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Q.

 If the bond energies of H–H, Br-Br and H-Br are 433, 192 and 364 kJ mol–1 respectively, then ΔH0 for the reaction :
H2(g) + Br2(g) → 2HBr(g) is

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a

+ 261 kJ

b

– 261 kJ

c

+ 103 kJ

d

– 103 kJ

answer is D.

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Detailed Solution

\begin{array}{l} \Delta H = \sum B{E_{reac\tan ts}} - \sum B{E_{products}}\\ \,\,\,\,\,\,\,\,\, = B{E_{H - H}} + B{E_{Br - Br}} - 2B{E_{H - Br}}\\ \,\,\,\,\,\,\,\,\, = 433 + 192 - 2(364)\\ \,\,\,\,\,\,\,\,\,\, = - 103KJ \end{array}

 

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