What is the correct ionic compound formula for NH4 and N?

Ithink iit might be

Given data,temperature, t = 230°c = 503 kconsider the chemical reaction,ch₃oh(l) + 1/2 o₂(g)  → ch₂o(g) + h₂o(l)calculate the enthalpy change for the reaction: δh⁰rxn =  σnδh⁰f(products) -  σmδh⁰f(reactants)substitute the values from the table: δh⁰rxn = { -115.9 + (-285.8)} - {0+(-238.7)}              = -163.0 kjcalculate the entropy change for the reaction.δs⁰rxn = (219 + 70) - (127.2 + 1/2 x 205.2) = 59.2 j/kconsider the expression for free energy: δg⁰rxn =  δh⁰rxn - t  δs⁰rxnsubstitute the values in the above equationδg⁰rxn = -163 x 10³ - (503 x 59.2)              = -192777.6 j              = -192.777 kj

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2poclâ‚(g) ⇜ 2 pclâ‚(g) + oâ‚‚(g) â†gâ° = (2 mol)î”gfâ°(pclâ‚) – (2 mol)î”gfâ°(poclâ‚) = (2 mol)(–270 –(–502) kj/mol) â†gâ° = + 464 kj = â†hⰠ– tî”sâ° î”sâ° = 179 j/k ⇒ â†hâ° = â†gâ° + (298 k)(179 j/k) = –411 kj assuming that î”hâ° and î”sâ° do not depend on temperature, the temperature at which â†gâ° = 0 is where t = â†hâ°/î”sâ° = (–411 kj)/(0.179 j/k) = 2294 k. at that temp. keq = 1 for the reaction.