Chemical Thermodynamics Question 363
Question: The enthalpy of combustion of $ H_2(g), $ to give $ H_2(g) $ is $ -249kJmo{{l}^{-1}} $ and bond enthalpies of $ H-H $ and $ O=O $ are $ 433kJmo{{l}^{-1}} $ and $ 492kJmo{{l}^{-1}} $ respectively. The bond enthalpy of $ O-H $ is
Options:
A) $ 464kJmo{{l}^{-1}} $
B) $ -464kJmo{{l}^{-1}} $
C) $ 232kJmo{{l}^{-1}} $
D) $ -232kJmo{{l}^{-1}} $
Show Answer
Answer:
Correct Answer: A
Solution:
- $ H_2(g)+\frac{1}{2}O_2(g)\to H_2O(g)-h=-249 $
Let the bond enthalpy of $ O-H $ is x. Then $ \Delta H=\sum B.E. $ of reactant $ -\sum B.E. $ of product $ -249=433+\frac{1}{2}\times 492-2x $
$ \rightarrow $ $ x=464,kJ,mo{{l}^{-1}} $
BETA
