# Enthalpy of Formation

N2H4(g) + H2(g)--> 2 NH3(g) H1 = –1876kJ

3 H2(g) + N2(g)--> 2 NH3(g) H2 = –922 kJ

The H.f for the formation of hydrazine: 2 H2(g) + N2(g)--> N2H4(g) will be______kj/mol

I am very confused here. I thought I would just add H1 and H2 and divide by the molar mass of N2H4, but that is not correct. Please Help. Thank You!

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1. I got the answer to be 95.4 kj/mol
I reversed the 1st equation to get an overall 187.6 kj = h1
then I just added that to H2 and I got the answer.
But I still don't really understand the processes. If someone could explain that to me I would truly appreciate it.

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2. You have to reverse the direction of one reaction and change the sign of H.

3 H2+ N2--> 2 NH3 H2 = -922 kJ/mole
2NH3 -> N2H4 + H2 H = +1876 kJ/mole

NOW add, and cancel out terms that appear on both sides.

N2 + 2H2 -> N2H4 H = -954 kJ

There is another problem. Your heat of formation of NH3 does not agree with accepted data. It should be -10.97 kJ/per mole of NH3, or twice that much for the reaction as written (which forms two moles). Also, the standard form of N2H4 at room temperature is a liquid, and your reaction involves the gas.

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3. For what it is worth, the correct heat of formation of N2H4 (liq.) is +12.10 kcal/mole and for N2H4(g) it is +22.79 kcal/mole. Those values are at 298 K. They are a few kcal/mole different at 0 K. Multiply by 4.18 for kJ/mole. That gives +95.3 kJ/mole for the N2H4(g) heat of formation, which agrees well with your second answer. You seem to have misplaced decimal points in your first version of the question.

Those values come from the JANAF Thermochemcial Tables of the U.S National Bureau of Standards, 2nd edition.

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