Week 14 © Pearson Education Ltd 2009 This document may have been altered from the original Explain and use the term: rate of reaction. Deduce the rate.

Week 14

© Pearson Education Ltd 2009This document may have been altered from the original

• Explain and use the term: rate of reaction.

• Deduce the rate of a reaction from a concentration–time graph.

Rates of Reaction

• Rate = speed of reaction• = change in concentration of reactant or product• time for the change to take place• = moles.dm-3 (USUALLY)

• sec

• Examples of familiar reactions:• Marble chips + acid• Sodium thiosulphate + acid• Magnesium + acid• How would you measure the rate of these

reactions?

Measuring Rates of Reaction

• Marble chips + acid?• Loss in mass per time?• Gas given off per time? Units?• Time taken for reaction to finish?

• Thiosulphate + acid?• Time taken to obscure the cross• Use a data logger and light sensor to

measure arrival of precipitate.

Measuring Rates of Reaction

• Magnesium + acid?• Time taken to stop fizzing?• Gas given off per time?• Mass change over time?

• Which would you choose?• How would these methods give rate?• Which would give the most accurate value

for the rate of reaction?• What would the graphs look like?

Rate Graphs

• Draw a sketch graph for each of the previous suggested methods of measuring rate.

• How can rate be obtained from each?• Rate is generally obtained from the slope

of a graph – a tangent to the concentration/time graph at regular intervals.

• All graphs of this type show that the rate is fastest at the beginning of the reaction- why?

Factors Affecting Rate of Reaction

• Concentration• Temperature• Light• Catalyst• Reactions slow down as they proceed

because the concentration goes down, the number of particles per volume goes down as they are used up and so collision frequency goes down.

• The number of effective collisions goes down and so does rate of production of products.

Measuring decrease in concentration of reactant.

• Rate fastest at the beginning so gradient steepest. This is rate at t=o – the initial rate.

• Rate falls off so gradient decreases.• When the reaction is over the graph goes

flat.• E.g. Mass loss of marble chips over time.

Week 14

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Concentration change of a reactant during a reaction

Measuring Increase in Concentration of Product

• Gradient changes exactly as before but the curve has a positive not a negative slope.

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Measuring an initial rate

Monitoring Reactions

• Continuous data is recorded from an experiment where the change in concentration continues to the end of a reaction e.g. gas given off during marble chip and acid or mass change in the same reaction.

• Clock reactions take place when several separate experiments with different concentrations are carried out e.g. iodine clock or acid/thio reaction.

Week 14

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• Plot a concentration–time graph from experimental results.

• Deduce the rate of a reaction from a concentration–time graph.

Week 14

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Drawing tangents to a concentration–time graph

Week 14

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• Explain and use the terms: order and rate constant.

• Deduce a rate equation from orders.

The Rate Equation

• Since rate decreases as concentration of reactants decreases we can express this as an equation- the rate equation (or rate law).

• Rate α [reactant]n

• So• Rate = k x [reactant]n

• where k = the rate constant (which links rate with concentration of reactants)

• and n = the order of reaction with respect to the given reactant.

Order of Reaction

• The order of a reaction is the power to which the concentration of a reactant is raised in the EXPERIMENTALLY determined rate equation.

• Order has NOTHING to do with the balancing of an equation – the stoichiometry and everything to do with experimental data.If we carry out experiments we can find out how changing the concentrations of reactants changes rate.

Orders

• Zero order• If the rate is order 0 with respect to

reactant A then rate = k [A]0

• This means that the rate is unaffected by changing the concentration of A.

• If the order is 1 then the concentration and rate are directly proportional, doubling the concentration doubles the rate etc.

• If the order is 2 then the rate is proportional to the concentration increase squared.

Overall Order

• In the reaction:

• A + B + C = products• Rate = k [A]2[B][C]0

• Rate α [A]2[B] since the concentration of C has no effect on rate.

• The overall order is the sum of the individual orders so the overall order in this case is 3.

Simple Examples

• E.g.What do these mean?

• 2N2O5(g)→2N2O4(g) + O2(g)

• Rate =k[N2O5]

• 2HI(g) → H2(g) + I2(g)

• Rate = k[HI]2

• CH3CHO(g) →CH4(g) + CO(g)

• Rate = k[CH3CHO]2

• H2O2 +2HI→2H2O + I2• Rate = k [H2O2][HI]

• Rate is FIRST ORDER with respect to each reactant so is 2 OVERALL.

Units

• The rate constant k has units determined by the overall order of the reactants in the rate equation.

• E.g. rate = k[A] k = rate• [A]• = mol.dm-3.s-1

• mol.dm-3

• = s-1

More Units

• Work out the units for reactions with overall

• orders 0,2 and 3and the form:

• Rate = k [A] 0 etc

Experimental Determination of Order of Reaction and Rate

Constant• Table of data• Variation of the initial reaction rate for a

reaction:

• NO2(g) + CO(g) → NO(g) + CO2(g)

Experiment Rate/mol.dm3.s-1

[NO2]/mol.dm-3

1 0.010 0.15

2 0.040 0.30

3 0.16 0.69

4 0.32 0.90

Determining Order (1)

• Rate clearly depends on [NO2].

• These can usually be done ‘by inspection’ -mental arithmetic but may be more reliably done as follows:

• Rate(expt 1) α k[NO2,exp 1]n

• Rate(expt 2) α k[NO2.exp 2]n

• THEN:

• Rate(expt2) = ( [NO2,expt 2]) }n

• Rate(expt1) ([NO2,expt 1])

Determining Order

• Substituting:

• 0.04 = 0.3 n

• 0.01 0.15• 4=2n

n=2• The rate is second order with respect

to[NO2].

• The RATE CONSTANT is obtained by substituting.

Calculating k

• k = rate

• [NO2]2

= 0.04 (0.3)2

= 0.44 units? mol-1.dm3.s-1

More Examples

Expt Conc [A]Mol.dm-3

Conc[B] mol.dm-3

Initial rate

1 0.5 1.0 2.0

2 0.5 2.0 8.0

3 0.5 3.0 18.0

4 1.0 3.0 36

5 2.0 3.0 72

Calculate the orderand the value of k.

Calculation

• Let the rate equation be:• Rate = k[A]m[B]n

• i) Compare expt 4 and 5 where [B]is a constant.

• Rate(exp 5) = [A expt 5] m

• Rate(exp 4) [A expt 4] • 72 =2.0 m• 36 1.0• 2 = 2m

• m = 1

continued

• ii) Compare exp 1 and 2 where [A] is const.• Rate(expt2) = [B expt 2] n

• Rate(expt1) [B expt 1]• 80 = 2.0 n

• 20 1.0• 4 = 2n

• n=2• The reaction is FIRST ORDER with respect

to A and SECOND ORDER with respect to B. RATE = k[A][B]2 overall order 3.

Calculating k

• k = rate• [A][B]2

• = 72• (2.0)(3.0)2

• 72• 2x9• 72 units?• 18• = 4.0 • dm6.mol-2.s-1

Week 15

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• Explain and use the terms: order and half-life.

• Deduce the half-life of a first-order reaction from a concentration–time graph.

• State that the half-life of a first-order reaction is independent of the concentration.

Graphs

• Sometimes we can’t use inspection of experimental data to determine the order of a reaction.

• E.g. Mg + 2HCl → MgCl2 + H2

Rate/mol.dm-3s-

1

[HCl]mol.dm-3

0.004 0.50

0.011 0.80

0.022 1.10

0.034 1.40

• To find order of reaction plot rate against [HCl]n where n =1,2 or 3

• The shape of the graphs tell us the order.• A straight line through the origin shows a

direct relationship between rate and (concentration)n

• Once the order has been established the rate constant can be obtained by substituting a pair of values into the rate equation.

Other Graphs

• Concentration/time and concentration/rate graphs are always diagnostic of the order of the reaction in question and these should be known.

• Half Life• Half life is the time taken for the

concentration of a reactant to decrease to half its original value.

• For first order reactions half life is a constant, regardless of initial concentration.

• Also called exponential decay.

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Measuring half-lives

• Zero Order• A straight line (from high on the

concentration axis down to the right hand side of the horizontal axis.)

• The rate is always the same regardless of concentration.

• As the reaction proceeds there is a costant decline in concentration.

• Second Order• The concentration time graph approaches

zero much more slowly than a first order reaction.

• As concentration goes down half life increases.

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Other concentration–time graphs: (a) zero-order, (b) first-order, (c) second-order

(a) (b) (c)

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• Deduce the order (0, 1 or 2) with respect to a reactant from a rate–concentration graph.

• Determine, using the initial rates method, the order (0, 1 or 2) with respect to a reactant.

Rate Concentration Graphs• These are also diagnostic and should be

learned.• Zero Order• Rate is unaffected by changes in

concentration.• A plot of rate against concentration gives a

horizontal line• First Order• Concentration and rate are directly

proportional.• Graph is a straight line through the origin.

• Second Order• Concentration doubles – rate raised to

power 2.• Concentration increased by 3 times then

rate is cubed.• Result is a curve with an increasing

gradient.

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Orders and rate–concentration graphs: (a) zero-order, (b) first-order, (c) second-order

(a) (b) (c)

Week 16

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The rate constant, k, increases with increasing temperature

The Effect of Temperature on Rate Constant and Rate of Reaction

• Remember Boltzmann?• At higher T a greater proportion of

molecules have higher energies.• So more have energy equal to or greater

than activation energy.• So for a reaction A + B → C + D• Rate = k[A]m[B]n

• If the concentration of A and B are kept constant then the value of k must increase with increasing temperature.

Effect of a catalyst on reaction rate

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Week 16

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• Propose a rate equation that is consistent with the rate-determining step.

• Propose the steps in a reaction mechanism from the rate equation and the balanced equation for the overall reaction.

Reaction Mechanisms

• A reaction mechanism is a series of steps that, together, make up the overall reaction.

• The rate determining step is the slowest step in the reaction mechanism of a multi-step reaction.

• The overall reaction can be no faster than the rate determining step.

• Equations say NOTHING about mechanisms, only rate experiments can do this.

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The two molecules of NO2

Examples

• Example 1.

→ CH3CH=CH2

This is the high temperature conversion of cyclopropane to propene.

By experiment

Rate = k[cyclopropane]

• The reaction is first order so the reaction must be single step with only one molecule of cyclopropane involved.

• Energy must shake the (unstable and strained) molecule apart.

• This is a UNIMOLECULAR STEP.• (The molecularity is the number of species

involved in the RDS).

Example 2

• Base hydrolysis of 2-bromo-dimethylpropane:

• (CH3)3CBr + OH- → (CH3)3COH + Br-

• Rate = k [(CH3)3CBr]

• Mechanism has a slow first step (RDS) which is independent of the hydroxide ions.

• Proposed mechanism:

• slow

• (CH3)3CBr → (CH3)3C+ + Br-

• fast

• (CH3)3C+ + OH- → (CH3)3COH

• The slow step governs the rate of reaction – the Rate Determining Step.

• In this example the RDS is UNIMOLECULAR and the fast step is BIMOLECULAR.

• A useful way to check if a proposed mechanism is possible is to remember that intermediates which appear on both the left and right hand sides of the equations must cancel out, leaving only the reactants and products from the balanced equation for the overall reaction.