Doc Brown's Advanced A Level Chemistry

Advanced A Level Chemistry

 Kinetics Part 6

5.3 Rate data, orders of reaction and rate expressions (rate equation) for more advanced kinetics analysis

Experimental techniques i.e. examples of how to obtain rate data, explaining and deducing orders of reactants/reactions, calculating rate constants and deducing their units, with some exemplar rates questions involving deducing and using orders of reactants and rate expressions including graphical analysis.

Advanced A Level Kinetics Index

5.3 Obtaining rate data, interpreting rate data, orders of reaction and rate expressions

5.3a. 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.

      • 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 peroxodisulfate 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 thiosulfate 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 thiosulfate ...

        • 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, peroxodisulfate or an added catalyst (e.g. transition metal ion) can be varied.

    • (c) doc b How to measure–express initial rates is introduced in the GCSE Notes on Rates of Chemical Reaction


5.3b. 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 (rate equation):

    • 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. (usually mol dm–3)

      • a, b, c represent the individual orders 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, however coincident it may seem, beware!

    • 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 peroxodisulfate 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 a

rate 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 B

x, 2y

concns of A and B

2x, y

concns of A and B

2x, 2y

concns of A and B

x, y/2

concns of A and B

x/3, y

concns of A and B

2x, y/3

concns of A and B

x/2, y/2

r = k [A][B]

1st order with respect 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 B, overall 3rd order

'initial' x, y

r = 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]n, for an overall order of n ...

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

    • n=0, zero order, rate = k(conc'n)0, rate = k,

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

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

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

    • n=2, 2nd order, rate = k(conc'n)2, k = rate/(conc'n)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(conc'n)3, k = rate/(conc'n)3, so ...

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


5.3c. 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.

    • In the zero order graph the gradient is constant as the rate is independent of concentration, so the graph is of a linear descent in concentration of reactant.

    • 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 2nd order graph tends to 'decay' more steeply than 1st order BUT that proves nothing!

    • The idea is that somehow you test for the order with an appropriate linear graph ... read on ...

doc b Connect the graphs with the following:

  • n=0, zero order, rate = k(concn)0, rate = k, units of  k = mol dm–3 s–1 for overall zero order, easily obtained from the gradient.

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

doc b

  • 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.

      • [1] to [3] are the essential linear plots to test for, i.e. deduce, the order of reactant or reaction.

    • [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.

doc b

  • 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.


5.3d. Simple exemplar rates questions to derive rate expressions

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.

These examples do NOT involve graphs directly, but a 'graphical' section of examples has been added in section 5.3e. I've made the numbers quite simple to follow the logic of the argument. I've also shown how to calculate the rate constant.

These example calculations below are based on the initial rate of reaction analysis - so we are assuming the variation of concentration with time for each experimental run has been processed in some way e.g. by graphical analysis, to obtain an initial rate of reaction.

The graph on the left illustrates the initial rate method for the formation of product. The gradients A and B would be for two different concentrations of a reactant, the concentration for A would be greater than the concentration of B. The initial rate is taken as the positive tangent - gradient for the curve at the point 0,0. The same argument applies if you imagine the graph inverted and you were following the depletion of a reactant. Then you would get two negative gradients one steeper than the other for the greater concentration.

Reminder [x] means concentration of x, usually mol dm-3


Example 1.

The table below gives some initial data for the reaction: A  +  B  ===>  products

  initial [A]/mol dm-3 initial [B]/mol dm-3 initial rate/mol dm-3 s-1
run (i) 0.1 0.1 3.0 x 10-4
run (ii) 0.2 0.1 6.0 x 10-4
run (iii) 0.2 0.2 6.0 x 10-4

From runs (i) and (ii), keeping [B] constant, by doubling [A], the rate is doubled, so 1st order with respect to reactant A.

From runs (ii) and (iii), keeping [A] constant, by doubling [B], the rate is unchanged,  so zero order with respect to reactant B.

Therefore the reaction is 1st order overall, and the rate expression is ...

rate = k [A]   ([B] can be omitted, anything to the power zero is 1)

To calculate the rate constant, rearrange the rate expression and substitute appropriate values into it.

Therefore, using run (iii) ...

k = rate constant = rate / [A] = 6.0 x 10-4/0.2 = 3.0 x 10-3 s-1

In reality the results would be not this perfect and you would calculate k for each set of results and quote the average!


Example 2.

The table below gives some initial data for the reaction: A  +  B  ===>  products

  initial [A]/mol dm-3 initial [B]/mol dm-3 initial rate/mol dm-3 s-1
run (i) 0.1 0.1 3.0 x 10-4
run (ii) 0.2 0.1 3.0 x 10-4
run (iii) 0.2 0.2 1.2 x 10-3

From runs (i) and (ii), keeping [B] constant, by doubling [A], the rate is unchanged, so zero order with respect to reactant A.

From runs (ii) and (iii), keeping [A] constant, by doubling [B], the rate is quadrupled,  so 2nd order with respect to reactant B.

Therefore the reaction is 2nd order overall, and the rate expression is ...

rate = k [B]2   ([A] can be omitted, anything to the power zero is 1)

To calculate the rate constant, rearrange the rate expression and substitute appropriate values into it.

Therefore, using run (iii) ...

k = rate constant = rate / [B]2 = 1.2 x 10-3/0.22 = 3.0 x 10-2 mol-1 dm3 s-1

In reality the results would be not this perfect and you would calculate k for each set of results and quote the average!


Example 3.

The table below gives some initial data for the reaction: A  +  B  ===>  products

  initial [A]/mol dm-3 initial [B]/mol dm-3 initial rate/mol dm-3 s-1
run (i) 1.0 1.0 2.0 x 10-3
run (ii) 2.0 1.0 4.0 x 10-3
run (iii) 2.0 2.0 8.0 x 10-3

From runs (i) and (ii), keeping [B] constant, by doubling [A], the rate is doubled, so 1st order with respect to reactant A.

From runs (ii) and (iii). keeping [A] constant, by doubling [B], the rate is doubled,  so 1st order with respect to reactant B.

Therefore the reaction is 2nd order overall, and the rate expression is ...

rate = k [A] [B]

To calculate the rate constant, rearrange the rate expression and substitute appropriate values into it.

Therefore, using run (iii) ...

k = rate constant = rate / ([A][B] = 8 x 10-3/(2.0 x 2.0) = 2 x 10-3 mol-1 dm3 s-1

In reality the results would be not this perfect and you would calculate k for each set of results and quote the average!


Example 4.

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

Expt. no [A]/mol dm–3 [B]/mol dm–3 initial 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 for the reaction as a whole.

(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.


5.3e More on graph work and deducing orders of a reaction by data – graphical analysis

Example 1. The inversion (hydrolysis) of sucrose

sucrose + water  ===>  glucose + fructose

In simple terms: C12H22O12  +  H2O   ====>  C6H12O6  +  C6H12O6

Some rate data for the inversion of sucrose is given below.

A linear graph, the gradient of the graph of concentration versus time does not change, therefore a zero order reaction.

Therefore the rate expression is ...

rate = k (mol dm–3 s–1)

A non-linear graph of concentration versus time would suggest first or second order kinetics.

This zero order reaction occurs when the enzyme (invertase) concentration is low and the substrate (sucrose) concentration is high. The maximum number of enzyme sites are occupied, which is itself a constant at constant enzyme concentration.

For more details see section 7.3 Enzyme kinetics


Example 2. Analysing a single set of data to deduce the order of reaction

The data below are for the hydrolysis of 2–chloro–2–methylpropane in an ethanol–water mixture.

The water concentration is effectively constant.

The reaction is: 2–chloro–2–methylpropane + water  ===>  2–methyl–propan–2–ol + 'hydrochloric acid'

(CH3)3CCl  +  H2O  ====> (CH3)3COH  +  H+  +  Cl

A graph is drawn of (CH3)3CCl concentration versus time.

 

From the graph the gradient (relative rate) was measured at 6 points.

Since the gradient (rate) changes with concentration, it cannot be a zero order reaction.

gradient (CH3)3C–Cl concentration

 (conc'n)2

gradient = rate Rate in
points [RX] in mol dm–3 [RX]2 rate = d[RX]/dt mol dm–3 s–1
(0) 0.0 0.000 0 0.00000
(1) 0.1 0.010 0.112/175 0.00064
(2) 0.2 0.040 0.2/150 0.00133
(3) 0.3 0.090 0.3/177 0.00169
(4) 0.4 0.160 0.29/122 0.00238
(5) 0.5 0.250 0.2/75 0.00267

 

A graph is then drawn of rate versus concentration of (CH3)3CCl (RX)

The graph is 'reasonably linear' suggesting it is a 1st order reaction.

Therefore the rate expression is ...

 rate = k[(CH3)3CCl]

 

To put this graph in perspective, a 2nd order plot is done below of rate versus [RX]2. (see data table above)

Wherever you draw a straight line, the data does not express itself as a linear plot and cannot be a 2nd order reaction.

 

Nucleophilic substitution by water/hydroxide ion [SN1 or SN2, hydrolysis to give alcohols] for a wider ranging mechanistic and kinetic analysis of the hydrolysis of haloalkanes.


Example 3. Kinetics of the thermal decomposition of hydrogen iodide

A single set of reaction rate data at a temperature of 781K

Reaction: 2HI(g) ==> H2(g) + I2(g)

 

 for 1st order plot

 for 2nd order plot

graph gradient
time/s [HI]/mol dm-3 [HI]2/mol2 dm-6 RATE/mol dm-3s-1
0 0.02000 0.00040000  
500 0.01736 0.00030140 0.000004640
1000 0.01536 0.00023590 0.000003600
1500 0.01376 0.00018930 0.000002890
2000 0.01247 0.00015550 0.000002360
2500 0.01140 0.00013000 0.000001980
3000 0.01049 0.00011000 0.000001680
3500 0.00972 0.00009450 0.000001430
4000 0.00906 0.00008208  

The concentration of hydrogen iodide was measured every 500 seconds for 4000 seconds.

 

A plot of HI concentration versus time (above) was curved showing it could not be a zero order reaction with respect to the concentration of HI.

From the [HI] versus time graph (above) the tangent angle (gradient = rate) was accurately measured from 500 s to 3500 s at 500 second intervals (results tabulated above).

 

The rate of reaction was is then plotted against HI concentration to test for 1st order kinetics. This proved to be a curve - compare the blue rate data curve with the black 'best straight line' (courtesy of Excel!), so not 1st order either.

 

The third graph is a plot of HI decomposition rate versus [HI] squared, and, proved to linear - the blue data line was pretty coincident with a black 'best straight line'.

This proved that the decomposition of hydrogen iodide reaction is a 2nd order reaction.

Therefore the rate expression is ...

rate = k [HI]2

For more on this reaction see KINETICS section 7.4

Advanced A Level Kinetics Index


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) Experimental techniques to get rate data and how to derive rate expressions how to deduce orders of reactions with some worked out questions involving deducing and using orders of reactants and rate expressions.

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