If the bond energies of H-H- Br-Br and H-Br are 433- 192 and 364 kJ mol -1 respectively- Then - Ho for the reaction H2g-Br2g- 2HBrg is -

The correct option is D 103 kJ Δ H^o=∑ bond energy (reactants) ∑ bond energy (products) =433+192 2×364= 103 KJ ...

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Bond Order and Enthalpy

If the bond e...

Question

If the bond energies of $H - H$ , $B r - B r$ and $H - B r$ are $433, 192$ and $364$ kJ mol $^{- 1}$ respectively. Then $Δ H^{o}$ for the reaction $H_{2 (g)} + B r_{2 (g)} \to 2 H B r_{(g)}$ is :

- 261

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+ 103

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+ 261

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- 103

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Solution

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Similar questions

H - H

B r - B r

H - B r

433

192

364 k J m o l^{- 1}

Δ H^{\circ}

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

m o l^{- 1}

Δ H^{\circ}

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

{m o l}^{- 1}

△ H^{0}

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

H - H

B r - B r

H - B r

433, 192

364 {kJ mol}^{- 1}

Δ H^{o}

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

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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