Chemical Thermodynamics Question 364
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 $
BETA
