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If the bond energies of H – H Br – Br and H – Br are 433, 192 and $364 KJ {mol}^{- 1}$ respectively. Then ${ΔH}^{0}$ for the reaction $H_{2_{(g)}} + B r_{2_{(g)}} \overset{}{\to} 2 {HBr}_{(g)}$ is :

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– 262 KJ

+ 103 KJ

+ 261 KJ

– 103 KJ

answer is D.

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Detailed Solution

One mole of $H_{2}$ and one mole of ${Br}_{2}$ reacts to give 2 moles HBr.

${ΔH}^{0} {= ΔH}_{H-H} {+ ΔH}_{Br-Br} {-ΔH}_{H-Br}$

${ΔH}^{0} = 433 + 192 - 2 (364) = 625-728=103$

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