A) \[-\text{261 kJ}\]
B) \[+\text{1}0\text{3 kJ}\]
C) \[+\text{261 kJ}\]
D) \[-\text{1}0\text{3 kJ}\]
Correct Answer: D
Solution :
For reaction \[{{H}_{2}}(g)+B{{r}_{2}}(g)\xrightarrow[{}]{{}}2Hbr(g)\Delta {{H}^{o}}=?\] On the basis of bond energies of\[{{H}_{2}}\], \[Br,\]and \[HBr\], \[\Delta H\] of above is calculated as follows: \[\Delta H=-[2\times \]bond energy of \[HBr-\](bond energy of\[{{\text{H}}_{\text{2}}}+\] bond energy of\[C{{l}_{2}})]\] \[\Delta H=-[2\times (364)-(433)+192]kJ\] \[=-[728-(625)]\,kJ\]\[=-103\,kJ\]You need to login to perform this action.
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