if I have the value of ΔG°f and ΔS° how do I find ΔG° (kJ) and ΔH° (kJ) of a reaction?

Question

This is the information that I have if anyone needs it:
2NO(g) + O2(g) ----> 2NO2(g)

NO(g) ΔG°f=86.7 (kJ/mol) ΔS°=211 (J/mol)
NO2(g) ΔG°f=51.8 (kJ/mol) ΔS°=240 (J/mol)
NOCl(g) ΔG°f=66.3 (kJ/mol) ΔS°=264 (J/mol)
N2O(g) ΔG°f=103.6 (kJ/mol) ΔS°=220 (J/mol)

All taking place at 25°C

Answer 1

Somewhere in this course, you learned that DG for a reaction (sorry, i can't type the deltas) can be calculated from the sum of the DGf of the products - the sum of DGf of the reactants. So, for this rxn, DG° will be 2(51.8 kJ) -[2(86.7 kJ) + 0]. You can calculate DS° in the same manner. You also know that DG°= DH°-TDS°. You can use this equation to get to DH°.

These equations and probably examples should be in your text.

Answer 2

that cant be it because at the very top the left side is not equal to the right