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Temperature of Equilibrium

  1. Apr 11, 2007 #1
    [tex]N_2 (g) + 3H2(g)[/tex] <--> [tex]2NH3(g)[/tex]

    [tex]\Delta H^{\circ} of NH3 = -46.2 kJ/mol[/tex]
    [tex]\Delta G^{\circ} of NH3 = -16.7 kJ/mol[/tex]

    At what temperature can [tex]N_2, H_2, and NH_3[/tex] gases by maintained at equilibrium each with a partial pressure of 1 atm?

    The solution my book uses is to solve for T in the equation [tex]\Delta G = \Delta H^{\circ} - T\Delta S^{\circ}[/tex] with [tex]\Delta G = 0[/tex]

    Is this relationship true?

    Also, how can you be sure that at that temperature, the pressures will all be 1 atm?

    I thought [tex]\Delta G = \Delta G^{\circ} + RT \ln Q[/tex]?

    If reactants/products are all 1 atm, then ln Q = 0, and [tex]\Delta G^{\circ}[/tex] must equal zero, which it clearly does not, thus there shouldn't exist a temperature where this is possible.
     
  2. jcsd
  3. Apr 11, 2007 #2
    Careful, at equilibrium delta G is zero, not delta G naught

    At equilibrium


    delta G naught=-RT lnKeq

    you are given delta G naught, you know R, what is Keq expression in terms of pressure and temperature?
     
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