Chemical Thermodynamics Question 482

Question: Use the bond enthalpies in the table to determine $ \Delta H{}^\circ $ for the formation of hydrazine, $ N_2H_4(g) $ from nitrogen and hydrogen in standard state according to the equation : $ N_2(g)+2H_2(g)\xrightarrow{{}}N_2H_4(g) $
Bond Bond enthalpies
$ N-N $ $ 159kJmo{{l}^{-1}} $
$ N=N $ $ 418kJmo{{l}^{-1}} $
$ N\equiv N $ $ 941kJmo{{l}^{-1}} $
$ H-H $ $ 436kJmo{{l}^{-1}} $
$ H-N $ $ 389kJmo{{l}^{-1}} $

Options:

A) $ \Delta H{}^\circ =-425kJ $

B) $ \Delta H{}^\circ =-98kJ $

C) $ \Delta H{}^\circ =+98kJ $

D) $ \Delta H{}^\circ =+711,kJ $

Show Answer

Answer:

Correct Answer: C

Solution:

  • $ \Delta H{{{}^\circ } _{rxn}}=\Delta H{{{}^\circ } _{f}}[N_2H_4]=[(941+2\times 436) $

$ -(159+4\times 389)] $

$ \Delta H{}^\circ =+98,kJ $



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