For the following reaction run at 800 K, a mixture at equlibrium contains #P_(N_2)=0.0040# atm, #P_(H_2)=0.063# atm, and #P_(NH_3)=0.48# atm. Calculate #\DeltaG^o# at 800 K?

Based on the determined equilibrium constant, #DeltaG^@ = -"82.13 kJ/mol"#.

You don't need to calculate #Q#, because you're at equilibrium already and can calculate #K_P# right away (in implied units of #"atm"#).

#K_P = (P_(NH_3)^2)/(P_(N_2)P_(H_2)^3)#

#= ("0.48 atm"//"1 atm")^2/(("0.0040 atm"//"1 atm")("0.063 atm"//"1 atm")^3)#

#= 2.304 xx 10^5#

(FYI, this equilibrium constant makes no physical sense, because this reaction requires excess reactants in the industry to push it forward and it becomes LESS favorable at higher temperatures. #K_P = 5.38 xx 10^(-6)# at #"823.15 K"#. Thus, #K_P ~~ 10^(-5)# at #"800 K"#.)

For now we'll go with this one... In general,

#DeltaG = DeltaG^@ + RTlnQ#

where #Q# is the reaction quotient, i.e. the NOT-YET-equilibrium constant. #DeltaG# is the change in Gibbs' free energy, and #@# indicates standard conditions (#"1 M"# concentrations, #"1 atm"# surroundings, at a certain temperature).

At equilibrium, recall that #DeltaG = 0# AND #Q = K#. Thus,

#DeltaG^@ = -RTlnK#

For gas-phase reactions, we define #K -= K_P# at standard conditions, so

#DeltaG^@ ("gas-phase") = -RTlnK_P#

Therefore,

#color(red)(DeltaG^@) = -("0.008314 kJ/mol"cdot"K")("800 K")ln(2.304 xx 10^5)#

#= color(red)(-"82.13 kJ/mol")#

In reality, for this reaction at #"800 K"#, #DeltaG^@ ~~ "76.57 kJ/mol"#.