Substitution Reactions Of Halogenoalkanes Watch

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#1
Report Thread starter 10 years ago
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Hi,

I am currently in the middle of my practicals for my salters chemistry investigation. The title of my investigation is investigating the substitution reactions of halogenoalkanes. I have completed the practical using the tertiary halogenoalkane 2-methyl-2-bromopropane. My aim is to calculate the order with respect to each reactant and therefore the rate equation for the reaction. I have decided to measure the initial rate of the reaction to do so. Firstly I react the halogenoalkane with potassium hydroxide using ethanol as a common solvent at room temperature. After 30s I quenched the reaction using Sulphuric acid and then did a back titration with Sodium Hydroxide. I have then tried to calculate the initial rate but have had many problems. I am hoping it is an error in my calculation that someone maybe able to help me with! The data is as follows:

15 cm cubed of 0.02M Potassium Hydroxide solution in ethanol
20cm cubed of ethanol
0.1 cubed 0.1M 2-methyl-2-bromopropane
Quenched with 50cm cubed of 0.01 molar Sulphuric Acid
Indicator = 5 drops of phenol red
Titre of 0.1M Sodium Hydroxide = 8.9cm cubed.

And another set of data is:

5 cm cubed of Potassium Hydroxide solution in ethanol
0.6 cm cubed of 0.1 Molar 2-methyl-2-bromopropane
20.4 cm cubed of ethanol
Quenched with 20 cm cubed of 0.01M Sulphuric Acid
Titre of 4 cm cubed using 0.1 Sodium Hydroxide
Indicator = 5 drop of phenol red

Any help with the calculation or advice for this experiment would be greatly appreciated!
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Freya77
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#2
Report 10 years ago
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Did you vary the time at which you quenched and then titrated? In the first set of data I would assume that as quite a large volume of potassium hydroxide is used it's in excess and so it's conc. remains fairly constant so the rate would be k=[2-methyl-2-bromopropane]^x. Then I would plot a graph of time of quenching vs volume of sodium hydroxide needed to neutralise. Then I would vary the volume of 2-methyl-2-bromopropane (maybe double it for example - but then remember to keep the total volume the same so concentration of the halogenoalkane is effectively being changed) then again plot a graph of time of quenching vs. volume of sodium hydroxide needed to neutralise. I would then compare these two gradients and see how it corresponds to the change in volume of the halogenoalkane.

Sorry if this is not of any use to you; I did this prac a long time ago! Good luck with it.
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