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Homework Help: Equilibrium in reaction of C2H5CO2H(l) + C3H7OH(l)

  1. Feb 9, 2008 #1
    C2H5CO2H(l) + C3H7OH(l) ---> C2H5COOC3H7(l) + H2O(l)

    18.5 mL propanoic acid are mixed with 19.0 mL n-propanol to produce n-propyl propanoate when heated and distilled.

    At room temperature, the equilibrium constant, Keq is approximately 3. Calculate the number of moles of reactants and products expected in the equilibrium mixture of this reaction. (Hint: Let X be the number of moles of each product formed at equilibrium. Substitute into the equilibrium equation using the moles of each reactant and product at equilibrium each divided by the total reaction volume. Leave the equilibrium concentrations as fractions. You can then take the square root of both sides of the equation after canceling like terms, i.e. the volumes.)

    Other Data:

    Boiling Point of n-propyl propanoate = 122 degrees C
    Volume of water collected = 3.31 mL
    Weight of product (C2H5COOC3H7) = 7.4846 g

    I have no idea of how to start this problem, can anybody give me a push in the right direction please?
  2. jcsd
  3. Feb 10, 2008 #2
    I can give you a push in what I think is the right direction but the question
    as stated is not clear to me. From the information given, you can easily
    estimate the concentrations of all substances in the mixture at a particular
    temperature. i.e. if the mixture were refluxed at some temperature T:
    Let n(p) be the number of moles of water at equilibrium at temperature T.
    Then the number of moles of ester is also n(p), the number of moles of
    acid and alcohol is x(ac) = n(ac) - n(p) and x(al) = n(al) - n(p), where
    n(ac) and n(al) are the number of moles of acid and alcohol, repsectively,
    that are available (given as 18.5 mL and 19 mL respectively---to compute
    the number of moles n(ac) and n(al) you need the density
    at room temperature and the molecular weights). Given any temperature T,
    you first need to estimate the equilibrium constant at that T, K(T).

    d(log K)/d(1/T) = -Delta HoR ,

    where "log" is the natural logarithm, Usually, one can assume that the
    enthalpy of reaction is constant over the temperature range; if this is
    not true, then you must express it as a function of T as well. Then,
    obtain n(p) by equating the equilibrium-constant-expression to K(T) and
    solving for the one unknown, n(p). That's it.

    But maybe I have not interpreted the question properly. First of all,
    I don't understand the reference to "volumes" in the hint. There is
    no big volume change on reaction, even in the gas phase, so volume should
    play a negligible role. Secondly, the question mentions distillation.
    Water and n-propyl alcohol both have boiling points near 100 C,
    whereas propanoic acid boils at about 141 C. Removal of water would
    drive the equilibrium in the direction of products but removal of
    alcohol would have the opposite effect.

    Finally, the number of moles of water collected is about 2.85 times the
    number of moles of product (propyl ester), but these should be equal.
    Was the entire mixture distilled? If not, there must be a considerable
    amount of product that was not distilled, or there are other reactions
    going on that were not mentioned.

    I conclude that I have misunderstood the question or have made some
    mistake in my rough calculations. Nevertheless, maybe you will get
    some idea of how to approach this better, and that would also
    constitute a "push".
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