Answer Enthalpy Change Atomisation Hydrazine: B 1720 kJ

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SUMMARY

The enthalpy change of atomization for 1 mol of gaseous hydrazine (N2H4) is definitively 1720 kJ, as indicated in option B. The reaction for the formation of hydrazine from its elements is represented as N2(g) + 2H2(g) → N2H4(g). The bond enthalpies provided include N≡N at 994 kJ/mol, H-H at 436 kJ/mol, N-N at 160 kJ/mol, and N-H at 390 kJ/mol. The calculation method involves the difference between the total bond breaking energy and the total bond making energy, leading to the conclusion that 1720 kJ is the correct answer.

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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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