Thermodynamics Question 95
Question: Given the bond energies $ N\equiv N,H-H $ and $ N-H $ bonds are $ 945,436 $ and $ 391kJmol{{e}^{-1}} $ respectively, the enthalpy of the following reaction $ N _{2}(g)+3H _{2}(g)\to 2NH _{3}(g) $ is
[EAMCET 1992; JIPMER 1997]
Options:
A) $ -93kJ $
B) $ 102kJ $
C) $ 90kJ $
D) $ 105kJ $
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Answer:
Correct Answer: A
Solution:
$ \underset{\begin{matrix} {} \ \text{Energy absorbed} \ \end{matrix}}{\mathop{\underset{945+3\times 436}{\mathop{N\equiv N+3H-H}}}}\xrightarrow{{}}\underset{\begin{matrix} {} \ \text{Energy released} \ \end{matrix}}{\mathop{\underset{2\times (3\times 391)=2346}{\mathop{\overset{H}{\mathop{\overset{|}{\mathop{\underset{H}{\mathop{\underset{|}{\mathop{2N}}}}-}}}}H}}}} $ Net. energy released = 2346 - 2253 = 93 kJ i.e. ΔH = - 93 kJ .
BETA
