUID/SOLUTION/GAS HAVE A RANGE OF KINETIC ENERGIES

• AND THIS HAS CONSEQUENCES WHEN CONSIDERING KINETIC FACTORS SUCH AS THE VARIATION OF THE SPEED OF REACTION AT DIFFERENT TEMPERATURES.

• In any gas or liquid the particles are in total random motion in all directions with a huge range of kinetic energies (or velocities).

• At room temperature there are about 1028 particle collisions per cm3 every second which means on average an individual particle undergoes over 109 (1000 million) collisions per second!

• This is a 'strange' world where dimensions are incredibly small and 'event' times are correspondingly short!

• In fact the lifetime of an 'intermediate' in a reaction mechanism might be as little as 10–9 of a second!

• The distribution of the translational kinetic energy (KE) of the particles is derived from the statistical mathematics of Maxwell–Boltzmann and the KE distribution curve for a given 'population' of molecules is shown in the graph above.

• The average KE is just to the right of the peak because, although there is a lower limit of zero for KE, theoretically there is no upper limit, it just depends on how hot the gas/liquid is.

• The peak of the curve equals the most probable KE in this unsymmetrical distribution.

• Although there is virtually no chance of a particle having a KE of zero because of the collision frequency, though there is always a chance of a small proportion of the particle population having a KE way above the average AND it is this small fraction of high KE molecules which collide with enough KE to break open bonds i.e. to allow the reactants to overcome the activation energy and form products.

• An understanding of the statistical nature and shape of the Maxwell–Boltzmann particle KE distribution graph is crucial to a higher level understanding of the effect of (i) temperature change, and (ii) a catalyst, on the speed of a chemical reaction, especially as little as 1 collision in 104 to 1011 can leads to product formation! Most collisions are NOT 'fruitful'!

• Computer simulations of kinetic particle theory – Maxwell Boltzman Distribution of particle speeds/KE's

1b. The Effect of Increasing temperature and Activation Energy

• WHY DOES THE RATE/SPEED OF A REACTION INCREASE WITH TEMPERATURE?

• WHAT HAS THIS TO DO WITH THE DISTRIBUTION OF PARTICLE KINETIC ENERGIES AND THE ACTIVATION ENERGY?

• When the temperature is raised the added 'heat energy' shows itself in the form of increased particle kinetic energy. In the graph above, two distribution curves are shown for a lower/higher temperatures, T1/T2, and it is assumed that the area under the whole curve is the same for both temperatures i.e. the same number/population of molecules.

• Comparing lower temperature T1 with higher temperature T2, you can see that as the temperature increases, the peak for the most probable KE is reduced, and more significantly with the rest of the KE distribution, moves to the right to higher values so more particles have the highest KE values.

• Now, if we consider an activation energy Ea, the minimum KE the particles must have to react via e.g. bond breaking, the fraction of the population able to react at T1 is given by the blue area.

• However, at the higher temperature T2, the fraction with enough KE to react is given by the combined blue area plus the red area.

• Therefore, because of the shift in the distribution at the higher temperature T2, a greater fraction of particles has the minimum KE to react and hence a greater chance of a fruitful collision happening i.e. reactant molecule bonds breaking en route to product formation.

• In the diagram, for the sake of argument, a temperature rise from T1 to T2 results in the fraction of particles with a KE of >=Ea being doubled (area blue==>blue+red).

• For reactions with an activation energy in the range 50–100 kJmol–1 (i.e. most reactions), this results in an approximately  doubling of the reaction rate for every 10o rise in temperature i.e. where T2 = T1 + 10, because if you double the number of particles of KE >= Ea, you therefore double the chance of a fruitful collision and hence double the rate of reaction.

• So, a relatively small change in temperature e.g. 10o rise, can have quite a 'statistically' dramatic effect on the small, but significant population of the highest KE molecules, hence a significant change in reaction rate.

• The last point accounts for why a plot of rate versus temperature shows an 'exponential' or 'accelerating' curve upwards. Almost all reaction rates increase by a factor of 1.5 to 4.5 on doubling the temperature, but it does depend on the actual activation energy, so the "10o temperature rise effect" is a very rough rule of thumb when we say it "doubles the rate"!

• 2nd minor factor note:

• The rise in temperature does lead to an increase in collision frequency, and hence an increase in the possibility of a 'fruitful' collision and so increasing the speed of a reaction.

• However, this effect on the rate of reaction, is proportionally much smaller by a factor of 100–200x, compared to the increase in reaction speed due to the increase in the proportion of high KE molecules on increasing the temperature as described above.

• Computer simulations of kinetic particle theory – Maxwell Boltzman Distribution of particle speeds/KE's

1c. The effect of a catalyst

• HOW DOES A CATALYST AFFECT THE ACTIVATION ENERGY? AND HOW DOES THIS AFFECT THE REACTION KINETICS?

• A catalyst speeds up a reaction, but it must be involved 'chemically', however temporarily, in some way, and is continually changed and reformed as the reaction proceeds.

• Catalysts work by providing an alternative reaction pathway of lower activation energy, e.g. it can assist in endothermic bond breaking processes (see section 2. Catalytic mechanisms for some examples).

• If you consider the KE distribution curve above, at a fixed temperature, the green area shows the molecules which have sufficient KE to react and overcome the activation energy Ea1 for the un–catalysed reaction.

• However, in the presence of a catalyst, the lower activation energy Ea2, allows a much greater proportion of the molecules to have enough energy to react at the same temperature.

• This is shown by the combined green area plus the purple area and this increased fraction of molecules (increased area) considerably increases the chance of a 'fruitful' collision leading to product formation, so speeding up of the reaction.

• There are lots of examples of catalysts and mechanisms in Part 2

2. Catalytic mechanisms

2a. Introduction

• REMINDER OF WHAT A CATALYST IS AND HOW DO CATALYSTS WORK?

• ARE THERE DIFFERENT KINDS OF CATALYSTS?

• A catalyst is a substance that alters the rate of chemical reaction without itself being permanently chemically changed. Never state things like "it doesn't react, just speeds it up". It must take part in the reaction and it must change chemically, albeit on a temporary basis. A catalyst provides a different 'pathway' or mechanism that makes the bond breaking processes (or other electronic changes in the reactants) occur more readily. Catalysis also involves the formation of intermediates, not just a matter of an 'activated complex' or 'transition state'.

• e.g. for a transition metal the reactant molecules may be adsorbed and their bonds weakened, or, for a transition metal compound, it may involve a change in ligand or oxidation state or other bonding re–arrangement, but will return to is original state often via a 2–3 stage 'catalytic cycle'.

• In organic chemistry, e.g. in acid catalysis, the reactant molecule may be protonated to form some 'active' protonated intermediate e.g. a carbocation in the case of adding water to alkenes to form alcohols.

• By providing an alternative reaction pathway of lower activation energy Ea, compared to the uncatalysed reaction, e.g. see the diagram below for a simple exothermic reaction, although more realistically, it is usually a more complex cycle profile of at least two stages (see the 2nd reaction profile further down.

• A catalyst can be changed physically e.g. the granules can end up more powdery or the surface become roughened. This may be due to a heat effect from exothermic reactions or just side effect of regeneration in the catalytic cycle.

• e.g. in the laboratory preparation of oxygen from the MnO2(s) catalysed decomposition of hydrogen peroxide solution, the residual water seems stained brown due to very fine particles of MnO2(s).

Two primary modes of catalytic action – heterogeneous and homogeneous

2b. Heterogeneous catalysts and theory

• HETEROGENEOUS CATALYST THEORY:

• A HETEROGENEOUS CATALYST IS IN A DIFFERENT PHASE (often solid state') THAN THE REACTANTS (often gaseous or liquid/solution)

• The catalyst and reactants are in different phases (usually solid catalyst and liquid/gaseous reactant)

• The reaction occurs on the catalyst surface which may be the transition metal or one of its compounds, examples quoted above. The reactants must be adsorbed onto the catalyst surface at the 'active sites'.

• This can be physical adsorbed or 'weakly' chemically bonded to the catalyst surface. Either way, it has the effect of concentrating the reactants close to each other and weakening the original intra–molecular bonds within the reactant molecules and so allows a greater chance of 'fruitful' collision.

• The diagram above illustrates a typical heterogeneous catalysis e.g. hydrogenation of alkenes with hydrogen and a nickel catalyst.

• The strength of adsorption is crucial to having a 'fruitful' catalyst surface,

• The bonding to the catalyst surface (chemisorption/adsorption) must be strong enough to weaken reactant molecule bonds but weak enough to allow new bonds to form and the products to 'escape' from the catalyst surface (desorption). Typical examples to illustrate this idea ....

• If the catalyst–reactant bonding is too strong, most reactant/product molecules will be too strongly 'chemisorped' inhibiting reaction progress e.g. tungsten (W),

• If the catalyst–reactant bonding is too weak, many reactants are not chemisorped strongly enough to allow the initial bond breaking processes to happen e.g. gold (Au) and silver (Ag) tend to be more limited catalysts,

• but even silver, can act as a catalyst for some reactions.

• just right: copper (Cu), nickel (Ni), platinum (Pt), rhodium (Rh), palladium* (Pd) catalyse many reactions such as hydrogenation, redox reactions involving CO and NO etc.

• *Palladium can catalyse the spontaneous combustion/combination of hydrogen and oxygen at room temperature!

• In catalytic converters the very expensive metals Pt, Rh and Pd are used.

• Cu and Ni are cheaper alternatives but they are more vulnerable to catalytic poisoning of the active sites by traces of sulphur dioxide in the exhaust gases.

• Unfortunately it would appear that the cheaper alternative metals chemisorb the 'catalytic poisons' more strongly the more expensive Pt and Rh.

• Once poisoned, the catalyst in a converter cannot be regenerated and its a new costly converter!

• It is usual to use the catalyst in a finely divided form to maximise surface area to give the greatest and therefore most efficient rate of reaction.

• This means the catalyst must be physically supported. e.g. the platinum–rhodium metal is distributed on a temperature resistant ceramic support in catalytic converters of motor vehicle exhausts.

• Catalyst poisoning should be avoided if at all possible and in industry catalysts have to be replaced or 're–furbished' via suitable physical/chemical treatment. The poisoning inhibiting effect is caused by impurity molecules being strongly chemisorbed (chemically bonded) to the most active sites* of the catalyst surface. It considerably reduces the efficiency of the catalyst, especially as the most effective catalyst sites bind impurities the strongest, competing with the reactant molecules e.g.

• Sulphur poisons the iron catalyst in the Haber Process for making ammonia (probably chemisorbed to form iron sulphide).

• Lead poisons the platinum–rhodium surface in car exhaust catalytic converters, hence the need for 'un–leaded' fuel. Lead is strongly absorbed preferentially onto active sites.

• Active sites* Not all the surface of a catalyst is effective due to minute imperfections in crystal structure at the atomic level. The real crystal lattices are far from the geometrical perfection we present them as in our diagrams for teaching purposes. For catalytic surfaces, this appears to be desirable up to a point! It has been shown that an irregular catalytic surface is more effective than a 'perfect' plane one. These irregularities in the first few layers of atoms can be an atomic holes (vacancies), steps and terraces, and planes of atoms not quite in line with other layers of atoms (dislocations). One theory of 'active sites' suggests the substrate reactant molecules are more strongly bound, aiding bond scission, if surrounded by more of the catalyst atoms in a 'hole' or on a 'step/terrace', but the same effect applies to catalyst poisons too!

• Below is a diagram of a two stage reaction profile for a catalytic cycle (Ea = activation energy)

• This can apply to heterogeneous catalysis or homogeneous catalysis.

• Ea1 is the activation energy leading to the formation of an intermediate complex.

• Ea2 is the activation energy for the change of the intermediate complex into products.

• Ea3 is the activation energy of the uncatalysed reaction.

• In heterogeneous catalysis the catalyst and reactants are in different phases (usually solid catalyst and liquid/gaseous reactant) e.g.

• Case study 2b.1: The black insoluble powder, manganese(IV) oxide, MnO2(s),  catalyses the decomposition of hydrogen peroxide solution into water and oxygen.

• 2H2O2(aq)  ==> 2H2O(l) + O2(g)

• Case study 2b.2: Iron, Fe(s), catalyses the combination of nitrogen and hydrogen gases in the important industrial Haber synthesis of ammonia, important in the manufacture of nitric acid* and artificial fertiliser salts.

• N2(g) + 3H2(g) ==> 2NH3(g)

• * In the 1st stage of making nitric acid the ammonia is oxidised to nitrogen(II) oxide by mixing it with oxygen over a hot platinum catalyst (another heterogeneous catalysis) which is then reacted with oxygen and water to form nitric acid solution. See GCSE ammonia page.

• Case study 2b.3: Platinum/rhodium/palladium metals, Pt(s)/Rh(s)/Pd(s),on a ceramic support, catalyse the following reactions in car exhausts inside the catalytic converter.

• 2NO(g) + 2CO(g) ==> N2(g) + 2CO2(g)

• The NO and CO are adsorbed onto the catalyst surface, bonds broken and reformed prior to the products nitrogen and carbon dioxide leaving the catalyst surface in a similar way to the hydrogenation illustrated above. Remember that the bonding to the catalyst surface (chemisorption/adsorption) must be strong enough to weaken reactant bonds but weak enough to allow the products to 'escape' (desorption)

• The CO is from the inefficient combustion of the hydrocarbon fuel,

• CxHy + (x/2 + y/4)O2 ==> xCO + y/2H2O

• and the nitrogen(II) oxide is 'naturally' formed at high temperature in the engine (as it is in lightning strikes!).

• N2(g) + O2(g) ==> 2NO(g)

• These transition metal catalysts can also oxidise unburned hydrocarbons from inefficient combustion.

• CxHy + (x+y/4)O2 ==> xCO2 + y/2H2O

• Note: Pt, Rh and Pd are very expensive metals and copper and nickel are cheaper alternatives but they are vulnerable to catalytic poisoning by traces of sulphur dioxide in the exhaust gases. Once poisoned, the catalyst in a converter cannot be regenerated, so, its a new costly converter!

• Case study 2b.4: Nickel, Ni(s), catalyses the addition of hydrogen to an alkene double bond, e.g. in the hydrogenation of unsaturated vegetable oils to make more saturated margarine with a slightly higher softening point making it more spreadable.

• –CH=CH– + H2 ==> –CH2–CH2

• The diagram above illustrates this typical heterogeneous catalysis.

• Case study 2b.5: Solid heterogeneous catalysts are really important in the petrochemical industry e.g.

• Isomerisation: These reactions convert linear alkane vapours into branched alkanes of the same carbon number over a platinum–aluminium oxide (Pt/Al2O3) catalyst at 150oC. Branched alkanes have a higher octane rating than linear alkanes, so a better petrol fuel components.

• e.g. hexane ==> methylpentanes or dimethylbutanes

• Reforming: Converting straight chain alkane vapour into cyclic alkanes and aromatic hydrocarbons can be achieved by using a Pt/Al2O3 catalyst at 500oC. Aromatic hydrocarbons are important chemical feedstock to make many useful aromatic compounds.

• e.g. hexane ==> cyclohexane (+H2) ==> benzene (+3H2)

• Cracking: Catalytic cracking of vapourised hydrocarbons at e.g. 500oC using zeolites to make more lower alkanes for petrol or alkenes. Alkenes are important intermediates in making many useful compounds from anti–freeze to plastics.

• e.g. higher alkanes ==> lower alkanes and alkenes

• Note: Zeolites tend to become 'poisoned' with carbon–soot deposits in the high temperature cracking reactions and this blocks the adsorption of the hydrocarbons. However, in this case, the catalyst can be regenerated in a separate container through which very hot air is passed to burn off the carbon–soot deposits.

• Case study 2b.6: Vanadium(V) oxide, V2O5, is used as a catalyst in the 'Contact Process' in the production of sulphur trioxide for the manufacture of sulphuric acid.

• The catalysing of the conversion of sulphur dioxide into sulphur trioxide is explained via change in oxidation state changes.

• 2SO2(g) + O2(g) ==> 2SO3(g)

• The mechanism, somewhat simplified, it goes via the catalytic cycle ...

• (i) SO2 + V2O5 ==> SO3 + V2O4, then (ii) V2O4 + 1/2O2 ==> V2O5

• The vanadium changes oxidation state from +5 to +4 and back to +5 in the cycle.

2c. Homogeneous catalysis and theory

• A HOMOGENEOUS CATALYST IS IN THE SAME PHASE

• (usually a solution of a catalyst and a solution of the reactants – contrast with a heterogeneous catalyst)

• The catalyst and reactants are in the same phase (usually a solution), and so the catalysed reaction can happen throughout the bulk of the reaction medium.

• Catalysis can be due to temporary changes in the oxidation state and ligand(s) of a transition metal ion and results in a 'catalytic cycle'. In other words, the reaction occurs via some intermediate species and the original catalyst is reformed e.g.

• Case study 2c.1: Iron(II)/iron(III) ions catalyse the oxidation of iodide ions by peroxodisulphate

• The uncatalysed reaction is overall is ...

• (a) S2O82–(aq) + 2I(aq) ==> 2SO42–(aq) + I2(aq)

• However, this 'direct' uncatalysed reaction involves the collision of two highly repelling negative ions and so has a very high activation energy (Ea3 in the diagram above).

• BUT, the collision of an Fe3+ ion and an I ion involves a positive ion–negative ion attraction, reducing repulsion, so this interaction which has a much lower activation energy.

• Initially, the 1st step overall for the catalysed reaction is ... (Ea1 in diagram above)

• (b) 2Fe3+(aq) + 2I(aq) ==> 2Fe2+(aq) + I2(aq)

• Fe2+ is the 'intermediate', and in the 2nd step overall, it is oxidised to Fe3+ and the peroxodisulphate ion is reduced to sulphate ion ... (Ea2 in diagram above)

• (c) 2Fe2+(aq) + S2O82–(aq) ==> 2SO42–(aq) + 2Fe3+(aq)

• So, the iron(III) ion is regenerated in the cycle, showing the iron(II/III) ions act in a genuine catalytic cycle but remember it cannot be simply two steps, the above must represent the summations of at least four steps.

• Note 1: It doesn't matter whether you start with the iron(II) or iron(III) ion, catalysis will occur because the peroxodisulphate would oxidise some Fe2+ to Fe3+ (reaction b) and the Fe3+ then oxidises the iodide!

• Note 2: If you added up the two equations (b + c) of the cycle you get equation (a) showing the overall reaction change.

• Note 3: The full mechanism must be quite complex e.g. at least 4 steps because the chances of three particles colliding in the right way (a termolecular collision) and with sufficient frequency is unlikely. Most mechanisms proceed by bimolecular collisions, whatever the overall order of the reaction!

• The rate expression (explained below in section 3.)for the uncatalysed reaction is:

•   rate = k[S2O82–(aq)][I(aq)], so what will it be for the catalysed?

• maybe rate = k[S2O82–(aq)][I(aq)][Fe2+(aq)] ?

• or [rate = k[Fe2+(aq)][I(aq)] ?

• or rate = k[S2O82–(aq)][Fe2+(aq)]?,

• depending on the activation energies of the steps, but I don't know on this one???

• Case study 2c.2: The autocatalysis by Mn2+ ions when the oxidising agent potassium manganate(VII), KMnO4, is used to titrate the ethanedioate ion, C2O42–, or the acid H2C2O4 / (COOH)2, (old names 'oxalate/oxalic acid').

• The titration, particularly at the start, is too slow, so the initial mixture is heated to 60oC before starting the KMnO4 addition. One reason why the initial reaction is slow, is because the two ions involved (C2O42– and MnO4) are both negative, so on collision the repulsion force factor is high, so the activation energy is high giving low kinetic activity.

• However, as the titration proceeds, the decolourisation of the added manganate(VII) speeds up as the conical flask is swirled. The reason for this, is that one of the reaction products, the Mn2+ ion, can itself act as a catalyst for this redox reaction, hence the 'auto–catalysis'

• The overall balanced redox reaction is ... based on ... 2 x Mn(+7) ==> 2 x Mn(+2) and 10 x C(+3) ==>10 x C(+4)

• 2MnO4(aq) + 16H+(aq) + 5C2O42–(aq) ==> 2Mn2+(aq) + 8H2O(l) + 10CO2(g)

• or 2MnO4(aq) + 6H+(aq) + 5H2C2O4(aq) ==> 2Mn2+(aq) + 8H2O(l) + 10CO2(g)

• The intensely purple coloured  manganate(VII) ion is reduced by the ethanedioate ion to the very pale pink (virtually colourless) manganese(II) ion in the presence of dilute sulphuric acid.

• The intense colour of the manganate(VII) ion is not due to the splitting of the 3d electron levels by the 'field splitting' ligands, and the subsequent absorption of visible light photons, causing electronic promotion from the lower to higher upper 3d sub–level, which is the usual case for transition metal complexes. The MnO4 ion is known as a charge transfer complex ion, because the colour is due to the electronic transitions of oxygen's 2p electrons. These are temporarily raised from a full oxygen 2p orbital to the vacant 3d or 4s levels of the Mn(VII) ion.

• Initially, on face value, the reactant collision is between two anions (–) which will have a high activation energy due to extra repulsion as well as the outer electron–outer electron repulsion that exists between any two atoms/molecules/ions, hence the slow start to the reaction. However, in the acid medium, the organic species is likely to be mainly in the form of the unionised acid H2C2O4 but the activation energy is obviously high until the first traces of the catalyst Mn2+ are formed, when the lower energy pathway kicks in!

• In the titration, the first few drops of manganate(VII) seem to take a relatively long time to 'decolourise' (from purple to colourless), but as the titration proceeds, the decolourisation becomes faster because the Mn2+(aq) complex ions act as a catalyst in the oxidation of the ethanedioate ion and recent research has shown that colloidal manganese(IV) oxide, MnO2,is involved too (MnO2 in bulk is insoluble). If you carry out the reaction without the addition of acid you actually see brown–black MnO2 formed as a precipitate, which is a neutral or alkaline solution reduction product of MnO4.

• At GCE AS–A2 level this is often quoted as an example of auto–catalysis BUT it is an extremely complicated reaction. The reaction is still a subject of research and the latest theories involve at least nine mechanism steps and the formation of colloidal manganese(IV) oxide, MnO2, and complexes of MnO2 combined with ethanedioic acid. This reaction is seriously complicated!

• Incidentally, if you mix potassium manganate(VII), acidified ethanedioic acid and manganese(II) sulphate to deliberately speed up the reaction apparently you see a red MnIII complex formed, maybe [Mn(C2O4)3]3– which will be a part of one pathway in the multi–catalytic cycle.

• Well, make of it what you can! but to sum up for GCE AS–A2 level, and not undergraduate level purposes the sequence in terms of the initial/intermediate reactants and products is basically ... (the equations are NOT meant to be balanced or the full complex structures always shown, oxidation states are also shown in the compounds/complexes as Roman numeral superscripts)

• H2C2O4(aq) + MnO4(aq) ==> Mn2+(aq) + CO2(g) + H2O(l)

• Initially there is a very slow uncatalysed reaction to give the initial trace of the Mn2+ based catalyst, presumably by a different, but equally complex sequence, such as that described for the auto–catalytic cycle below, involving, amongst others, Mn(III) complexes.

1. C2O42–(aq) + Mn2+(aq) ==> [MnIIC2O4](aq)

• 1. The formation an Mn(II)–complex with ethanedioate ion ligand which is the basis of the faster catalysed route of lower activation energy. This reacts with the manganate(VII) ion, followed by the reduction of Mn(III) and Mn(IV) species by the ethanedioate ion or a complex of it.

2. [MnIIC2O4](aq) + MnVIIO4(aq) ==> 2MnIVO2(colloidal) + CO2(g)

• ==> MnIII/IV complexes like [MnIII(C2O4)3]3–/[MnIVO2.xH2O]

• ==> Mn2+(aq) + CO2(g) + H2O(l) + [MnIII(C2O4)2](aq)

• 2. The formation of colloidal manganese(IV) oxide (as a hydrated complex) as well as 'soluble' Mn(III) complexes is also involved in producing the final reaction products. However, the bulk formation of MnO2 is suppressed by acidification of the solution, if you don't swirl fast enough in the titration, you get a black–brown stain of MnO2 (colloidal or precipitate).

• All these reactions involve changes in ligand and oxidation state of Mn so characteristic of transition metal chemistry. A change in ligand, or other chemical state, can change the relative half–cell potential (or stability) for the interchange of two oxidation states of the same metal ion, in this case involving MnII/III/IV/VII species.

3. [MnIII(C2O4)2] ==> [MnIIC2O4] + 2CO2 + e

• 3. Shows, in principle, one possible reduction of Mn(III) to Mn(II) in which, by electron transfer within the complex, an ethanedioate ion is oxidised to carbon dioxide (e loss) and the MnIII–ethanedioate complex is reduced back to the MnII complex (e gain) first formed in step (1), so completing the autocatalytic cycle.

• The concentration of Mn2+ or other MnIII or MnIV species increases as the reaction proceeds, so this increase in 'catalytic species' accounts for the observed increase in rate of the manganate(VII) 'decolourisation' as the titration proceeds, at least until near the end, when the reactant/catalytic intermediate concentrations are then becoming quite low.

• Case study 2c.3: Cobalt(II) ions catalyse the oxidation of the 2,3–dihydroxybutandioate ion (acid/salt, old name 'tartaric/tartrate') to water, methanoate ion and carbon dioxide with hydrogen peroxide solution. The likely scheme of events is outlined below, the equations are NOT meant to be balanced.

• Starting with the pink hexa–aqa Co2+ ion, which is a Co(II) complex

• and the carboxylate ion, OOCCH(OH)CH(OH)COO (bidentate 2– anionic ligand)

• [Co(H2O)6]2+(aq) ==> [Co(OOCCH(OH)CH(OH)COO)3]4–(aq)

• the pink Co(II) complex changes ligand from water to the organic acid, but no change in oxidation state or co–ordination number, and I don't know its colour?, but it perhaps it doesn't exist long enough to be seen?

• [Co(OOCCH(OH)CH(OH)COO)3]4–(aq) ==via H2O2==>  [Co(OOCCH(OH)CH(OH)COO)3]3–(aq)

• the Co(II)–acid complex is oxidised by the hydrogen peroxide to a Co(III) –acid complex which is green,

• [Co(OOCCH(OH)CH(OH)COO)3]3–(aq) ==> [Co(H2O)6]2+(aq),H2O(l),HCOO(aq),CO2 (aq/g)

• the green Co(III) complex then breaks down to give the products,

• and you see the bubbles of carbon dioxide and the 'return' of the pink hexa–aqa Co2+ complex ion.

• In the above sequence, the change in ligand affects the relative stability of the oxidation states. The CoII–acid complex is stable as regards 'breakdown', but is readily oxidised to the CoIII–acid complex, which is NOT stable to breakdown.

• Transition metal ions are often at the 'heart' of many biological catalysts.

• Case study 2c.4: Protein built enzymes form complexes with the metal ions, the protein acting as a multi–dentate ligand,.

• e.g. transition metal ions like Fe2+/Fe3+, Cu2+, Co2+

• and non–transition metal ions like Zn2+, Mg2+, Ca2+

• e.g. the enzymes peroxidase/catalase rapidly decompose hydrogen peroxide.

• The reaction involves a protein–Fen+ complex undergoing ligand and oxidation state changes to break the hydrogen peroxide down into water and oxygen,

• H2O2(aq) ==> 2H2O(l) + O2(g)

• and the enzymes are at least a thousand times more effective than, the 'magic' black powder catalyst, manganese(IV) oxide, MnO2(s), used in the laboratory preparation of oxygen. We have a long way to go to beat enzymes using completely 'synthetic' catalysts.

• Biotechnology is developing lots of organic synthesis reactions which all go via enzyme catalysts.

• Note: If the enzyme is immobilised on some support, and the solution of substrate molecules moves over the surface, it is strictly speaking a case of heterogeneous catalysis?

• Other examples of homogeneous catalysis NOT involving transition metal ions.

• Case study 2c.4: Esterification: Acids, proton donors (H+), catalyse the formation of an ester from a carboxylic acid and alcohol, they also catalyse the reverse reaction of hydrolysis.

• RCOOH + R'OH RCOOR' + H2O

• Case study 2c.5: Friedel–Craft reactions: The alkylation and acylation of aromatic hydrocarbons is catalysed by aluminium chloride.

• Case study 2c.6: The 'catalytic cycle' of ozone destruction by chlorine atoms from CFC's is described on another page.

• It is an example of a gaseous phase homogenous catalysis.

3. Obtaining data, rate expressions and orders of reaction

3a. Examples of obtaining rate data

• A BRIEF REVIEW OF METHODS OF COLLECTING RATE DATA

• The speed or rate of reaction is the rate of removal of reactant or the rate of formation of product.

• Experimental results can be obtained in a variety of ways depending on the nature of the reaction e.g.

• Collecting a gaseous product in a gas syringe or inverted burette.

• The initial rate is based on the tangent from the 0,0 origin, over the first few minutes?, which is usually reasonably linear at the start. See graphs on GCSE rates page.

• e.g. the decomposition of hydrogen peroxide by MnO2(s) , other transition metal compounds (insoluble–heterogeneous or soluble–homogeneous) and of course enzymes.

• 2H2O2(aq) ==> 2H2O(l) + O2(g)

• or a metal reacting with acids, and you can study the effects of a catalyst e.g. adding Cu2+(aq) ions to a zinc–acid mixture, though I'm not sure easy it is to get good quantitative results for advanced level coursework?

• i.e. 2H+(aq) + M(s) ==> M2+(aq) + H2(g)

• Titrating an acid or alkali where one is produced or used up e.g.

• The hydrolysis of a tertiary chloroalkane produces hydrochloric acid which can be titrated with standardised alkali NaOH(aq), or the chloride ion produced can be titrated with silver nitrate solution, AgNO3(aq).

• R3C–Cl + 2H2O ==> R3C–OH + Cl + H3O+

• The hydrolysis of primary halogenoalkanes with sodium hydroxide can be followed by titrating the remaining alkali with acid. You may need to use aqueous ethanol as a solvent since the halogenoalkane is insoluble in water and a large volume of reactants, so that sample aliquot's can be pipetted at regular time intervals.

• e.g. RCH2Br + NaOH ==> RCH2OH + NaBr

• or RCH2Br + OH ==> RCH2OH + Br

• Hydrolysis of an ester with (i) sodium hydroxide or (ii) dil. hydrochloric acid can be followed by titrating (i) the remaining alkali with standardised acid HCl(aq) or (ii) using standard alkali to measure the total acid present – a blank is done to check on the amount of acid catalyst present and is subtracted from the total titration.

• (i) RCOOR' + NaOH ==> RCOONa+ + R'OH

• (ii) RCOOR' + H2O ==> RCOOH + R'OH

• The iodination of a ketone such as colourless propanone, can be followed by making colorimetric measurements of the iodine colour intensity diminishing as the reaction proceeds.

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

• With a suitably chosen filter, and 'lowish' iodine concentrations, the absorbance is proportional to concentration, though a calibration graph should always be produced to check the relationship.

• The oxidation of iodide to iodine by potassium peroxodisulphate can be followed by a method known as the 'iodine clock'.

• It assumes the initial rate is reasonably constant in the first few minutes and only a few % of the reactants are used up during the 'reaction time'.

• A small and constant amount of sodium thiosulphate and starch solution is added to the reaction mixture.

• The first amount of iodine formed from the reaction by ...

• S2O82–(aq) + 2I(aq) ==> 2SO42– (aq) + I2(aq)

• is removed by reaction with the sodium thiosulphate ...

• 2S2O32–(aq) + I2(aq) ==> S4O62–(aq) + 2I(aq)

• so the starch does NOT turn blue, but immediately the next bit of iodine is formed the starch will turn blue.

• Therefore it is possible to get a reaction time for producing the same amount of iodine each time. The concentration of iodide, peroxodisulphate or an added catalyst (e.g. transition metal ion) can be varied.

3b. Rate expressions and orders of reaction

• WHAT CAN WE DO WE WITH THE DATA FROM CHEMICAL KINETICS EXPERIMENTS?

• CAN WE DERIVE A MATHEMATICAL EXPRESSION TO DESCRIBE THE KINETICS OF A GIVEN REACTION?

• From experimental results you need to know how the speed of a reaction varies with respect to individual reactant concentrations. Only then is it possible to derive a rate expression, which summarises what controls the speed of a particular reaction in terms of the relevant concentrations, which is not necessarily all the reactants!

• The rate at which a reaction depends on the concentrations of the reactant(s)/catalyst can be expressed in the form of a rate expression:

• for the reaction: fA + gB + hC etc. ==> products

• rate = k [A]a[B]b[C]c etc.

• depending on the number of reactants/catalyst involved and the orders of reaction a/b/c etc. may or may not, coincide with the molar ratios f/g/h etc. in the balanced equation and 1's are not shown by mathematical convention in either the equation or rate expression.

• k = the rate constant, which is constant at a constant temperature.

• [..] represents the concentrations of reactants A, B or C etc.

• a, b, c represent the individual order of reaction with respect to each reactant/catalyst etc.

• An individual order of reaction is the power to which the concentration term is raised in the rate expression. Powers of 1 are not shown by mathematical convention.

• The overall order of the reaction is a + b + c etc.

• Orders of reaction can only be determined by rate experiments, never from a balanced equation.

• The orders of a reaction may or may not be the same as the balancing numbers of the balanced equations.

• The orders of reaction are a consequence of the mechanism of the reaction and can only be found from rate experiments and they cannot be predicted from the balanced equation.

• For example, many reactions occur via a single bimolecular collision of only two reactants and no catalyst e.g.

• A + B ==> products, giving, by experiment, the rate expression,

• rate = k[A][B], 1st order for both reactants, 2nd order overall.

• Here, the coincidence is not surprising, the chance of a 'fruitful' collision is directly dependent on both reactants initially colliding, its often the slowest step even in a multi–step mechanism and if there are no other kinetic complications, the orders of reaction do match the numbers of the balanced equation

• e.g. the alkaline hydrolysis of esters

• RCOOR' + OH ==> RCOO + R'OH (reactant ratio 1:1)

• rate = k2[RCOOR'][OH(order ratio 1:1)

• For many other reactions, particularly those involving multi–step mechanisms and intermediates, things are not so simple, and the orders for each reactant do not necessarily match completely the integer ratio of the balanced equation.

• e.g. the acidified oxidation of bromide by bromate(V) ions

• BrO3(aq) + 5Br(aq) + 6H+(aq) ==> 3Br2(aq) + 3H2O(l) (reactant ratio 1:5:6)

• rate = k[BrO3(aq)][Br(aq)][H+(aq)]2, 4th order overall! (order ratio 1:1:2)

• and sometimes, despite a multi–step mechanism, the rate expression can be quite simple,

• e.g. the oxidation of iodide by peroxodisulphate goes in at least two steps with a reactant ratio of 1:2,

• S2O82–(aq) + 2I(aq) ==> 2SO42–(aq) + I2(aq), but the rate expression

• rate = k[S2O82–(aq)][I(aq)], is overall 2nd order (1+1) for the uncatalysed reaction.

• We can examine theoretically the effect of changing concentration on the rate of reaction by using a simplified rate expression of the form for a single reactant ..

• rate r = k [A]a

• the constant k, will include the rate constant k and any other constant reactant/catalyst concentration terms,

• [A] is the variable concentration of a reactant A,

• a = the order with respect to reactant A,

• and the 'theoretical results' are shown in the table below for an initial concentration x.

• The numbers in bold show the factor change in concentration and its effect on the rate.

 Example relative concentration of the reactant [A] relative rate r for orders 0, 1 and 2 (the rate factor change) a = 0 (zero order) a = 1 (1st order) a = 2 (2nd order) Ex. 1 initially x r = k (x)0 = k r = k (x)1 = kx r = k (x)2 = kx2 Ex. 2 2x r = k (2x)0 = k r = k (2x)1 = 2kx r = k (2x)2 = 4kx2 Ex. 3 3x r = k (3x)0 = k r = k (3x)1 = 3kx r = k (3x)2 = 9kx2 Ex. 4 x/2 r = k(x/2)0 = k r = k (x/2)1 = kx/2 r = k (x/2)2 = kx2/4 Ex. 5 x/3 r = k (x/3)0 = k r = k (x/3)1 = kx/3 r = k (x/3)2 = kx2/9
• This shows that for ...

• zero order, the rate is completely independent of reactant concentration,

• first order, the rate doubles if a concentration doubled or the rate is reduced to 1/3rd if the concentration is reduced by a factor of 3, etc., i.e. the rate is proportional to concentration.

• and for second order, the rate quadruples if the concentration is doubled, the rate is cut to 1/4 if the concentration is halved, or the rate increases 9x if a concentration is tripled etc. i.e. the rate is proportional to concentration squared.

• We can now examine theoretically, the effect of changing individual concentrations on the rate of reaction of a more complicated rate expression of the form ..

• rate r = k[A]a[B]b

• k = rate constant k

• [A] and [B] are the concentration of a reactants A and B,

• a and b = are the individual orders with respect to reactant A and B,

• the overall order = a + b,

• and exemplar 'theoretical results' are shown in the table below, with the concentration and rate factor changes shown in bold.

• The initial concentration of [A] = x, and that of [B] denoted by y and the resulting changes in rate (factor in bold) are 'computed' for selected changes in concentrations x and y for [A] and [B]

 Example of arate expression Examples of rate r for relative concentrations in terms of x initially for [A] and y initially for [B]  (the rate factor change) concns of A and B initial x, y concns of A and Bx, 2y concns of A and B2x, y concns of A and B2x, 2y concns of A and Bx, y/2 concns of A and Bx/3, y concns of A and B2x, y/3 concns of A and Bx/2, y/2 r = k [A][B] 1st order wrt to both A and B, overall 2nd order 'initial' x, y r = kxy 'baseline' relative rate r = kx2y r = 2kxy double B, doubles rate r = k2xy r = 2kxy double A, doubles rate r = k2x2y r = 4kxy double both A and B, rate quadrupled r = kxy/2 r = kxy/2 B halved, rate halved r = k(x/3)y r = kxy/3 reduce A by 1/3rd, rate reduced by 1/3rd r = k2x(y/3) r = 2/3kxy double A,  1/3B, rate reduced by 2/3rds r = k(x/2)(y/2) r = kxy/4 halve both A and B, rate reduced by 1/4 r = k [A][B]2 1st order A and 2nd order, overall 3rd order 'initial' x, yr = kxy2 'baseline' relative rate r = kx(2y)2 r = 4kxy2 double B, quadruple rate r = k2xy r = 2kxy2 double A, double rate r = k2x(2y)2 r = 8kxy2 double both A and B, rate inc. 8x r = kx(y/2)2 r = kxy2/4 halve B, 1/4 rate r = k(x/3)y2 r = kxy2/3 reduce A by 1/3rd, rate reduced 1/3 r = k2x(y/3)2 r = 2/9kxy2 double A, 1/3 B, rate drops by 2/9 r = k(x/2)(y/2)2 r = kxy2/8 both A and B halved, rate red. by 1/8th
• The units of k, the rate constant. This frequently causes problems!

• I'm assuming the rate = change in concentration/time taken

• so units of rate = mol dm–3 s–1,

• so you will have to adapt the arguments below if using other units of time or concentration.

• So, rate = d[A]/dt, with rate units of mol dm–3 s–1,

• [A] = (concn) = concentration of A in mol dm–3.

• Just using the simple expression rate = k[A]a, for an overall order of a ...

• (but the same arguments apply for rate = k[A]a[B]b[C]c etc.)

• n=0, zero order, rate = k(concn)0, rate = k,

• so units of  k = mol dm–3 s–1 for overall zero order.

• n=1, 1st order, rate = k(concn), k = rate/(concn), so ...

• units of k = mol dm–3 s–1 / mol dm–3 = s–1 for overall 1st order.

• n=2, 2nd order, rate = k(concn)2, k = rate/(concn)2, so ...

• units of k = mol dm–3 s–1 / (mol dm–3)2 = mol–1 dm3 s–1 for overall 2nd order.

• n=3, 3rd order, rate = k(concn)3, k = rate/(concn)3, so ...

• units of k = mol dm–3 s–1 / (mol dm–3)3 = mol–2 dm6 s–1 for overall 3rd order.

3c. Deducing orders of reaction

• HOW DO WE DEDUCE ORDERS OF REACTION?

• In rate investigations, each reactant/catalyst concentration should be individually investigated to determine its individual order and ultimately to derive the full rate expression for the reaction. Any other reactant/catalyst concentration must be kept constant.

• The graph below show typical changes in concentration (or amount of moles remaining) of a reactant with time, for zero, 1st and 2nd order. The 2nd order graph tends to be a bit steeper than 1st order BUT that proves nothing! See the second graph/plot diagram. In the zero order graph the gradient is constant as the rate is independent of concentration. For the 1st/2nd order graphs, the initial gradient can be taken as the initial rate as the gradient gradually decreases as the rate is concentration dependent and the rate decreases as the concentration decreases.

• The second graph diagram shows how the 'initial rate'* data can be plotted against concentration (c) or concentration2 (c2) etc. to obtain an unambiguous linear plot from which the order may be deduced.

• * The rate data must be the speed of the reaction in its early stages, otherwise errors will arise due to a significant changes in the amounts or concentrations of the reactants as they are used up.

• Some possible graphical results are shown above.

• [1] shows the results for zero order, where the rate is independent of concentration.

• [2] shows the results for 1st order, where the rate is directly/linearly proportional to concentration.

• [3] shows the results for 2nd order, where the rate is proportional to concentration squared.

• [4] means it cannot be 1st order, order could be <1?

• [5] means it cannot be 2nd order, order could be <2 e.g. 1st? check via plot [2].

• [6] means it cannot be 1st order, order could be  >1 e.g. 2nd? check via plot [3].

• [7] means it cannot be 2nd order, suggests order >2.

• Of course [4] to [7] could simply represent inaccurate data!

• There is another graphical way of showing the order with respect to a reactant is 1st order, but it requires accurate data showing how the concentration or moles remaining of a reactant changes with time within a single experiment (apart from repeats to confirm the pattern).

• The rate of radioactive decay is an example of 1st order kinetics. The decay curve should show that the time taken for the remaining radioisotope or count rate to halve is always the same time interval – the half–life!

• A plot of concentration/moles of reactant remaining versus time, should show the same pattern as for radioactive decay. The graph below shows what happens to a reactant with a half–life of 5 minutes.

• In other words 100% ==> 50% ==> 25% etc. at 5 minute time intervals, but the y axis could be concentration or moles left etc.

• It is the constancy of the half–life which proves the 1st order kinetics.

• The mathematics of 1st order rate equations (units).

• rate = –d[A]/dt = k1[A]

• on integration we get: k1t = ln[Ao] – ln[A]

• rate (mol dm–3 s–1)

• k1 = 1st order rate constant (s–1)

• [Ao] = initial concentration of A at t=0 (mol dm–3)

• [A] = concentration of A at time t (mol dm–3)

• A graph of ln[A] versus t will have a gradient of –k1

• and an intercept of ln[Ao].

• The half–life = t1/2 = 0.693/k1 (s) and is independent of the initial concentration.

• Note: You can get k from the half–life graph above since ..

• k1 = 0.693/t1/2

• The above equations can be expressed in amounts of A e.g. in number of moles of A.

• If a = initial moles of A, x = moles of A reacted,

• so (a–x) = number of moles A remaining at time t

• rate = –d(a–x)/dt = k(a–x)

• on integration we get: kt = ln(a) – ln(a–x)

• A plot of ln(a–x) versus t has a gradient of –k,

• and t1/2 = 0.693/k (s)

• It is also possible from the type of graph of concentration versus time, to measure the initial gradient, which gives the initial rate of reaction (applies to any order of reaction analysis).

From the point of view of coursework projects the detailed analysis described above is required, but quite often in examination questions a very limited amount of data is given and some clear logical thinking is required. So simplified rate data questions and their solution is given below.

3d. Simple exemplar rates questions

A FEW EXAMPLES OF PROBLEM SOLVING IN CHEMICAL KINETICS

From the point of view of coursework projects the detailed analysis described above is required, but quite often in examination questions a very limited amount of data is given and some clear logical thinking is required. So simplified rate data questions and their solutions avoiding graphical analysis are given below.

Example Q1 The following rate data was obtained at 25oC for the reaction: A + 2B ==> C

 Expt. no [A]/mol dm–3 [B]/mol dm–3 rate of formation of C mol dm–3s–1 1 0.10 0.05 0.02 x 102 2 0.10 0.10 0.04 x 102 3 0.05 0.10 0.01 x 102 4 0.10 0.20 0.08 x 102

(a) Deduce the order of reaction with respect to reactant A.

Compare expts.  2 and 3. [B] is kept constant, but on halving A the rate is reduced by a factor of 1/4 (1/2 x 1/2) so the rate is proportional to [A]2. Therefore the order with respect to A is 2 or 2nd order. You can think the other way round i.e. from expt. 3 to 2, [A] is doubled and the rate quadruples.

(b) Deduce the order of reaction with respect to reactant B.

Compare expts. 1 and 2. [A] is kept constant but doubling [B] doubles the rate, so the reaction is directly proportional to [B]. Therefore the order with respect to B is 1 or 1st order. Similarly, comparing expts. 1 and 3 (2x [B] gives 2x rate) or comparing expts. 1 and 4 (4x [B] gives 4x rate).

(c) What is the overall order of the reaction between A and B?

Total order = 2 + 1 = 3, 3rd order overall.

(d) Write out the full rate expression.

rate = k [A]2 [B]

(e) Calculate the value and units of the rate constant.

rearranging rate = k [A]2 [B] gives k = rate / [A]2 [B]

and the units will be mol dm–3s–1 / (mol dm–3)2(mol dm–3) = s–1 / (mol dm–3)2 = mol–2 dm6 s–1

so, using the data from expt. 1 (or any set, assuming data perfect) gives

k = 0.02 x 102 / (0.12 x 0.05) = 4 x 103 mol–2 dm6 s–1

or expt. 3 gives 0.01 x 102 / (0.052 x 0.1) = 4 x 103 etc. check the same for expts. 2 and 4. (hope they work out ok)

Its not a bad idea to repeat the calculation with another set of data as a double check!

(f) What will be the rate of reaction if the concentration of A is 0.20 mol dm–3 and the concentration of B is 0.30 mol dm–3?

You just substitute the values into the full rate expression:

rate = k [A]2 [B] = 4 x 103 x 0.22 x 0.3 = 0.48 x 102 mol dm–3 s–1

Note: The reacting mole ratio is 2 : 1 BUT that does not mean that the orders are a similar ratio (since here, it happens to be the other way round for the individual orders). Orders of reaction can only be obtained by direct experiment and their 'complication' are due to complications of the actual mechanism, which can be far from simple.

Example Q2

REMINDER before studying the two advanced kinetic pages ...

YOU SHOULD KNOW ALL THE RATES IDEAS IN THE GCSE/IGCSE NOTES

Advanced Level Theoretical Physical Chemistry of chromium – A level Revision notes to help revise for GCE Advanced Subsidiary Level AS Advanced Level A2 IB Revise AQA GCE Advanced Level Chemistry OCR GCE Advanced Level Chemistry Edexcel GCE Advanced Level Chemistry Salters AS A2 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: 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 Questions

Website content copyright © Dr W P Brown 2000–2012 All rights reserved on revision notes, puzzles, quizzes, worksheets, x–words etc. * Copying of website material is not permitted * I do not personally endorse the adverts – but they do pay for the site!

Alphabetical Index for Science Pages Content A B C D E F G H I J K L M N O P Q R S T U V W X Y Z

SITE HELP SEARCH – ENTER SPECIFIC WORDS/FORMULA etc.