question_answer

If the bond energies of \[H-H,\,\,Br-Br\] and\[H-Br\] are 433, 192 and 364 kJ \[mo{{l}^{-1}}\] respectively, then \[\Delta {{H}^{0}}\] for the reaction\[{{H}_{2}}(g)+B{{r}_{2}}(g)\xrightarrow[{}]{{}}2HBr(g)\] is:           [AIPMT (S) 2004]

A)  \[-261\text{ }kJ\]

B)                  \[+103\text{ }kJ\]        

C)  \[+261\text{ }kJ\]        

D)       \[-103\text{ }kJ\]

Correct Answer: D

Solution :

For reaction

On the basis of bond energies of, and ,  of above is calculated as follows:

bond energy of (bond energy

of bond energy of