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