# Lesson 9 – Boltzmann Distribution

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Lesson 9 – Boltzmann Distribution
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 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. Why does increasing the pressure of a solution of HCl and Na2S2O3 NOT increase the rate of reaction? What needs to happen for two molecules to react with each other? What do we call the theory which explains these conditions? What do we call the minimum energy molecules need to have in order to react? What effect does temperature have on the rate of reaction? Name two methods of measuring the rate of a chemical reaction. What is a catalyst? How does a catalyst work?

Starter - quiz 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. Because changing the pressure of a solution has little effect on concentration. They need to collide in the correct orientation and with sufficient energy. Collision theory Activation energy Increases the rate of reaction by increasing the energy the molecules have therefore increasing the chance of successful collisions Timing of - mass lost, gas collection, change in colour/precipitation. A substance that increases the rate of a reaction without being used up by it 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