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Homework Help: MCAT Preparation Question - Solutions

  1. Jan 6, 2009 #1
    Hi folks,

    I'm currently studying for the MCAT. I did register with my MCAT prep book's web site but you have to be a part of the class to post in the forums... I can only view the topics and not make posts myself.
    I'll be posting MCAT questions from time to time that I feel the book gives unsatisfactory explanations for/explanations I disagree with.

    Anyway....The question is as follows:

    81. NaCl dissolves spontaneously in water. Based upon the following reaction: NaCl (s) --> Na+(g) + Cl-(g)

    deltaH for reaction = +786 kJ/mol

    the heat of hydration for NaCl must be:

    A. negative with a magnitude less than 786.
    B. negative with a magnitude greater than 786.
    C. positive with a magnitude greater than 786.
    D. Nothing can be determined about the heat of hydration without more information.

    I answered A to this question, rationalizing that because the dissolution reaction is spontaneous it must be negative. B and C just seem like silly answers to me.

    The book claims the answer is D. Their explanation is as follows:
    "The change in entropy is positive in solution formation and Gibbs free energy is negative in a spontaneous reaction. From eqn:

    deltaG = deltaHsol - T(deltaS), we see that the heat of solution may be either positive or negative in this case. The heat of hydration is the separation of water molecules (which requires energy) and the formation of bonds between the ions and water molecules (which releases energy). Thus, the value of the heat of hydration could be either positive or negative. The actual heat of hydration is -783 kJ/mol making the heat of solution +3 kJ/mol."

    Doesn't their explanation implicitly support answer A? Or am I missing something?


  2. jcsd
  3. Jan 6, 2009 #2


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    You are missing something.

    A reaction or process is spontaneous if the value of the Gibbs free energy is negative.
    The Gibbs free energy is defined as

    delta G = delta H - T(delta S)

    You are only given the heat of dissociation. You have two processes... one is dissociation and one is hydration and you know nothing about the corresponding entropy values, so you don't have enough information to say anything about the value of the heat of hydration.
    Last edited: Jan 6, 2009
  4. Jan 6, 2009 #3
    Hmm... so let me see if I understand this:

    - We know that the delta G value is some negative number.
    - We also know that the overall delta S for the reaction will be some positive number, because entropy will increase in a dissolution.

    wait... *lightbulb*

    So the deltaH that we get out of the number-crunching for this reaction is really the summed delta H for both dissociation and hydration.

    deltaHsol = deltaH of dissociation + deltaH of hydration.

    We know the deltaH of dissociation but with what we are given we cannot solve for deltaG and hence have no way of deriving the heat of hydration.

    ... And we can't solve for deltaG because we know neither temperature nor entropy... and hence can't get deltaH either?

  5. Jan 6, 2009 #4


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