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Bond Order and Enthalpy

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 $△ H^{0}$ 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
$H - H$ bond energy- $433 K J {m o l}^{- 1} = △ H$
$B r - B r$ bond energy- $192 K J {m o l}^{- 1} = △ H$
$H - B r$ bond energy- $364 K J {m o l}^{- 1} = △ H$
$H_{2_{(9)}} + {B r}_{2_{(9)}} ⟶ 2 H {B r}_{(9)}$
${△ H}^{0} = - 2 (364) + 433 + 192$
$= - 103 K J$

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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)

H - H

B r - B r

H - B r

433, 192

364

^{- 1}

Δ H^{o}

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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