This is why the reaction is spontaneous. Spontaneity is determined by the sign of the Gibbs free energy, ΔG, not the entropy change, ΔS. ΔG is determined from both the enthalpy change, ΔH, and the entropy change, ΔS.
......... ΔG = ΔH - TΔS
ΔH = ΣΔHf(products) - ΣΔHf(reactants)
ΔH = 4ΔHf(H2O(g)) - ΔHf(N2O4)
ΔH = 4(-241.8 kJ) - (+9.3kJ) = -976.5 kJ
Suppose the reaction occurs at 400K....
ΔG = ΔH - TΔS
ΔG = -976.5 kJ - 400K(0.120 kJ/K) = -1024.5 kJ
As you can see, a negative ΔH and a positive ΔS always produce a negative ΔG, and a spontaneous reaction at all temperatures.
Perhaps you are asking why ΔS is positive even though the moles of reactants and products, and the phases are the same. ΔS is determined from the entropies of the reactants and products.
Looking at the number of moles of reactants and products and their phases often gives a good approximation of the change in entropy of a reaction, but it is not foolproof.
Suppose the product was liquid water and not water vapor. Then the calculations would look like this, and the values of ΔS would be a negative value.
Answers & Comments
Thermodynamics ......
This is why the reaction is spontaneous. Spontaneity is determined by the sign of the Gibbs free energy, ΔG, not the entropy change, ΔS. ΔG is determined from both the enthalpy change, ΔH, and the entropy change, ΔS.
......... ΔG = ΔH - TΔS
ΔH = ΣΔHf(products) - ΣΔHf(reactants)
ΔH = 4ΔHf(H2O(g)) - ΔHf(N2O4)
ΔH = 4(-241.8 kJ) - (+9.3kJ) = -976.5 kJ
Suppose the reaction occurs at 400K....
ΔG = ΔH - TΔS
ΔG = -976.5 kJ - 400K(0.120 kJ/K) = -1024.5 kJ
As you can see, a negative ΔH and a positive ΔS always produce a negative ΔG, and a spontaneous reaction at all temperatures.
Perhaps you are asking why ΔS is positive even though the moles of reactants and products, and the phases are the same. ΔS is determined from the entropies of the reactants and products.
ΔS = ΣΔS(products) - ΣΔS(reactants)
ΔS = (4ΔS(H2O(g)) + ΔS(N2)) - (ΔS(N2O4) + 4(ΔS(H2)))
ΔS = (4(188.7J/K) + 191.5 J/K) - (304.2 J/K + 4(130.6 J/K)
ΔS = 119.7 J/K ≈ 120 J/K
Looking at the number of moles of reactants and products and their phases often gives a good approximation of the change in entropy of a reaction, but it is not foolproof.
Suppose the product was liquid water and not water vapor. Then the calculations would look like this, and the values of ΔS would be a negative value.
ΔS = ΣΔS(products) - ΣΔS(reactants)
ΔS = (4ΔS(H2O(l)) + ΔS(N2)) - (ΔS(N2O4) + 4(ΔS(H2)))
ΔS = (4(69.9 J/K) + 191.5 J/K) - (304.2 J/K + 4(130.6 J/K)
ΔS = -355.5 J/K
Indeed, if the phase of water was liquid and not gas, then there would have been a decrease in entropy. But there is simply no escaping the numbers.