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Salt Dissolution Help

  1. Apr 27, 2006 #1
    Well, today in my chemistry lab, my teacher let us out early without getting the class to answer a few questions, and now I'm stuck on my lab report! We were supposed to have a class discussion, but we didn't. I was wondering if anyone knew how to answer the following questions:

    If delta G and delta H for a reaction are both negative what will the sign of delta S be and why?

    If delta G is negative and delta H is positive for a reactoin, what will the sign of delta S be and why?

    I do know that delta G = delta H - T delta S, but that's about it. Any help is greatly appreciated. Thanks!
     
  2. jcsd
  3. Apr 27, 2006 #2

    mrjeffy321

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    A negative value for ∆G means that the process is spontaneous.

    As you said,
    ∆G = ∆H - T*∆S

    If ∆G is negative, then that means that either T*∆S is positive and larger in magnitude than ∆H (thus making the enture equation negative), and/or that ∆H is negative and so large in magnitude that even a negative value for T*∆S will not make ∆G positive.

    If ∆G is negative and ∆H is negative then ∆S could be either positive or negative.
    For example, lets say ∆H was -10 kJ and ∆G was -15 kJ, solving for T*∆S, we find that it equals 10 kJ.
    -15 kJ = -5 kJ - 10 kJ
    But then if we say that ∆G = -10 kJ and ∆H = -20 kJ, then we would need a negative value for T*∆S, -10 kJ,
    -10 kJ = -20 kJ - (-10 kJ)
    -10 kJ = -20 kJ + 10 kJ

    If ∆G is negative and ∆H is positive, then we know the T*∆S must be positive (substracting a [large] positive value from a positive ∆H makes ∆G negative).
     
  4. Apr 27, 2006 #3
    Thank you! I also noticed one more question...

    Does the sign of delta H determine the sign of delta G? Why or why not?
     
  5. Apr 27, 2006 #4

    mrjeffy321

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    The sign of ∆H does not directly determin the sign of ∆G.

    Remember, when ∆G is negative, the reaction is spontaneous. When ∆H is negative the reaction is exothermic and when ∆H is positive, the reaction is endothermite.
    It is pretty easy to think of examples of exothermic, spontenous reactions (a match burning, Iron rusting, ....), but it may not be as easy to think of endothermic spontaneous reactions...but they do exist.
    For example, cold packs use a spontaneous process which occurs when Ammonium Nitrate is dissolved in water...this process endothermic, thus is gets cold (hense the name "cold pack").
    So there are examples of both endo- and exo- thermite spontaneous reactions, in both cases, the sign of ∆H did not effect the fact that ∆G was negative.
     
  6. Apr 27, 2006 #5
    So to answer the question I had above, I should just put that it doesn't directly determine it because when ∆H is negative the reaction is exothermic and when ∆H is positive, the reaction is endothermic?
     
  7. Apr 27, 2006 #6

    mrjeffy321

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    I would say something to the effect of,
    The sign of ∆G is not soley dependent upon the sign of ∆H since both endothermic (∆H > 0) and exothermic (∆H < 0) spontenous reactions (∆G < 0) can be seen to occur.
     
  8. Apr 27, 2006 #7
    Okay, I think this makes sense...thank you!
     
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