At standard conditions, if the change in the enthalpy for the following reaction is $- 109 {kJmol}^{- 1}$
$H_{2 (g)} + {Br}_{2 (g)} \to 2 {HBr}_{(g)}$
Given that bond energy of $H_{2}$ and ${Br}_{2}$ is $435 {kJmol}^{- 1}$ and $192 {kJmol}^{- 1}$ , respectively, what is the bond energy (in $kJ {mol}^{- 1}$ ) of $HBr ?$

Question

At standard conditions, if the change in the enthalpy for the following reaction is $- 109 {kJmol}^{- 1}$

$H_{2 (g)} + {Br}_{2 (g)} \to 2 {HBr}_{(g)}$

Given that bond energy of $H_{2}$ and ${Br}_{2}$ is $435 {kJmol}^{- 1}$ and $192 {kJmol}^{- 1}$ , respectively, what is the bond energy (in $kJ {mol}^{- 1}$ ) of $HBr ?$

Aicharya · Accepted Answer

Δ H = Σ ( B · E ) Reactants - Σ ( B . E ) Products - 109 = B · E ( H - H ) + B · E ( Br - Br ) - 2 × B · E ( H - Br ) - 109 = 435 + 192 - 2 × B · E ( H - Br ) B . E ( H - Br ) = 435 + 192 + 109 2 = 368 KJ / mol

At standard conditions, if the change in the enthalpy for the following reaction is - 109 kJmol - 1 H 2 ( g…

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