A) \[-261\text{ }kJ\]
B) \[+103\text{ }kJ\]
C) \[+261\text{ }kJ\]
D) \[-103kJ\]
Correct Answer: D
Solution :
For 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 \[HBr,\text{ }\Delta H\]of above is calculated as Follows \[\Delta H=-[2\times \]bond energy of\[HBr-\](bond energy of Ha + bond energy of\[C{{l}_{2}})\]] \[\Delta H=[-2\times (364)-(433)+192]\text{ }kJ\] \[=-[728-(625)]\text{ }kJ=-103\text{ }kJ\]You need to login to perform this action.
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