Questions:
- Give an equation for the reaction between phenol and bromine.
- What is the use of methyl orange in this experiment?
The methyl orange is used as an indicator for the reaction. Bromine will be produces during the reaction and it will react with phenol rapidly. Once all the phenol is used up, further bromine produced will bleach the methyl orange immediately. So the rate of the reaction can be calculated from the time for the indicator to be decolorized.
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Based on your results, is it advisable to perform the experiment at high temperatures such as 80℃?
From the above results, the time of the reaction is shorter as the experiment temperature increase. So it is not advisable to perform the experiment at high temperatures as the time of reaction is too short, it is not easy for us to record.
- Why is it not necessary to know how far the reaction has proceeded at the point where the methyl red is decolourized?
In the experiment, the number of mole and the molarity of each chemicals used are the same. And the appearance of the pink colour only determines that there are some content of the reaction has processed. As the result, it is only necessary for us to compare the time required for the reaction mixture to reach the point the decolourization occurs at different temperature.
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The Arrhenius equation can be represented as:
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Can substitute k in this equation? Why?
can substitute k in this equation. As in the above experiment, the initial rate can be represented by due to the constant concentration of the reactants. Also the rate constant k is proportional to the rate of the reaction when the temperature changes. Therefore, can substitute k.
- Derive an equation relating ln k and 1/T.
ln k = ln A – Ea/RT
ln k = ln A – (Ea/R)(1/T)
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Determine Ea by plotting a suitable graph. (Given R=8.3JK-1 mol-1 )
Slope=［-4.727387819-(-3.044522438)］/(3.289473684*10-3-3.086419753*10-3)
=-8287.78
∵ lnk = lnA – Ea/RT
The slope=-Ea/R
∴-8287.78=-Ea/R
Ea=8287.78*8.314
=68904.6 J mol-1
=68.9046 kJ mol-1
- Explain why the reaction rate can be affected by temperature.
As the temperature increase, the numbers of vibration increase, the speed of the particles increase, then the frequency of collision increases. As the result, the frequency of effective collision also increases. Therefore the reaction rate will increase vice versa.
