Bond | Bond enthalpies |
\[N-N\] | \[159\text{ }kJ\text{ }mo{{l}^{-1}}\] |
\[N=N\] | \[418\text{ }kJ\text{ }mo{{l}^{-1}}\] |
\[N\equiv N\] | \[941\text{ }kJ\text{ }mo{{l}^{-1}}\] |
\[H-H\] | \[436\text{ }kJ\text{ }mo{{l}^{-1}}\] |
\[H-N\] | \[389\text{ }kJ\text{ }mo{{l}^{-1}}\] |
A) \[\Delta H{}^\circ =-425\text{ }kJ\]
B) \[\Delta H{}^\circ =-\text{ }98\text{ }kJ\]
C) \[\Delta H{}^\circ =+\text{ }98\text{ }kJ\]
D) \[\Delta H{}^\circ =+\text{ }711\,kJ\]
Correct Answer: C
Solution :
[c] \[\Delta H{{{}^\circ }_{rxn}}\,\,=\Delta H{{{}^\circ }_{f}}\,\,[{{N}_{2}}{{H}_{4}}]=[(941+2\times 436)\]\[-(159+4\times 389)]\] \[\Delta H{}^\circ =+98\,kJ\]You need to login to perform this action.
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