Section 1

Controlling the rate

Collision Theory

LI To learn about the collision theory (a)

S.C. By the end of this lesson you should be able to

• Using the “Collision Theory” explain the effects of concentration, particle size, temperature, and collision geometry on reaction rates.

•Calculate the average reaction of a reaction.

• State that where there is a fixed endpoint in a reaction rate = 1/ t (s-1)

(A) Collision Theory

The collision theory states that for a chemical reaction to occur, reactant molecules must

(i) collide with enough energy (ii) collide in the correct orientation.

Increasing the surface area or decreasing the particle size- the smaller the particles, the greater the surface area, the greater the chance of successful collisions and so increases the rate.

Effect of Surface Area

Workbook activity

Calculating the average rate of reaction

The rate of a chemical reaction may be expressed in terms of the changes in concentration or volume or mass over a period of time. This is called the average rate of reaction.

Experiment

Method A - change in volume

Method B – change in mass

in quantity

in timeAverage reaction rate =

What to do

You are going to follow the rate of the reaction by:

A Measuring the volume of gases produced over time

B Measuring the loss of mass over time

Measuring rate of reactionTwo common ways:

1) Measure how fast the products are formed

2) Measure how fast the reactants are used up

How can we follow the reaction?

If we use a container fitted with a delivery tube we could measure the amount of gas produced. How?

What to do - Method A

Measure 25 cm3 of 2 mol l-1 HCl into a conical flask fitted witha stopper and a delivery tube

Set up an inverted measuring cylinder of water to collect the gas

Add 2g marble chips to the acid

Measure the volume of gas every 10 seconds

Repeat with 2g crushed marble chips

What to do – Method A

Record your results in the table.

Plot a graph of volume vs time using the same axes for both sets of data

rate = change in volume ( the unit is cm3s-1) time interval

Calculate the rate for the 1st and 2nd 25 seconds for each set of results

What to do –Method B

Weigh out 2 g marble chips

Measure 25 cm3 1 mol l-1 HCl into a conical flask

Place on balance and zero it.

Now add the marble chips to the acid and take mass readings every 10

seconds

Repeat using crushed chips

What to do – Method B

Record your results in a the table.

Plot a graph of volume vs time using the same axes for both sets of data

rate = change in mass ( the unit is gs-1) time interval

Calculate the rate for the 1st and 2nd 25 seconds for each set of results

Swap results

time (s) 0 10 20 30 40 50 60 70 80

volume (cm3) C

volume(cm3) P

Method A results

Plot the results on a graph with time on the x axis and volume on the y. Use the same set of axes for both sets of results.

0 14 26 36 44 50 50 50 50

0 22 40 48 50 50 50 50 50

Sample results

Rate over 1st 25 seconds (cm3 s-1)

rate over 2nd 25seconds (cm3 s-1)

Whole chips (C)

32-025-0=1.3

50-3250-25=0.72

Ground chips (G)

45-025-0=1.8

50-4550-25=0.2

Work out the rate of reaction over the first 25 seconds and the second 25 seconds using the formula

rate = change in volume = _____________ cm3 s-1

time interval

Work out the rate of reaction over the first 25 seconds and the second 25 seconds using the formula

rate = change in volume = _____________ cm3 s-1

time interval

Time (s)

Volume of gas cm3

Method B results

Plot the results on a graph with time on the x axis and mass on the y. Use the same set of axes for both sets of results.

Method B results

Plot the results on a graph with time on the x axis and mass on the y. Use the same set of axes for both sets of results.

time (s) 0 20 40 60 80 100 120 140 160

Mass (g)C

mass(g)G

2.0 1.4 1.0 0.7 0.5 0.35 0.3 0.25 0.25

2.0 0.7 0.4 0.3 0.25 0.25 0.25 0.25 0.25

Sample results

Work out the rate of reaction over the first 25 seconds and the second 25 seconds using the formula

rate = change in mass The answer will have the units g s-1

time intervalRate over 1st 25 seconds (g s-1)

rate over 2nd 25seconds (g s-1)

Whole chips (C)0.8-225-0=0.05

0.35 -0.850-25=0.018

Ground chips (G)

0.3-225-0=0.068

0.25-0.350-25=1x10-3

Loss in mass (g)

Time (s)

Increasing concentration- increasing the number of particlesincreases the number of collisions and so increases the rate.

Effect of Concentration

2 mol/l 5 mol/l

Workbook activity

Using the graph below calculate the average rate between: (i)0-20 s (ii) 10-20 s (iii) 40-50 s

Average rate = change in concentrationchange in time

The answer will have the units moll-1s-1

Workbook activity

You will carry out the reaction using a series of dilutions of thepotassium iodide solution. This will be diluted by replacing some of

thevolume with water.

Effect of concentration – the chemical clock challenge

Your aim is to find out how changing the concentration of potassiumiodide affects the rate of reaction.

In the earlier experiments you measured the average rate of reaction over a period of time. Sometimes it is easier to make comparisons by calculating the relative rate of a reaction using the formula Relative Rate = 1/ t (s-1) time = 1/ rate (s) This allows a comparison of the rate under different conditions to be compared.

Effect of concentration –the chemical clock challenge

Experiment

2I- (aq) + H2O2 (aq) + 2H+ (aq) 2H2O (l) + I2 (aq)

The reaction mixture stays colourless as the iodine molecules are converted back to iodide molecules by the thiosulphate ions.

Once all the thiosulphate ions have been used, a blue black colour appears suddenly as iodine reacts with starch.

t being a measure of how long it takes for the blue/black colour to form. (when excess I2 forms)

Relative Rate =1

tUnits s-1

+ 2S2O32- (aa) 2I- (aq) + S4O6

2- (ag)I2 (aq)

Effect of concentration –the chemical clock challenge

1) Using syringes measure out 10cm3 sulphuric acid 0.1moll-1

10cm3 sodium thiosulphate 0.005moll-1

1cm3 starch solution25cm3 potassium iodide solution 0.1mol l-1

Into a dry 100cm3 beaker

2) Measure out 5cm3 of hydrogen peroxide 0.1moll-1

into a syringe. Add it to the mixture as quickly as possible and start the timer.

3) Stop the clock when the mixture suddenly turns dark blue.

4) Repeat, using 20 cm3 of potassium iodide solution and 5cm3 of water.

Effect of concentration –the chemical clock challenge

Volume of water (cm3)

Volume of 0.5 mol l-1 KI (aq)

(cm3)

Relative conc

KI

Time (s)

Rate (1/t)(s-1)

0.0 25.0 1

5.0 20.0 0.810.0 15.0 0.6

15.0 10.0 0.420.0 5.0 0.2

Effect of concentration –the chemical clock challenge

RESULTS - Plot a graph showing the relative concentration of potassium iodide x axis and the rate of reaction (1/t) on the y axis.

Workbook activity

Your challenge is to create a series of solutions that will change

colour in time to music

http://www.youtube.com/watch?v=rSAaiYKF0cs

(Daniel Radcliffe)

Effect of concentration –the chemical clock challenge

Listen to the song and identify points where you want to have a colour change come in

Time them accurately. Allocate times to each group.

Look at your results and check that these are times you can achieve

Calculate the rate that each time requires (rate = 1/t)

Read off the required concentration from your graph

Effect of concentration –the chemical clock challenge

Use the relative concentration to help you work out the volume of water and KI(aq) needed to make up 100 cm3 of the required concentration.

Good luck!

Effect of concentration –the chemical clock challenge

How does changing the temperature affect the rate of reaction between oxalic acid and potassium permanganate?

5(COOH)2 (aq) + 6H + (aq) + 2MnO4 2-

(aq) 2Mn 2+ (aq) + 10 CO2 (aq)

8H2O (l)

What colour change takes place?

purple to colourless

This reaction is self indicating. No indicator is needed.

Effect of Temperature

Temperature (oC)

Time (s) Relative rate 1/t (s-1)

40

50

60

70

Plot a graph of 1/time on the vertical (y) axis and average temperature on the horizontal (x) axis.

Work out the rise in temperature required to double the rate of reaction.

Workbook activity

LI To learn about reaction profiles (b)

S.C. By the end of this lesson you should be able to

•Describe the term enthalpy•Explain the terms activated complex and activation energy•Use potential energy diagrams to identify whether a reaction is exothermic or endothermic•Use potential energy diagrams to calculate the enthalpy change for a reaction•Use potential energy diagrams to calculate the activation energy for a reaction•Explain why it is essential for chemists to predict the quantity of heat absorbed or released in an industrial process

(B) Reaction Profiles

Every substance contains stored energy known as enthalpy (H).

Reaction profiles (potential energy diagrams) can be used to show the energy pathway for a chemical reaction. The profile below shows:

A – activation energy for the forward reaction

B – activation energy for the reverse reaction

C – enthalpy change

ENTHALPY

H = H products- H reactants

Has the symbol H

The H can be positive or negative

The units for H kJmol-1

Activated Complex

reactants activated complex products

A highly energetic and unstable arrangement of atoms formed between reactants and products. It is an intermediate in the reaction and only exists for a short period of time.

Exothermic Reactions

ΔH =

ΔH is always negative for exothermic reactions. Exothermic reactions include combustion, neutralisation and respiration.

Workbook activity

Endothermic Reactions

ΔH =

ΔH is always positive. Endothermic reactions include photosynthesis which takes in energy in the form of light.

Workbook activity

Reaction Temp before

mixing/oC

Temp after mixing/oC

Endothermicor

exothermic

10cm3 NaOH + 10cm3

HCl

10cm3

NaHCO3 + 4 spatulas

citric acid

10cm3

CuSO4 + spatula of Zn powder

10cm3 H2SO4

+ Mg ribbon

Workbook activity

For industrial processes it is essential that chemists can predictthe quantity of heat taken in or given out as this will influence the design of the process.

Workbook questions

Runaway reactions such as those causing the disasters in Bhopal and Seveso occur when the rate at which a chemical reaction releases energy exceeds the capabilities of the plant to remove heat. http://news.bbc.co.uk/1/hi/world/south_asia/8392093.stm

ACTIVATION ENERGY AND THE REACTION PATHWAY

Activation energy (EA)

The activation energy (EA) is the minimum kinetic energy required by colliding molecules for a reaction to occur.

The activation energy in the above graph is 40 kJmol-1. (Start to peak)

Workbook activity

COLLISION GEOMETRY

Favourable geometry

Unfavourable geometry

LI To learn about catalysts (c)

S.C. By the end of this lesson you should be able to

•name the two different types of catalysts

•describe how catalysts affect the activation energy of a chemical reaction

•draw potential energy diagrams to show the effect of a catalyst on the activation energy

(C) Catalysts

Heterogeneous - example

Homogeneous - example

Enzymes are biological catalysts, and are protein moleculesthat work by homogeneous catalysis.

When the catalyst and reactants are in different states

When the catalyst and reactants are in the same state

Workbook activity

How heterogeneous catalysts work

A catalyst increases the rate of reactionand can be recovered at the end.

The catalytic mechanism involves the reactant particles being adsorbed onto the surface of the catalyst.

Scholar animation

For an explanation of what happens click on the numbers in turn, starting with

How a heterogenous catalyst works

Adsorption (STEP 1)Incoming species lands on an active site and forms bonds with the catalyst. It may use some of the bonding electrons in the molecules thus weakening them and making a subsequent reaction easier.

How a heterogenous catalyst works

Adsorption (STEP 1)Incoming species lands on an active site and forms bonds with the catalyst. It may use some of the bonding electrons in the molecules thus weakening them and making a subsequent reaction easier.

Reaction (STEPS 2 and 3)Adsorbed gases may be held on the surface in just the right orientation for a reaction to occur.This increases the chances of favourable collisions taking place.

How a heterogenous catalyst works

Desorption (STEP 4)There is a re-arrangement of electrons and the products are then released from the active sites

Adsorption (STEP 1)Incoming species lands on an active site and forms bonds with the catalyst. It may use some of the bonding electrons in the molecules thus weakening them and making a subsequent reaction easier.Reaction (STEPS 2 and 3)Adsorbed gases may be held on the surface in just the right orientation for a reaction to occur.This increases the chances of favourable collisions taking place.

How a heterogenous catalyst works

Catalysts at Work

When the surface activity of a catalyst has been reduced, i.e. poisoned, the catalyst will stop working.

However, catalysts can be regenerated.

e.g. burning the carbon (soot) off the catalyst used duringcracking in the petrochemical industry.

Practical on homogeneous/heterogeneous catalysis

Workbook questions

Potential energy graphs and catalysts

Catalysts lower the activation energy needed for a successful collision.

P.E.

Reaction path

1

3

2

Activation energy EA without a catalyst

Activation energy EA with a catalyst

A catalyst does not affect the potential energies of reactants and products ie. start point and end point the same.

Potential energy graphs and catalysts

Workbook activity

LI To learn about Temperature and kinetic energy (d)

S.C. By the end of this lesson you should be able to

•give a definition for temperature

•use energy distribution diagrams to explain the effect of temperature and catalyst on reaction rate

(D) Temperature and Kinetic Energy

Workbook activity

distribution of the kinetic energy of particles

No ofcollisionswith a givenK.E.

Kinetic energy

EAShaded area represents the no.ofsuccessful collisions - those with K.Egreater than the EA

Temperature is a measure of the average kinetic energy of the particles in a chemical.

Workbook activity

No ofmolecules

Kinetic energy

As the temperature increases the particles gain kinetic energy, so more particles have the required EA so there are more successful collision.

index

EA

T2

T1

Workbook activity

No. of collisions with K.E. more thanactivation energy - successful collisions

Catalysed reactionEA is reduced - shaded area increasesso no. of successful collisions increases.

No ofcollisionswith a givenK.E.

Kinetic energy

EA

EA

Un-catalysed reaction

Catalysts lower the EA

Workbook activity

Temperature and energy