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?