Lesson 10 – Practical and Arrhenius

Module 3
Lesson 10 – Practical and Arrhenius
Describe qualitatively, using the Boltzmann distribution,
the effect of temperature changes on the proportion of
molecules possessing a certain energy.
Conduct an experiment to investigate the effect of
temperature on rate of reaction.
Manipulate the results of the experiment to calculate the
activation energy using a graphical method and the
Arrhenius equation.
Starter homework – peer assess
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]
• In order to react molecules need to collide in the correct orientation and
with sufficient energy [1 mark]
• The Boltzmann distribution shows the number of molecules with a certain
energy. The number of molecules with the Ea is shown as the shaded area
[2 marks]
• Diagram [2marks]
• 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. [1 mark]
• A catalyst reduces the activation energy
(shown illustrated OK) [1 mark].
• This increases the proportion of molecules
possessing the activation energy / area under
curve increases [1 mark].
• The catalyst is an enzyme which becomes
denatured at >40°C. [1 mark].
• The activation energy returns to the
uncatalysed value and rate falls. [1 mark]
The Reaction
Dilute hydrochloric acid reacts with sodium
thiosulphate solution to give a precipitate of sulphur
Na2S2O3 (aq) + 2HCl (aq)  2NaCl (aq) + H2O (l) + SO2 + S (s)
• Using fixed concentrations of 0.5M HCl and 0.1 M Na2S2O3,
investigate the effect of temperature on the rate of
• You should investigate 5 different temperatures with three
repeats across the whole group.
• Your temperatures should span approximately (but
recorded accurately) 5°, 10°, 20°, 30°, 40°
• To do the repeats have all your boiling tubes containing
your reactants in the water bath/trough at the same time.
• Plot three graphs and bring to next lesson
• Time taken (t) against temperature (T)
• Rate (1/t) against temperature (T)
• ln(1/t) vs. 1/T
Extension – The rate equation
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
Extension - Arrhenius
Rate = k[a]x[b]y
For constant [a] and [b] ie concentration kept constant
Rateµk µ 1
K = AexpEa/RT
Lnk = lnA+Ea/RT
So plot ln (1/t) vs 1/T has gradient Ea/R
Further extension
Orders of reaction – experiment
varying concentration
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!!!!

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