Module 3 Lesson 10 – Practical and Arrhenius

Objectives Must Describe qualitatively, using the Boltzmann distribution, the effect of temperature changes on the proportion of molecules possessing a certain energy. Should Conduct an experiment to investigate the effect of temperature on rate of reaction. Could 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 Dilute hydrochloric acid reacts with sodium thiosulphate solution to give a precipitate of sulphur Na 2 S 2 O 3 (aq) + 2HCl (aq)  2NaCl (aq) + H 2 O (l) + SO 2 + S (s)

Practical Using fixed concentrations of 0.5M HCl and 0.1 M Na 2 S 2 O 3, investigate the effect of temperature on the rate of reaction. 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.

Plenary 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 Na 2 S 2 O 3 and HCl We can say that RATE = k[Na 2 S 2 O 3 ] m [HCl] n Where [Na 2 S 2 O 3 ] 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 K = AexpE a /RT Lnk = lnA+E a /RT So plot ln (1/t) vs 1/T has gradient E a /R

Further extension Orders of reaction – experiment varying concentration

RATE = k[Na 2 S 2 O 3 ] 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 1 st, 2 nd, 3 rd 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[Na 2 S 2 O 3 ] 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 = 2 nd order – doubling the concentration more than doubles the rate.

RATE = k[Na 2 S 2 O 3 ] 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!!!!