section 1 controlling the rate collision theory. li to learn about the collision theory (a) s.c. by...
TRANSCRIPT
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