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Heat of Reaction

If the bond e...

Question

If the bond energies of H – H, Br – Br and H – Br are 433, 192 and 364 kJ $m o l^{- 1}$ respectively, the $Δ H^{\circ}$ for the reaction ; $H_{2} (g) + B r_{2} (g) \to 2 H B r (g)$ is

– 261 kJ

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+ 103 kJ

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+ 261 kJ

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– 103 kJ

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Solution

The correct option is D – 103 kJ
$Δ H^{\circ} = \sum B . E (r e a c t a n t s) - \sum B . E (p r o d u c t s) = [B . E (H - H) + B . E (B r - B r)] - [2 B . E (H - B r)] = (433 + 192) - (2 \times 364) = 625 - 728 = - 103 k J$

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H - H, B r - B r

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H_{2} (g) + B r_{2} (g) \to 2 H B r (g)

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Δ H^{\circ}

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Use bond energies to estimate the enthalpy of formation of HBr(g).
BE(H-H) = 436 kJ/mol
BE(Br-Br) = 192 kJ/mol
BE(H-Br) = 366 kJ/mol

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