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Chemistry question - Entropy watch

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    So there are two equations

    Gibbs free energy = delta H - T delta S - equation 1.

    which is derived from

    Delta S total = Delta S system - (delta H/temperature) - equation 2

    Consider a reaction which is exothermic, and delta S system is positive. If temperature increases, will it make the reaction more or less spontaneous??

    For equation 1, since delta S is positive, increasing temperature will make Gibbs more negative, more spontaneous

    For equation 2, since delta H is negative, increasing temperature will make the whole thing less positive, less spontaneous...

    What's wrong with my reasoning? I can't figure this out!!
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    in ur explanation for equation 2 u said "less positive"

    so this is more negative, thus MORE spontaneous
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    Exactly.....if the Delta G is negative the reaction is spontaneous, thus it will occur, if its positive it will not occur.

    Also, at low temperature Delta G is approximately equal to Delta H.
    And at high temperature, Dealta G is approximately equal to T Delta S.
    Just makes the whole equation a little more feasible.....simplification...
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    (Original post by Ataxus)
    in ur explanation for equation 2 u said "less positive"

    so this is more negative, thus MORE spontaneous
    Less positive, I am referring to the S total not the G value

    If S is more positive, then it'll be more spontaneous won't it?
 
 
 

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