Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    0
    ReputationRep:
    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!
    Offline

    0
    ReputationRep:
    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.
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Would you like to hibernate through the winter months?
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Quick reply
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.