- #1
breez
- 65
- 0
[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.
[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.