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 $kJ{mol}^{- 1}$ ) $of HBr ?$

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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 $kJ{mol}^{- 1}$ ) $of HBr ?$

Quantrex Academy · Accepted Answer

Correct answer: A. 368

= 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) = 368KJ / mol

At standard conditions, if the change in the enthalpy for the...

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