KINETICS III.docx

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RATE EQUATIONS AND REACTION MECHANISMS

1. Reaction mechanism: Aseries of steps which are proposed to explain how the reactants change to theproducts.

2. A proposed mechanism mustagree with:(i) The balanced equation i.e.

when the steps are added together they should give the overallequation/reaction in terms of reactants and products.

(ii) The mechanism must agreewith the reaction kinetics or the rate expression.

3. In most mechanisms there isone slow or rate determining step and one or more fast steps which do notaffect the rate. The slow rate-determining step can be derived form therate expression . The partial order of any species appearing in the mechanismafter the rate determining step will be zero order confirming it has no effecton the rate.

Mechanism for the hydrolysis of halogenoalkanes There are two possible mechanisms for the hydrolysis of halogenoalkanes. The chemical equation does not give any information about the mechanism of the

reaction. The mechanism can be deduced by finding the order of reaction with respect to each

reactant. Only substances involved in (or before) the rate-determining step appear in the rate

equation.



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The order with respect to a reactant indicates the number of that particles of thatreactant involved in the rate-determining step.

Examples: S N1 and S N2 mechanisms of the hydrolysis of halogenoalkanes

(a) SN1 mechanism (Nucleophilicsubstitution Unimolecular)

The term unimolecular indicates that there is only one species involved in the slowrate determining step.

Consider the reaction between 2-bromo-2-methylpropane, (CH 3)3CBr, a tertiaryhalogenoalkane and OH - from aqueous sodium hydroxide solution:

(CH3)3CBr + OH- (CH3)3COH + Br-

Rate = k [ halogenoalkane]

The reaction is FIRST order with respect to (CH 3)3CBr and zero order with respectto the OH - . Only (CH3)3CBr is taking part in the slow rate determining step of thereaction. The OH - are involved in a relatively fast step following the rate determiningstep and do not affect the rate.

Energy profile diagram for a two-step reaction



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The activation energy fort he production of theintermediate is greater than the activationenergy fort he intermediate going to theproducts. This means that the first step is

rate-determining step.

The second activation energy isgreater, so the second step is therate determining step.

(Unstable) Intermediate product : formed in the slow rate determining step whichthen reacts further to form the products.

Reactions which involve a single intermediate are ususally referred to as two-stepreactions.

Note that each step has its own transition state and activation energy.

(b) SN2 mechanism: ( Nucleophilic substitution Bimolecular)

The term bimolecular refers to the molecularity of the reaction i.e. the number ofspecies involved in the slow rate-determining step.

Consider the reaction between bromoethane, CH 3CH2Br, a primary halogenoalkane andhydroxide ions, OH - from aqueous sodium hydroxide solution:

CH3CH2Br + OH- CH3CH2OH + Br-

Rate = k [CH3CH2Br] [OH -]

The reaction is FIRST order with respect to both CH3CH2Br and + OH-. Both ofthese species must be taking part in the slow rate-determining step of the reaction.The reaction must happen by a straightforward collision between them.

Energy profile diagram for a one step reaction



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Transition step : This often represents a condition where old bonds are half brokenand new ones are half formed. This transition state breaks down directly to the finalproducts without forming a ‘more stable’ intermediate, which could then be involved ina futher step.

There is one transition state for each step. The transition state is formed whenreacting particles have required activation energy E A. The activation energy is thedifference in energy between the reactants and the transition state.

(c) While the mechanismsproposed for the alkaline hydrolysis of primary and tertiary halogenoalkanes areconsistent with the experimentally determined orders of reaction and theoverall equation for the reaction, the reaction between propanone and iodine isnot clear.

Propanone will react with iodine in the presence of acid as follows:

CH3COCH3(aq) + I 2(aq) CH3COCH2I (aq) + HI(aq)

Experiments are performed to deduce the order with respect to each reactant.The relative concentrations are shown below.

ExperimentNo.

[CH3COCH3] [I 2] [H+] Relative rate

H+



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1 1 1 1 22 1 2 1 23 2 1 1 44 1 1 2 4

Deduce the order of reaction with respect to each of the substances given inthe table and give the rate equation for the reaction.

A suggested mechanism is as follows:

Propanone will react with iodine in the presence of alkali as follows:



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CH3COCH3(aq) + I 2(aq) CH3COCH2I (aq) + HI(aq)

Experiments performed to deduce the orders with respect to each reactantreveal that the reaction first order with respect to propanone, zero order withrespect to iodine and first order with respect to hydroxide ions, which do noteven appear in the equation.

The rate equation for this reaction therefore is:

Rate = k [CH3COCH3] [OH-

]

A likely explanation for the mechanism of this reaction is that iodine entersafter the rate-determining step.

A suggested mechanism is as follows:

Step 1: the base removes a proton from a —CH3 group, which is the ratedetermining step.

CH3COCH3 + OH- CH3COCH2- + H2O

Step 2 : the lone pair of electrons on the C - of the carbanion forms a bond withan iodine atom in I 2 and I 2 bond breaks.

CH3COCH2- + I2 CH3COCH2I + I -

Since the reaction is first order with respect to both propanone and hydroxide ionsand zero order with respect to iodine, the first step must be the rate determiningstep. This fact can be confirmed by carrying out the same reaction but using bromineor chlorine instead of iodine, where it is found that the overall rate of reaction is thesame.

Second or subsequent step is rate determining

OH-



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If the second or a subsequent step is rate determining, the derivation of the overallrate equation is more complex.

Consider the reaction:

A + 2b + C → D + E

Step 1 is a rapid step that is reversible:

A + B Int

where Int is an intermediate

Step 2 is the slowest step and hence is rate determining:

Int +B → X

Step 3 is faster than step 2:

C + X→ D + E

The overall rate is determined by the rate of the slowest step 2:

Overall rate = rate of step 2 = k[B] [Int]

Examples:

Finding the slowest step in the reaction mechanism

The rate expression for the reaction:

2A (g) + 3B (g) → C (g) + 2D (g)

is found to be Rate = k [A] 2 [B]



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(a) Does this reaction have a mechanism?

Yes, because if there were no reaction mechanism, the rate expression of the given reaction

would be:Rate = k [A] 2 [B]3

But, the given rate expression is different from the rate expression we find. In that casethe reaction has a mechanism.

b) If it has a mechanism, write the slowest step.

Rate = k [A] 2 [B]

So, 2A + B → products.

Finding the rate expression of a reactionThe mechanism proposed for the reaction:

2 ICl + H2 → 2 HCl + I2 is as follows:

(I) H 2 + ICl → HI + HCl (slow)

(II) HI + ICl → HCl + I2 (fast)

a) Which step is the rate-determining step?Step I

b) What is the rate expression for this reaction?

Rate = k [H 2 ] [ICl]

The rate expression can be written form the rate-determining step directly. The rate of the

reaction is directly proportional to the concentration of the reactants of the rate- determining step.



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c) If the concentrations of H 2 and ICl are doubled, how many times will the rate of thereaction change?Increase by factor 4

(Because the rate of reaction is directly proportional to the both of theconcentrations of H 2 and ICl.

d) If the concentrations of HI and HCl are tripled, how many times will the rate change?HI and HCl do not appear on the rate equation, so they do not cause any change.

Finding the rate-determining step

The rate equation for the reaction:

2NO (g) + Cl2 (g) → 2 NOCl (g)

is found to be Rate = k [NO] 2 [Cl2]

which of the following mechanisms will likely belong to this reaction?

A. 2NO (g) → N2O2 (slow)N2O2 (g) + Cl2 (g) → 2 NOCl (g) (fast)

B. Cl2 (g) → 2Cl (slow)

NO (g) + Cl (g) → NOCl (g) (fast)

C. NO (g) + NO (g) + Cl2 (g) → 2 NOCl (g)

Answer: C



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Finding the rate expression

The reaction,

2A + B + K → L + F + 120 kJ

has a mechanism with three steps.

The first two steps of the mechanism are as follows:

2A + B → C + D (fast)

C + E → K + L (fast)

a) Deduce the rate expression for the reaction.The sum of the reactions in the mechanism gives the net (overall) reaction. Therefore,the slowest step which is not given can be found by subtracting the given steps fromthe net equation.

2A + B + K → L + F (overall reaction)

-2A + B → -C + D (fast)

-C + E → -K + L (fast)

____________________________________________

K - E → - K - D + F or

2K + D → F + E which is the slowest step.

Therefore,

Rate = k [K] 2 [D]

b) How many times does the rate increase when the concentration of K and D aretripled?The rate of reaction is directly proportional to [K] 2 [D]

If only [K] is tripled, the rate increases 3 2 = 9 times

If only [D] is tripled, the rate increases 3 times

If both are tripled, the rate = (3) 2

x (3) = 27 times.



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THE ARRHENIUS EQUATION

The Arrhenius equation gives the relationship between the rate constant, k; thetemperature, T (in Kelvin) and the activation energy E A which is the minimumenergy that the reacting particles should have to collide to start the reaction.

The Arrhenius equation:

k = A e –EA

/RT

where: k = The rate constant (determined at various temperatures)

T = Temperature of reaction in kelvin.

EA= Activation energy

R = Gas constant



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A = The Arrhenius constant which depends on

Using the Arrhenius equation

Taking natural logs (ln) on both sides, we obtain a more important form of theexpression:

(a) The effect of a change of temperature

The fraction of the molecules able to collide successfully almost doubles byincreasing the temperature by 10 0C. This causes the rate of reaction to almostdouble. This is the value in the rule-of-thumb often used in simple rate of reactionwork.

(b) The effect of a catalyst

A catalyst will provide a route for the reaction with a lower activation energy. Asa result, there is a massive increase in the fraction of the molecules which areable to collide successfully, so the rate of reaction increases.

The rate constant is determined at a number of temperatures

The rate constant (ln k) is plotted against 1/T and from the graph a value for theactivation energy E A can be found.

ln k

ln A

1/T/K -1

lnk = lnA – E A/RT



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Note: Since gradient is negative, E A is always positive . Units of E A are kJ mol -1

Study question

The second-order rate constant for the reaction of 1-bromopropane withaqueous hydroxide ions was measured as a function of temperature.

CH3CH2CH2Br (aq) + OH- (aq) CH3CH2CH2OH (aq) + Br- (aq)

The results are shown in the table below:

Tempertaure /K 1/T/K -1 k/mol dm -3 s -1 lnk

298 1.4 X 10-4

308 3.0 X 10 -4

318 6.8 X 10 -4

328 1.4 X 10 -3

Plot a graph of lnk against 1/T. Measure the gradient of the line and hencecalculate the activation energy of the reaction.

2. The activation energy of the oxidation in acid solution of iodide ions by iodate(V)ions can be found by experiment:

Place 25 cm 3 of a solution of potassium iodate(V) in a beaker and add 5 dropsof starch solution.

Place 25 cm 3 of a solution of potassium iodide and 5 cm 3 of a solution of sodiumthiosulfate in another beaker.

Mix the two solutions, start a stop clock immediately and stir with athermometer. Read and record the temperature of the mixture.

Record the time when the solution turns intense blue.

Repeat the experiment with the same volumes of the solutions but at a highertemperature.

Repeat the experiment at two more different temperatures.



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(a) Write the ionic equation for the formation of the iodine.

(b) Write the ionic equation for the reaction of iodine with thiosulfate ions.

When all the sodium thiosulfate has been used up, the next iodine that isproduced forms an intense blue colour with starch.

The results are shown in the table below.

Time/s 1/Time/s -1 ln(1/Time) Temperature/K 1/Temp./K -1

135 29846 31314 333

Complete the table and then plot a graph of ln(1/Time) on the y-axis against 1/T onthe x-axis. Measure the gradient of the line and evaluate the activation energy. Youmay measure that the gradient of this line is the same as that of lnk against 1/T.

Activation energy and types of catalysts

Catalysts can be divided into two types:

- Homogeneous catalysts- Heterogeneous catalysts

Homogeneous catalysts

A homogeneous catalyst is in the same phase as the reactants, usually the gas phaseor in solution.

The rate of reaction depends on the concentrations of the catalyst and one of thereactants. An example is the oxidation of iodide ions by persulfate ions.

2I - (aq) + S2O82- (aq) → I 2 (aq) + 2SO 4

2- (aq)

This is a slow reaction because it requires two negative ions to collide. The reaction iscatalysed by iron (ııı) ions. The catalysed, route is oxidation of iodide ions by iron(ııı)



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ions, followed by reduction of persulfate ions by iron (ıı) ions and the regeneration ofthe catalyst of iron(ııı) ions:

2Fe 3+(aq) + 2I - (aq) → I 2(aq) + 2Fe2+ (aq)

2Fe 2+ (aq) + S2O82- (aq) → 2Fe 3+(aq) + 2SO 42- (aq)

Heterogeneous catalysts

A heterogeneous catalyst is in a different phase from the reactants. Many industrialprocesses use this type of catalyst; they have the advantage that the catalyst can be

separated from the equilibrium mixture by simple physical means. For example, in theHaber process the solid iron catalyst catalyses the reaction between two gases,hydrogen and nitrogen.

Heterogeneous catalysts are frequently transition metals- for example iron, platinumand nickel.

The catalyst has active sites on its surface that rapidly become saturated byreactants, which are then slowly converted into products. These leave the metalsurface, thus allowing more of the reactant to be adsorbed. This means that the rate

of a reaction is not altered by an increase in pressure of the gaseous reactants.

Exercise:

The second-order rate constant for the reaction of 1-bromopropane with aqueous hydroxideions was measured as a function of temperature.



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CH3CH2CH2Br (aq) + OH- (aq) → CH3CH2CH2OH (aq) + Br- (aq)

The results are shown in the table below:

Temperature/ 0C Temp/K 1/T /K -1 k (mol-1dm3s-1) ln k

25 298 0.00336 1.4 x 10 -4 - 8.9

35 308 0.00325 3.0 x 10 -4 - 8.1

45 318 0.00314 6.8x 10 -4 -7.3

55 328 0.00306 1.4 x 10 -3 -6.6

Measure the slope (gradient) of the line and hence calculate the E a.

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KINETICS III.docx