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Free energy of formation

  1. Apr 17, 2008 #1
    [SOLVED] free energy of formation

    1. The problem statement, all variables and given/known data

    Consider the reaction
    2NH3(g) --> N2(g) + 3H2(g)
    If the standard molar free energy of formation of NH3(g) at 298 K is –16.45 kJ·mol–1, calculate the equilibrium constant for this reaction at 298 K.
    The correct answer is 1.71E-6

    2. Relevant equations
    DG= - RTlnK


    3. The attempt at a solution
    so, if the equation is for two moles NH3(g) then one would multiply the given DG (-16.45) by 2. Also, I converted from KJ to J giving me -32900. The equation given is for the reverse process of formation, so, do I reverse the sign of DG? I tried it both ways and got the wrong answer. In fact, I get huge numbers compared to the correct answer.
    I used 8.3145 for R and 298 for T.
     
  2. jcsd
  3. Apr 17, 2008 #2
    Hey,

    Seems your right on track. Not sure what happend to your calculation if you tried reversing the free energy.

    dG= 2*-16450 j/mol
    = 32900 j/mol(reversed)

    32900 j/mol = -8.314*298K * ln K
    K=1.71e^-6
     
  4. Apr 17, 2008 #3
    haha I was pressing the anti log on my calculator by accident instead of e^x. Thank you for your help. You know youve been studying to long when...
     
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