Can Equilibrium Be Achieved at 1 atm for N_2, H_2, and NH_3?

[breez]

N_2 (g) + 3H2(g) <--> 2NH3(g)

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

At what temperature can N_2, H_2, and NH_3 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 \Delta G = \Delta H^{\circ} - T\Delta S^{\circ} with \Delta G = 0

Is this relationship true?

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

I thought \Delta G = \Delta G^{\circ} + RT \ln Q?

If reactants/products are all 1 atm, then ln Q = 0, and \Delta G^{\circ} must equal zero, which it clearly does not, thus there shouldn't exist a temperature where this is possible.

 

[gravenewworld]

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?