For this particular reaction, #DeltaG_"rxn" = "+9.4 kJ/mol"#.
#CO_((g)) + 2H_(2(g)) rightleftharpoons CH_3OH_((g))#, #K_p = 2.26 * 10^(4)#
The equation you're going to use is
#DeltaG_"rxn" = DeltaG^0 + RT * ln(Q)#, where
#DeltaG^0# - the standard free energy change;
#R# - the gas constant - #"8.314 J mol"^(-1)"K"^(-1)#
#T# - the temperature at which the reaction takes place - expressed in Kelvin;
#Q# - the reaction quotient - expresses the relative amounts of rectans and products present at a particular point in a reaction.
For the given pressures, you need to see if the reaction is at equilibrium or not. To do this, calculate #DeltaG^0# and #Q# before solving for #DeltaG_"rxn"#.
So, at equilibrium, #DeltaG_"rxn" = 0#, which implies that
#DeltaG^0 = -RT * ln(K_p)#
#DeltaG^0 = -8.314"J"/("mol" * cancel("K")) * (273.15 + 25)cancel("K") * ln(2.26 * 10^(4))#
#DeltaG^0 = -24852.6"J/mol" = -"24.85 kJ/mol"#
Now calculate #Q#
#Q = P_(CH_3OH)/(P_(CO) * P_(H_2)^(2)) = 1.0/(1.0 * 10^(-2) * (1.0 * 10^(-2))^(2)) = 10^(6)#
Because #Q>K_p#, the reaction will favor the reactants, i.e. you have more product present at these conditions than you'd have at equilibrium.
As a result, the forward reaction will no longer be favored, and you can expect #DeltaG_"rxn"# to be positive #-># the reverse reaction will become spontaneous.
Plug these values into the main equation and solve for #DeltaG_"rxn"#
#DeltaG_"rxn" = -"24.85 kJ/mol" + 8.314"J"/("mol" * cancel("K")) * (273.15 + 25)cancel("K") * ln(10^(6))#
#DeltaG_"rxn" = -"24.85 kJ/mol" + "34.25 kJ/mol"#
#DeltaG_"rxn" = color(green)(+"9.4 kJ/mol")#
