To calculate the AG for the reaction at 547 K for a mixture of 2.3 bar NH3(g), 3.0 bar O₂(g), 4.9 bar H₂O(g), and 3.6 bar NO(g), you can use the following equation:
AG = ∑nG°(products) - ∑nG°(reactants)
where ∑n is the sum of the stoichiometric coefficients of the reactants and products, and G° is the standard Gibbs free energy of the species.
The standard Gibbs free energy of a species can be calculated using the following equation:
G° = G°(298 K) + RTln[P/P°]
where G°(298 K) is the standard Gibbs free energy of the species at 298 K, R is the gas constant (8.314 J/mol*K), T is the temperature in kelvins, P is the pressure of the species, and P° is the standard pressure (1 bar).
Plugging in the values, we get:
AG = (6 H₂O + 4 NO - 4 NH₃ - 5 O₂)
= (6 H₂O - 4 NH₃ - 5 O₂ + 4 NO)
= (6 H₂O - 4 NH₃ + 4 NO - 5 O₂)
= (6 H₂O + 4 NO - 4 NH₃ - 5 O₂)
= 6 G°(H₂O) + 4 G°(NO) - 4 G°(NH₃) - 5 G°(O₂)
= 6(-242.2 kJ/mol) + 4(-90.3 kJ/mol) - 4(-45.9 kJ/mol) - 5(-497.1 kJ/mol)
= (-1453.2 kJ/mol)
Therefore, AG for the reaction at 547 K for a mixture of 2.3 bar NH3(g), 3.0 bar O₂(g), 4.9 bar H₂O(g), and 3.6 bar NO(g) is -1453.2 kJ/mol.
Since AG is negative, the reaction