Chemical Reactions: Kinetics

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Rate of Reaction

University of Lincoln presentation

http://creativecommons.org/licenses/by-nc-sa/2.0/uk/

http://chemistryfm.blogs.lincoln.ac.uk/




Why the difference?

Is it the enthalpy change (Heat of combustion) ?

Paraffin wax 42 MJ kg-1

Petrol 45 MJ kg-1

Is it the temperature?Yellow/white – 1300oCPale orange/yellow – 1100oC

What is it then?



Approximately how long will a 2 litre pool of petrol burn for?

Important values: Petrol density = 0.8 kg litre-1

Heat of combustion is 45 MJ kg-1

2 litres of petrol has a mass of 1.6 kg (from the density)

Total energy available from 1.6 kg petrol

= 1.6 kg x 45 MJ kg-1 = 72 MJ

2 litre petrol pool is a 1 MW fire (this is a measured value)

1 MW = 1 MJ s-1 so at this rate it would take 72 s

ΔHmq



How do ignitable liquids burn?



2 litre petrol bomb takes about 10s to burn. What is the rate of heat release? 72 MJ in 10 s = 7.2 MW2 litre petrol fully evaporated takes about 1 s to burn. What is the rate of heat release?72 MJ in 1s = 72 MW

Conclusion: Same total energy available but released at a faster rate



How long to burn a 1.6 kg candle?

• 1.6 kg paraffin wax at 42 MJ kg-1 can release 67.2 MJ

• Candle flame has a heat release rate of 80 W (80 Js-1)

s840000Js80

J67.2x10Js80MJ67.2

time(s) 1

6

1



A candle bomb?

• NASA are researching the paraffin rocket!!

• How can this work?• Increase rate of combustion

– Increase concentration of the oxidant; use 100% oxygen

– Paraffin as small liquid droplets

• Study of the rates of reaction - kinetics



Factors affecting the rate of a chemical reaction

1. Concentration (hydrogen peroxide demo)

2. Pressure (gases)

3. Temperature (glowstick)

4. Surface area (dust explosion)

5. Catalysis



Measuring Reaction Rate

• Use a characteristic of the products or the reactants that can be used as a measure of amount.– Volume of gas– Change in mass– Absorption of light

• rate of decrease of reactant or rate of increase of a product

DCBA

tD

tC

tB

tA

RateΔ

Δ

Δ

Δ

Δ

Δ

Δ

Δ



H2O2(l) H2O(l) + ½O2(g)

0

50

100

150

200

0 50 100 150Time (s)

Am

ount

of H

2O2 re

mai

ning

(x1

05 m

ol)



Calculating Rate of Reaction

The gradient of tangent to the curve is the rate of reaction

What happens to the reaction rate with time?

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0 50 100 150

Time (s)

Con

cent

ration

of H

2O2

(mol

dm

-3)

Concentration = 0.3 mol dm-3 s-1

Rate = gradient = 0.0068 mol dm-3 s-1

Concentration = 0.1 mol dm-3 s-1

Rate = gradient = 0.0023 mol dm-3 s-1



A Mathematical Relationship

• Select two other points on the curve and calculate the rate of reaction at that concentration of H2O2

• Plot a graph of Rate of Reaction as a function of H2O2 concentration



Rate plot for the decomposition of hydrogen peroxide

0

0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35

Hydrogen Peroxide concentration (mol dm-3)

Rat

e of

Rea

ctio

n (m

ol d

m-3 s

-1)



What does the graph show?• Graph is a straight line through the origin

• The two variables have a linear mathematical relationship

• We can say: Rate of Reaction is directly proportional to Hydrogen Peroxide concentration

122OHRateα 122OHkRate

− Easy to predict what happens to reaction when [H2O2] is changed

− [H2O2] x2 Rate x 2

− First Order with respect to H2O2

− k is the rate constant; first order reaction has units of s-

1 when the rate of reaction is measured in mol dm-3 s-1. Show this by rearranging the rate equation and why are the units of rate important.



Rate of reaction can be measured from the rate that oxygen gas is produced.

50

40

30

20

10

0

Yeast suspension +hydrogen peroxide solution

Inverted burette

Water


0

5

10

15

20

25

30

0 50 100 150 200

Time (s)

Vol

um

e O

2 /cm

2


Vary the starting concentration and measure the initial rate

[H2O2]= 0.40 mol dm-3

[H2O2]= 0.32 mol dm-3[H2O2]= 0.24 mol dm-3[H2O2]= 0.16 mol dm-3[H2O2]= 0.08 mol dm-3



Initial Rate can be measured

0

5

10

15

20

25

30

0 50 100 150 200Time (s)

Vo

lum

e O

2/c

m-3

Initial gradient 0.51cm3s-1



Plot Initial Rate as a function of starting concentration

0

0.1

0.2

0.3

0.4

0.5

0 0.1 0.2 0.3 0.4

Conc of Hydrogen peroxide (mol dm-3)

Rate

of re

act

ion

(cm

2 (O

2)s-1

)



Summary– Decomposition of H2O2 can be followed by

measuring the decrease in H2O2 concentration or the volume of O2

evolved.– Rate of reaction can be calculated from

the progress curve at different times or initial rate measurements.

– Plots of rate as a function of reagent concentration can be used to determine the mathematical relationship

– Order of reaction can be determined– Rate equation can be written



For you to do: Initial rate data

[H2O2]/mol dm-3 Rate/cm3 O2 s-1

0.08 0.1

0.16 0.215

0.24 0.32

0.32 0.41

0.4 0.51

Determine the order of reaction with respect to hydrogen peroxide and calculate the value of the rate constant.



General Rate equations

kAkRate 0

nAkRate

2AkRate

Zero order

Units of k ?

First order

Units of k ? AkAkRate 1

Second order

Units of k ?



Further Analysis of data

• Logarithms can be very useful• Plot of log rate as a function of log

concentration (p439 Housecroft)

nAkRate

AnloglogkAloglogkAlogklogRate nn

Gradient is n; Intercept is log k

Use this method on the initial rate data in slide 21 to determine order and the value of k



Half-lifeTime taken for the concentration of reactant A at time t, [A]t to fall to half its value.

kt21

ln ktAA

ln0

t

0.693kt k

0.693t

A constant half-life for a first order reactionProgress curve and measure t½ at several different

points.



Constant half-life

0

50

100

150

200

0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180Time (s)

Am

ount

of H

2O2 re

mai

ning

(x1

05 m

ol)

27s 27s

26s

Going from 200 x 10-5 mol to 100 x 10-5 mol takes 27s





Using Luminol to detect blood stainsExponential decay curveFirst order write rate equationCalculate half-life and why is it important Video clip or demo



Reactions with more than one reactant

A + B → products

mn BAkRatee.g. C12H22O11 + H2O → C6H12O6 + C6H12O6

sucrose glucose fructose

OHOHCkRate 2112212

First order with respect to each reactant Second order reaction (sum of orders in rate equation)



Determining order and rate equations

Difficulties with more than one reactant?

•Initial rate method

•Isolation method

Experimental Design

Principle

Vary one concentration and keep other(s) constant while measuring rate.



An Example Reactionperoxodisulfate (VI) and iodide

ions

22

42

82 I2SO2IOS

Design the experiment

1. initial rate method (vary each concentration)

2. Plot a graph of log rate as a function of log initial concentration for each reactant. Gradient of each line is order of reaction for each reactant.

3. k is determined by rearranging the rate equation.

11282 IOSkRate

Task: Determine the rate equation and a value for k



Iodine clock data from experiment



Collision theory

• Molecules have to collide if they are to react – increasing frequency of collisions?

• Increasing concentration increases the frequency of collisions

• Increasing pressure increases frequency of collisions

• Increasing temperature increases frequency of collision

• But not just about rate of collisions – how do we explain slow reactions?



Activation Energy (Enthalpy)

Ea

• Energy of the collision must be above a certain value for reactants to react

• Why? Energy is needed to break bonds (remember bond enthalpies)

• This then creates reactive species to make new bonds

• The minimum energy required for a collision to result in chemical reaction is Ea



Only those molecules with sufficient energy can react

Num

ber

of

mole

cule

s w

ith

kineti

c energ

y E

Kinetic energy (E)

Activation enthalpy Ea =50kJ mol-1



Increasing Temperature increases Rate of Reaction

Num

ber

of

mole

cule

s w

ith

kineti

c energ

y E

Kinetic energy (E)

Activation enthalpy Ea =50kJ mol-1

Number of molecules with energy greater than 50kJ mol-1 at 300 K

Number of molecules with energy greater than 50kJ mol-1 at 310 K

300 K

310 K



Back to petrol

• Petrol vapour reacts with oxygen (air)

• But not spontaneous at room temperature

• Needs ignition. What does ignition do?

– Provides energy to break bonds (endothermic)

– Creates reactive species (free radicals)

– Self-sustaining (can remove ignition source and it carries on). Why????

– Energy released from the reaction breaks more bonds and the reaction continues



Combining Activation Energy and enthalpy

Draw a diagram for an endothermic reaction

Both can be shown on an enthalpy level diagram

Add Ea to the diagram

A+B

C+D

ΔHPosi

tive

en

thalp

yN

egati

ve

en

thalp

y

Reaction coordinate

Exothermic reaction



Rate equations and Temperature

RT

Ea

Aek

k is the rate constant; A is the pre-exponential factor; Ea is the activation energy; R is the molar gas constant (8.314 J mol-1 K-1); T is the absolute temperature (Kelvin).

increase temperature increase k increase rate

decrease Ea increase k increase rate

The Arrhenius equation

How does it work?

mn BAkRate

RT

EAk AlnlnIt might be easier to do

this



The well known ‘rule of thumb’• Reaction rate doubles if

temperature is increased by 10 oC

Temp/K Ea/kJ mol-1 A/L mol-1 s-1 k/L mol-1 s-1

313 54 8.7 x 106 8.5 x 10-3

323 54 8.7 x 106 1.6 x 10-2

Check the values of k by calculating them from the Arrhenius equation using the other values in the table

Calculate k at 333 K. What is happening to the value of k? How will this affect the rate of this reaction?



An experiment to determine Ea

• Determine order and rate equation for the reaction

• Measure the rate of reaction at different temperatures keeping the initial concentrations the same

• Calculate k at the different temperatures

RTE

lnAlnk ARTEa

Aek

Plot lnk against 1/T: gradient = -EA/R; intercept = lnA



Calculating Ea

Temperature/K

k/dm3 mol-1 s-1

296 2.9 x 10-3

302 4.2 x 10-3

313 8.3 x 10-3

323 1.9 x 10-2

Use the data below to calculate a value for the activation energy for this reaction



How do we explain catalysis?



What are catalysts?

Definition and some examples; reactions and catalystsHydrogen peroxide , metals and natural substancesEnzymesGases on metal surfacesWhat is a different reaction route?


A catalyst provides an alternative path for the reaction with a lower

activation enthalpyEnth

alp

y

Progress of reaction

Reactants

Products

Activation enthalpy of catalysed reaction

Activation enthalpy of uncatalysed reaction

Uncatalysed reaction

Catalysed reaction


Acknowledgements

• JISC• HEA• Centre for Educational Research and

Development• School of natural and applied sciences• School of Journalism• SirenFM• http://tango.freedesktop.org

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