Chemistry 2

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Supposed you stoppered the flasks after your titration to determine the equilibrium constant and kept the solutions for another day so that the equilibrium was re-established. What would happen to the amount of ethanol present? What would be the relationship of the original equilibrium constant you calculated to the new one?

Okay so this is my attempt at the answer, but I'm really not sure if this is right:

Because ethanol is more volatile than water, some of it, after a day, will evaporate, so we'll have less products, so the equilibrium constant will be different from the one we have and so the equilibrium will be towards the right to compensate for the evaporated ethanol, so K < 1.

Is this correct?

Thank you!

  • Chemistry 2 -

    I'm in the dark as to what you did.

  • Chemistry 2 -

    oh this is just a theoretical question. It's about the hydrolysis of ethyl acetate, it gives ethanol and acetic acid as the products

  • Chemistry 2 -

    It appears to me that if you stoppered the flask there is no question that the ethanol could NOT evaporate.

  • Chemistry 2 -

    oh okay, well in that case there's no change? Nothing happens to the ethanol and we get the same K constant that we had previously?

  • Chemistry 2 -

    Supposed you (stoppered) the flasks

  • Chemistry 2 -

    ethyl acetate + water ---> ethyl alcohol + acetic acid

    Once stoppered and allowed to reach equilibrium again. It will be in the reverse direction

    ethyl alcohol will react with acetic acid to form water and ethyl acetate

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