Lesson 9 – Boltzmann distribution

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Module 3
Lesson 9 – Boltzmann Distribution
Objectives
Must
Recall the shape and axes labelling for a Boltzmann
distribution.
Should
Describe qualitatively, using the Boltzmann distribution,
the effect of temperature changes on the proportion of
molecules possessing a certain energy.
Could
Explain the effect of the proportion of molecules
possessing greater than or equal to activation energy on
rate of reaction.
Starter - quiz
1.
2.
List 4 factors which can increase the rate of a chemical reaction.
Explain why increasing surface area of a solid increases the rate of
reaction.
3. Why does increasing the pressure of a solution of HCl and Na2S2O3
NOT increase the rate of reaction?
4. What needs to happen for two molecules to react with each
other?
5. What do we call the theory which explains these conditions?
6. What do we call the minimum energy molecules need to have in
order to react?
7. What effect does temperature have on the rate of reaction?
8. Name two methods of measuring the rate of a chemical reaction.
9. What is a catalyst?
10. How does a catalyst work?
Starter - quiz
1.
2.
Concentration, pressure, surface area, temperature, catalyst
Increasing surface area means a greater number of reactant molecules are
able to collide with the surface increasing the number of successful
collisions.
3. Because changing the pressure of a solution has little effect on
concentration.
4. They need to collide in the correct orientation and with sufficient energy.
5. Collision theory
6. Activation energy
7. Increases the rate of reaction by increasing the energy the molecules have
therefore increasing the chance of successful collisions
8. Timing of - mass lost, gas collection, change in colour/precipitation.
9. A substance that increases the rate of a reaction without being used up by it
10. Providing an alternative route with lower activation energy
Maxwell-Boltzmann
Maxwell-Boltzmann
• The total area under the graph represents the
total number of molecules
• The shaded area represents the number of
molecules with enough energy to react if they
collide.
The effect of temperature
• We would expect rate to increase with increasing
temperature because the number of collisions
increases.
• However, we also know that particles need a
minimum amount of energy to react when they
collide. We call this the ACTIVATION ENERGY.
• If particles do not have enough energy when they
collide they simply bounce off each other.
• Increasing temperature gives the particles more
energy and so more of them have enough energy
to react when they collide.
Maxwell Boltzmann
Maxwell Boltzmann
• When the temperature is higher the curve gets
broader and the shaded area gets MUCH bigger.
Many more molecules have enough energy to react
when they collide. This means that increasing the
temperature only by a few degrees could double the
rate.
Mark scheme
Homework – essay question
An experiment is conducted to measure the rate of hydrogen peroxide
decomposition. Using collision theory and the Boltzmann distribution
explain the following observations.
• When the temperature is raised by 20°C from 20°C to 40°C the rate
increases to more than four times the original rate.
• A catalyst is tested at 40°C which increases the rate of reaction still
further to double the previous rate
• Heating the mixture containing the catalyst still further to 50°C
caused the reaction rate to fall.
[10 marks]
Extension - Order of reaction
Rate of reaction is dependent on the
concentrations of reacting species
• Our reactants are Na2S2O3 and HCl
We can say that
• RATE = k[Na2S2O3]m[HCl]n
• Where [Na2S2O3] is the concentration of
thiosulphate in mol/l and
• [HCl] is the concentration of HCl in mol/l
RATE = k[Na2S2O3]m[HCl]n
• The small numbers ‘m’and ‘n’ are something
we call the ORDER of the reaction.
• The ORDER could be 1, 2, 3 etc. We call this
1st, 2nd, 3rd order.
• We can ONLY determine these by actually
doing the experiment.
• Plotting RATE (Y) against CONCENTRATION (X)
tells us the order by looking at the shape of
the graph.
RATE = k[Na2S2O3]m[HCl]n
• Rate = 1/time (Y AXIS)
• Straight horizontal line = 0 order changing the concentration
of this has NO effect
• Straight line upwards = 1st order doubling the concentration
doubles the rate.
• Curved line upwards = 2nd order – doubling the concentration
more than doubles the rate.
RATE = k[Na2S2O3]m[HCl]n
• If you have looked at changing the concentration
of thiosulphate you could tell me the ORDER with
respect to this IE ‘m’
• If you have looked at changing the concentration
of HCl you could tell me the ORDER with respect
to this IE ‘n’
• If you have done both you can give me the whole
RATE EQUATION for the reaction!!!!
AND IF YOU HAVE DONE THAT WELL DONE – IT’S
A-LEVEL CHEMISTRY
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