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Topic 11 – Kinetics Why do Reactions Occur?

Thermodynamics tells you about the relative stability of reactants and products,

e.g. H2 + O2 → 2 H2O

H2 should react with O2 (ΔH° = –286 kJ mol-1). At room temperature, the reaction is spontaneous

and K = 3.6 x 1041 - But no reaction occurs!

Thermodynamics vs. Kinetics

• Reactions that behave as thermodynamics predicts are said to be under thermodynamic

control.

• Reactions in which their rate is negligibly slow, but thermodynamics predicts should occur, are known to be under kinetic control.

The rate of a reaction is the speed with which the concentrations of the molecules present change.

The reaction rate is the change in concentration of a product or reactant per unit time. The rate is given by the gradient of a concentration vs time graph.

The rate of reaction depends on (in addition to other factors):

• Concentration of some or all of the molecules present

• Temperature

• The presence of a catalyst

reaction rate = - d[A]/dt

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e.g. decomposition of water

2H2O2 (aq) → 2H2O (l) + O2 (g) Start 1.000 M H2O2 After 10 s [H2O2] = 0.983 M

Δ[H2O2] = - 0.017 M Time interval Δt = 10.0 s Rate of reaction = Rate of change of [H2O2]

= Δ[ H2O2]/ Δ t

= - 0.017/10.0 = -1.7 x 10-3 M s-1 To avoid negative rates….

Rate is defined as -Δ[reactant]/Δt

The rate of removal of H2O2 is not constant.

The lower the concentration of H2O2 the slower the rate.

The rate is dependent on concentration

The Rate Constant The proportionality constant, k, is known as the rate constant.

• Rate constant = ΔRate / Δ[A] = gradient of plot below:

Concentraion

Time

highrate

low rate

!

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The rate of reaction at every point on the curve is directly proportional to [A] at that moment in time,

i.e. Rate = k[A] where k = rate constant.

The Rate Equation

Rate = k[A]

This relates the rate of reaction to the concentration terms that affect it.

For the general reaction: aA + bB → mM + nN

-d[A]/dt = k [A]x [B]y

• The rate equation can only be determined by experiment, not from the stoichiometric equation.

• There is no relationship between a and x, b and y ….

• Order of the reaction is given by x + y.

• x is the order with respect to A, y is the order with respect to B etc.

• The molecularity is given by the order of the reaction.

• Rate constant, k, depends on temperature.

Example

H2 (g) + 2 ICl (g) → 2 HCl (g) + I2 (s) Rate = -d[H2] / dt = k[H2][ICl] (by experiment)

This reaction is first order with respect to H2, first order with respect to ICl and second order overall.

Question: For the reaction: 2 NO2 (g) + F2 (g) → 2 NO2F (g) Rate = k[NO2][F2]

What is the order of reaction with respect to NO2, F2 and the overall order of reaction?

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How to Determine the Rate Equation...

Eg O2(g) + 2NO(g) → 2NO2(g)

[O2] / M [NO] / M Initial Rate / M s-1 1 1.10 x 10-2 1.30 x 10-2 3.21 x 10-3

2 2.20 x 10-2 1.30 x 10-2 6.40 x 10-3 3 1.10 x 10-2 2.60 x 10-2 12.8 x 10-3

4 3.30 x 10-2 1.30 x 10-2 9.60 x 10-3

Rate = k [O2]x [NO]y

Model: Keep one concentration constant, alter other concentration and see what happens to rate.

For the example above: Rate doubles when [O2] is doubled, therefore x =1 as 2 = (2)1

Rate quadruples when [NO] is doubled, therefore y = 2 as 4 = (2)2

Next, substitute values to find k:

Rate = k [O2]1 [NO]2

3.21 x 10-3 = k(1.10 x 10-2)(1.30 x 10-2)2

= 1.73 x 103 [O2][NO]2

Question: Determine the rate equation and value of the rate constant for the reaction:

NO2 (g) + CO (g) → NO (g) + CO2 (g)

[NO2] / M [CO] / M Initial Rate / M s-1

1 0.10 0.10 0.0050 2 0.40 0.10 0.080

3 0.10 0.20 0.0050

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Integrated Rate Equn

Analysing a curve is difficult; much better to convert the data to a straight line plot.

E.g. N2O5 → NO3 + NO2 Rate = - d[N2O5]/dt = k [N2O5]1

Integrate to give Ln[A]=Ln[A]o-kt and plot Ln[A] vs t

Order =

Molecularity

Differential Rate

Equation (rate/conc)

Integrated Rate

Equation (conc/time)

Plot

0 1

2

Rate = k Rate = k [A]

Rate = k [A]2

[A]=[A]o-kt Ln[A]=Ln[A]o-kt

1/[A]=1/[A]o+ kt

[A] vs t Ln[A] vs t

1/[A] vs t

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Half Life, t1/2

The half life of a reaction is the time required for the concentration to fall to half its initial value. For first order reactions (only) this is a constant t½= Ln 2 / k.

Eg First order radio active decay of iodine-131.

Example An antibiotic breaks down in the body with a first order rate constant of k = 1.9 x 10-2

min-1. How

long does it take for the concentration to drop to 1/8 th the initial value? t½ = Ln 2 / k = 0.693 / 1.9 x 10-2 min-1

= 36 min Three half lives = ½ x ½ x ½ = 1/8

Time of three half lives is 3 x 36 min = 109 min

Question The radioactive isotope 15O is used in medical imaging and has a half life of 122.2 seconds. What

percentage is left after 10 minutes? Use k = Ln2/t½ and Ln[A]=Ln[A]o - kt Note: Ln{ [A]/[A]o }= -k t and [A]/[A]o is amount left.

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[WS 38] The half life of a reaction, t1/2, is the time it takes for the amount of a single reactant to reach one half of its original value.

The figure below shows how the amount of oxiplatin in the blood decreases after a dose is injected.

Oxiplatin is a platinum based drug and its structure is shown on the right. It is designed as a second generation anti-tumour agent

with increased selectivity and fewer side effects than cis-platin

Questions: 1. (a) What is the amount of oxiplatin immediately after the injection?

(b) What is the amount of oxiplatin at t = 14 minutes? What is the ratio of the amount of the drug at t = 14 minutes to the amount at t = 0 minutes?

(c) What is the amount of oxiplatin at t = 28 minutes? What is the ratio of the amount of the drug at t = 28 minutes to the amount at t = 14 minutes?

012345678910

0 10 20 30 40 50time/min

amountinblood/mg

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(d) What is the amount of oxiplatin at t = 42 minutes? What is the ratio of the amount of the

drug at t = 42 minutes to the amount at t = 28 minutes?

2. Use the figure to estimate the amount in the blood at t = 10 minutes. Using this, estimate the amount that would be left after t = 24 minutes. Is this result consistent with your answer to Q1?

3. What is t1/2 for oxiplatin?

Integrated rate laws and half lives Commonly, reactions are zero, first or second order. For a reaction, A à B, this means:

Order Rate law Integrated rate law

0 ![!]!" = -k[A]0 = -k [A] = - kt + [A]0

1 ![!]!" = -k[A]1 = k[A] ln[A] = - kt + ln[A]0

2 ![!]!" = -k[A]2 !

[!] = kt + ![!]!

The integrated rate law allows you to predict the concentration of the reactant [A] at any time t if the initial concentration [A]0 is known.

Experimentally we almost always measure how the concentration of a reactant or product changes as the reaction progresses. We can use the integrated rate law to work out the order of a reaction.

For example, if a reaction is first-order, a plot of ln[A] vs t will be straight line with:

• gradient = -k

• y-intercept = ln[A]

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Questions: 1. Complete the first three columns of the table below by detailing the plot you would use to obtain a

straight line and its gradient and intercept for each reaction order.

2. From Model 1, the half life is is the time it takes for the amount of a single reactant to reach one half of its original value, so that [A] = [A]0 / 2. By substituting this into the integrated rate law, work out the

half life for each reaction order.

3. What is the order of the reaction in Model 1 with respect to oxiplatin?

Order Plot Gradient Intercept Half life

0

1 ln[A] vs t -k ln[A]

2

Reaction Mechanisms A balanced chemical equation describes the overall chemical reaction

e.g. 2 NO2 (g) + F2 (g) → 2 NO2F (g) A reaction mechanism is a series of ‘elementary’ reaction steps that add up to give a detailed

description of a chemical reaction.

e.g. NO2 + F2 → NO2F + F Step 1

NO2 + F → NO2F Step 2

2 NO2 + F2 → 2 NO2F Overall

A rate determining step in a reaction mechanism is an elementary process that is the slowest step in the mechanism

e.g. Step 1 NO2 + F2 → NO2F + F (slow)

• In simple reactions, the exponents in the rate equation are the same as the coefficients of the molecules of the rate determining elementary process. So in this case,

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Rate = k[NO2][F2]

• Note the overall reaction is: 2NO2(g) + F2(g) → 2NO2F(g)

Extra Notes

• Rate law (definition): an expression relating the rate of reaction to the rate constant and the concentrations of the reactants

• One can calculate the rate of reaction at any point given the values of k, x and y as well as [A] and [B]

• Reaction order: a term used to define the overall power of a rate law.

• Note that Rate = k[F2][ClO2] same as k [F2]1[ClO2]1 • Other example: If the rate law was Rate = k[A]2[B]1, then order = 3 (3rd order reaction)

Consider last example, reaction is zero order in A (ie. independent of the concentration of A). The reaction is zero order in A, but first order in B and first order overall

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Reaction Order Revisit substitution reactions:

Kinetics of a classic substitution reaction

• Rate = k[CH3Br][OH–]

• i.e. first order in CH3Br and first order in OH–.

Reaction Profile – SN2

Reaction Profile – SN1

CH3CH3CCH3

BrCH3CH3CCH3

OHOH Br+ +

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then

• Rate = k[(CH3)3CBr]

• i.e. first order in (CH3)3CBr only.

• Once again, there is no relationship between stoichiometry and the rate law. The rate law can only be determined experimentally.

• The second step is much faster than the first, and the overall rate of the reaction approximates the rate of the first step due to a “bottleneck”.

Extra Notes

• Experimental data normally obtained for rate equations using spectroscopic measurements (eg, UV-vis) or pressure measurements (for gases)

• When reaction involves more than one reactant: need to use isolation method.

• Isolation method: this is the procedure of fixing concentration of one reactant and altering

concentration of the other. (An every day method really: e.g., investigating cause of allergy in diet, may systematically remove one component while keeping others constants)

• Isolation method yields following:

• Rate a [A]2 and Rate a [B], combined gives Rate = k[A]2[B] .e., a third order reaction which is 2nd order in A and first order in B

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Example: NO2 + CO → NO + CO2

The experimentally determined rate equation is Rate = k[NO2]2

Show the rate expression is consistent with the mechanism:

2 NO2 ⇌ N2O4 fast equilibrium

N2O4 → NO + NO3 slow

NO3 + CO → NO2 + CO2 fast

NO2 + CO → NO + CO2 overall

Predict Rate = k2[N2O4] but N2O4 is not a reactant and we like the rate expression to be framed in terms of reactant concentrations.

but k1[NO2]2 = k-1[N2O4] or [N2O4] = k1/k-1 [NO2]2

so Rate = (k2 k1/k-1) [NO2]2 or Rate = k[NO2]2

Question: For the reaction: 2NO(g) + Br2(g) → 2NOBr(g) Proposed mechanism consists of two elementary reactions:

NO + Br2 ⇌ NOBr2 fast equilibrium

NOBr2 + NO → 2NOBr slow

Confirm this mechanism is consistent with the stoichiometry of the reaction and the observed rate law of:

Rate = k[NO]2[Br2]

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Kinetics Revision Questions [WS 37] What is the rate? The rate of a chemical reaction depends on how quickly the reactants are consumed or, alternatively, how fast the products are formed. The rate is always a positive number so, as

[reactant] decreases so:

!"#$ !" !"#$%&'()"# !" !"#$%#&% = - !!!"#$ !" !"#$%&'( !" !"#$%#&%!!!"#$ !" !"#$ = - ∆[!"#$%#&%]∆! (1)

Questions: 1. If time is measured in seconds, what are the units of the rate?

2. For the reaction, 3ClO-(aq) à 2Cl-(aq) + ClO3

-(aq)

time (s) [ClO-(aq)] / M [Cl-(aq)] / M [ClO3-(aq)] / M

0 1.20 0 0

100. 0.90 0.20 0.10

Based on this data, calculate

(a) rate of consumption of ClO- =

(b) rate of production of Cl- =

(c) rate of production of ClO3

- =

3. Use the chemical equation for the reaction to explain your answer to Q2.

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The rate of a reaction is the rate of consumption of reactant or the rate of production of product whose stoichiometric coefficient is 1 in the balanced equation. Questions: 1. What is the rate of this reaction?

2. What is the relationship between the rate of consumption of ClO- and the rate of the reaction?

3. What is the relationship between the rate of production of Cl- and the rate of the reaction?

The rate of a reaction depends on how much of each reactant there is: Exactly how the rate depends on the concentration of the reactants can only be determined experimentally. It has nothing to do with the chemical equation.

For the reaction, 2NO(g) + Br2(g) à 2NOBr(g)

Experiment [NO]initial / M [Br2]initial / M initial rate of reaction / M

s-1

1 0.10 0.10 0.078

2 0.20 0.10 0.312

3 0.20 0.30 0.936

Questions: 1. The only change between experiments (1) and (2) is that [NO]initial is doubled. What is the effect of this on the rate?

2. If rate is propotional to [NO]x, what is x?

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3. The only change between experiments (2) and (3) is that [Br2]initial is trebled. What is the effect of

this on the rate?

4. If rate is propotional to [Br2]y, what is y?

5. Using your answers to Q2 and Q4, complete the rate law:

rate = k[NO]x[Br2]y =

6. Between experiments (1) and (3), [NO]initial is doubled and [Br2]initial is trebled, what is the effect on the rate? Is this consistent with your rate law?

The rate constant The rate constant k is just the proportionality constant in the rate law. Its value depends on the

temperature but not on the concentrations. Questions:

1. Using your rate law, work out the value of k using the rate and the concentration for each of the experiments.

2. Using your rate law, write down the units of [NO]x[Br]y. The rate has units of M s-1 so what are the

units of k?

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Activation Energy The activation energy, Ea, of a reaction is the minimum amount of energy that the reacting

molecules must possess if the reaction is to be successful.

At higher temperatures, more molecules have the minimum required energy (Ea) and hence the faster the reaction.

S. Arrhenius (1859 – 1927) 1903 Chemistry Nobel prize

The Arrhenius equation describes the temperature dependence of the rate constant that is exponentially related to the activation energy, Ea (A is the pre-exponential factor, or the “A factor”).

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k = Ae-Ea/RT or Lnk = LnA –Ea/RT

where: T = temperature (K), R = universal gas constant (8.314 J K-1 mol-1) A = Arrhenius constant or pre-exponential factor (“reaction efficiency”)

Why does an activation barrier exist for chemical reactions?

• Molecules have to collide with sufficient energy for bond rearrangement to occur.

• Molecules have to collide with the correct orientation.

If the rate constant is measured, at least, at two different temperatures, the activation energy may

then be determined.

For example: at T1 Ln k1 = Ln A - Ea / RT1 and at T2 Ln k2 = Ln A - Ea / RT2

⎟⎟⎠

⎞⎜⎜⎝

⎛−

−==−

212

121

11lnlnlnTTR

Ekkkk a

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Example: If a reaction has a rate constant of 2.0 x 10-5 s-1 at 20 °C and 7.3 x 10-5 s-1 at 30 °C, what

is the activation energy?

Ln {(2.0 x 10-5)/ (7.3 x 10-5)} = - (Ea/8.314) {1/293 - 1/303}

Ea = 96 kJ mol-1 Question:

The rate constant of a particular reaction triples when the temperature is increased from 25 °C to

35 ˚C. Calculate the activation energy, Ea, for this reaction.

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Catalysis A catalyst provides an alternative reaction pathway of lower activation energy.

In this case more molecules have the minimum energy required for successful reaction and the reaction proceeds at a faster rate.

Four criteria must be met by catalysts:

• Catalysts increase the rate of reaction.

• Catalysts are not consumed by the reaction.

• A small quantity of catalyst affects the reaction rate for a large amount of reactant.

• The position of equilibrium for the reaction remains unchanged. Let’s look at the decomposition of H2O2.

2 H2O2(aq) → 2 H2O(l) + O2(g)

Catalyst Ea (kJ mol-1) Rel. rate of reaction

None 75.3 1

I– 56.5 2.0 x 103 Pt 49.0 4.1 x 104

Catalase 8.0 6.3 x 1011

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Enzyme Catalysis

• Enzymes are biological catalysts that act on specific reactants (substrates).

• Enzyme catalysis = homogeneous catalysis.

• All enzymes are proteins, but not all proteins are enzymes.

• The part of the enzyme tertiary structure that is responsible for the catalytic activity is known as the “active site”.

• Efficient: rate enhancements of 108 to 1020 possible.

• Specific.

• But low tolerance to temperature and pH changes.

e.g.

Nitrogenase converts N2 to NH3 at 1 atm and 25°C (compare with conditions of the Haber process!)