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Question: The bond dissociation energies of gaseous \(H_{2}\), \(Cl_{2}\) and \(HCl\) are 104, 58 and 103 kcal...

The bond dissociation energies of gaseous H2H_{2}, Cl2Cl_{2} and HClHCl are 104, 58 and 103 kcal respectively. The enthalpy of formation of HCl gas would be

A

– 44 kcal

B

44 kcal

C

–22 kcal

D

22 kcal

Answer

–22 kcal

Explanation

Solution

12H2+12Cl2→HCl\frac{1}{2}H_{2} + \frac{1}{2}Cl_{2} \rightarrow HCl

ΔH=ΣB.E.(Reactants)−ΣB.E.(Products)\mathbf{\Delta}\mathbf{H =}\mathbf{\Sigma}\mathbf{B.E}\mathbf{.}_{\mathbf{(}\text{Reactants}\mathbf{)}}\mathbf{- \Sigma}\mathbf{B.E}\mathbf{.}_{\mathbf{(}\text{Products}\mathbf{)}}

ΔH=[12B.E.(H2)+12BE(Cl2)]−B.E.(HCl)\mathbf{\Delta}\mathbf{H =}\left\lbrack \frac{\mathbf{1}}{\mathbf{2}}\mathbf{B.E.(}\mathbf{H}_{\mathbf{2}}\mathbf{) +}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{BE(C}\mathbf{l}_{\mathbf{2}}\mathbf{)} \right\rbrack\mathbf{-}\mathbf{B.E.(HCl)}

= 12(104)+12(58)−103\frac{\mathbf{1}}{\mathbf{2}}\mathbf{(104) +}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{(58)}\mathbf{-}\mathbf{103}

= 81−103=−22kcal\mathbf{81}\mathbf{-}\mathbf{103 =}\mathbf{-}\mathbf{22kcal}

ΔH=−22kcal.\mathbf{\Delta}\mathbf{H =}\mathbf{-}\mathbf{22kcal.}