Chemical Thermodynamics Question 214
Question: Given that
(i) $ {\Delta_{f}}H{}^\circ $ of $ N_2O $ is $ 82kJmo{{l}^{-1}} $
(ii) Bond energies of $ N\equiv N,N=N,O=O $ and $ N=O $ are 946, 418, 498 and $ 607kJmo{{l}^{-1}} $ respectively, The resonance energy of $ N_2O $ is:
Options:
A) $ -88kJ $
B) $ -66kJ $
C) $ -62kJ $
D) $ ~-,44,kJ $
Show Answer
Answer:
Correct Answer: A
Solution:
- $ N_2(g)+\frac{1}{2}O_2\to N_2O(g) $
$ N\equiv N( g )+\frac{1}{2}( O=O )\to \underset{\scriptscriptstyle\centerdot\centerdot}{\ddot{N}}=\overset{+}{\mathop{ }},=\ddot{\underset{\scriptscriptstyle\centerdot\centerdot}{O}}( g ) $
$ \Delta H_{f}^{{}^\circ }= $ [Energy required for breaking of bonds]-[Energy released for forming of bonds] $ =(\Delta {H_{N\equiv N}}+\frac{1}{2}\Delta {H_{O=O}}-(\Delta {H_{N=N}}+\Delta {H_{N=O}}) $
$ =( 946+\frac{1}{2}\times 498 )-( 418+607 ) $
$ =170kJ,mo{{l}^{-1}} $ Resonance energy $ =82-170 $
$ =-88kJ,mo{{l}^{-1}} $
BETA
