Answer Enthalpy Change Atomisation Hydrazine: B 1720 kJ

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The discussion centers on calculating the enthalpy change of atomization for gaseous hydrazine (N2H4). The options provided for the enthalpy change are 550 kJ, 1720 kJ, 1970 kJ, and 2554 kJ. The user attempts to derive the enthalpy change using bond enthalpies, noting the bond energies for N2, H2, N-N, and N-H. Despite calculating a value of 146 J/mol, this result does not match any of the given options, leading to confusion about the concept of atomization. Clarification on the definition and calculation method for atomization is sought.
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Homework Statement



Hydrazine was used as a fuel for the Messerschmidt 163 rocket fighter in World War II and for the
American Gemini and Apollo spacecraft .

What is the enthalpy change of atomisation of 1 mol of gaseous hydrazine?
A 550 kJ
B 1720 kJ
C 1970 kJ
D 2554 kJ
Use of the Data Booklet is relevant to this question


Homework Equations





The Attempt at a Solution



Is my equation correct?
N2(g) + 2H2(g) → N2H4(g)

The bond enthalpy for the N2(triple bond) in the data book let is 994. H2 is 436. N-N is 160 and N-H is 390 (if you need other bond enthalpies from the data booklet Il be happy to provide them)
I know that one of the ways to find the enthalpy change is Ʃ(bond breaking energy) - Ʃ(bond making energy). So i got 1866-1720= 146J/mol. Which is NOT IN THE OPTION!
 
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