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")
