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Gibbs free energy from partial pressures

  1. Apr 13, 2012 #1
    1. The problem statement, all variables and given/known data
    Consider the following reaction:
    CH3OH(g) <-> CO(g)+2H2(g)

    Calculate ΔG for this reaction at 298 K under the following conditions:
    PCH3OH=0.895atm
    PCO=0.115atm
    PH2=0.200atm


    2. Relevant equations
    ΔG=-R*T*ln(K)
    where R is the gas constant 8.314 J/molK, T is 298 in this case, and K is determined from the partial pressures.


    3. The attempt at a solution
    I calculated K by (0.2002)(0.115)/0.895 and found it to be 0.00514.
    This produced deltaG=-8.314*298*ln(0.00514)=-13.1 kJ, but this is incorrect according to the website. I'm really not sure what is going wrong; this is my last attempt on any of the pressure problems on this homework, so I'd like to get this figured out. My guess right now is that it's related to the # of moles not being used (as R suggests), but I'd appreciate input before I use my last attempt, especially so it's drilled into my brain after this assignment.

    Thanks for any help!
     
  2. jcsd
  3. Nov 12, 2017 #2
    The consistency in units is often the case

    Since the problem doesn't give Delta G naught, we calculate Delta G reaction assuming the reaction is at equilibrium (ΔG = 0)
    ΔG° reaction (kJ) = - R (J/°K.mol) x T(°K) x ln (Pp/Pr)​

    In this case, convert R = 8.314x10-3 (kJ/°K.mol)
     
  4. Nov 13, 2017 #3
    You neglected the minus sign. (Clue: if K < 1, ΔG0 must be positive.) Is +13.1 kJ/mol correct, according to your website?
     
  5. Nov 16, 2017 #4

    DrDu

    User Avatar
    Science Advisor

    The use of °K instead of K was discouraged in 1967.
     
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