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Homework Help: Thermo - Gibbs Free Energy & Entropy

  1. Apr 1, 2007 #1
    1. The problem statement, all variables and given/known data
    Consider fuel cell using methane as fuel. Reaction is

    CH^4 + 2O_2 -> 2H2O+CO_2

    Assume room temperature and atmospheric temperature
    Determine values of delta H (Helmholtz) and delta G (Gibbs) for this reaction for one mole of methane.

    Question instructed the use of the web to find thermodynamic tables with values of H and G for the chemicals in the reaction

    2. Relevant equations
    I haven't encountered this sort of question where substitution of "real" values is necessary. Hence I've used this site as a reference:

    http://members.aol.com/profchm/gibbs.html [Broken]

    I think I found H alright (-802.3kJ), but to find G I need entropy (delta S)

    3. The attempt at a solution

    dG = dH - T.dS

    dS = Sum of products (RHS) - Sum of reactants (LHS)
    = [2(188.7)+213.7]-[186.3 + 2(205)]
    = -5.2 (but isn't an entropy of less than zero impossible?)

    The problem lies in the uncertainty of me obtaining an negative delta S :confused:

    Thanks in advance for any hints/tips
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Apr 1, 2007 #2
    Actually, that formula uses [itex] \ \Delta G \ = \Delta H \ - \ T\Delta S_{internal}[/itex]

    and, [itex]\Delta S_{internal}+\Delta S_{surrounding}=\Delta S_{total} \geq 0[/itex]
    See here
    Last edited: Apr 1, 2007
  4. Apr 1, 2007 #3
    Thanks for your reply.

    So in this case the answer I obtained is correct?

    dS_total is > 0, but the entropy of the 'external' system (the "universe"?) balances the negative entropy of the internal system (the reaction and its components in the engine)

    ^Is this line of thinking correct^

    If it is, then:

    dG = -802.3 - (300K * (-5.2/1000))
    = -800.74 kJ

    ^Answer obtained^
    Last edited: Apr 1, 2007
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