Gibbs Free Energy help

  1. 1. The problem statement, all variables and given/known data
    The reaction is [tex]NH_{4}Cl(s)\rightarrow NH_{3}(g)+HCl(g)[/tex]
    [tex]\Delta H^{o}=+176 kJ and \Delta G^{o}=+91.2 kJ[/tex] at 298 K
    What is the value of [tex]\Delta G[/tex] at 1000 K?


    2. Relevant equations
    [tex]\Delta G=\Delta H-T\Delta S[/tex]
    The same applies if all 'deltas' are standard


    3. The attempt at a solution
    Well, I solved for standard change of entropy ([tex]\Delta S^{o}[/tex]) and came up with .284 kJ/K, which is the same when using a table of standard entropies. My problem is, I'm not sure where to go from this to find [tex]\Delta G[/tex], or even a way to link standard values to normal values for these. I tried plugging in the values for standard delta H and delta S with 1000K to find delta G, but something tells me that this is incorrect. Any help would be greatly appreciated!

    -Swerting
     
  2. jcsd
  3. Mapes

    Mapes 2,532
    Science Advisor
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    Why do you think your first approach is incorrect?
     
  4. I believe that my first approach is incorrect since standard values are at 298.15K, and this is asking for delta G at 1000K, implying that delta H and delta S have changed as well.
     
  5. Mapes

    Mapes 2,532
    Science Advisor
    Homework Helper
    Gold Member

    Do you know how to calculate changes in [itex]\Delta H[/itex] and [itex]\Delta S[/itex] with temperature? (Hint: it involves the heat capacity.) Alternatively, you could look up these values at 1000 K.
     
  6. I can calculate [tex]\Delta H[/tex] from changes in temperature, but unfortuneately, not [tex]\Delta S[/tex], nor could I find a table of entropies at 1000K.
     
  7. Alright, well, in the depths of the internet I finally found the answer explained, and apparently my using standard delta H and S were correct. The answers from the company that make the questions say that delta S doesn't change (thought says nothing about delta H) and so I just plug in the values and solve for delta G. Ah well, who would've guessed that values so dependant on temperature don't really change over an actual temperature change. :P
    Thank you very much for your help!

    -Swerting
     
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