The Bond dissociation Energies of $H_{2}$ , $C l_{2}$ , and $H C l$ are $104$ , $58$ and $103$ $K c a l$ respectively. The Enthalpy of formation of $H C l$ of $H C l$ gas will be

Q. Bond dissociation enthalpies of

H_{2 (g)}, C l_{2} a n d H C l_{(g)}

are 104, 58 and 103 kcal respectively. The enthalpy of formation of HCl is

Q. The bond dissociation energy of gaseous

H_{2}, C l_{2}

and HCl are 104, 58 and 103

k c a l m o l^{- 1}

respectively. The enthalpy of formation for

H C l

gas will be

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Bond dissociation energies of H2,Cl2 and HCI(g) are 104, 58 and 103 kcal mol−1 respectively, Calculate the enthalpy of formation of HCI gas.

The correct option is A -22 kcalBond enthalpies are given;12H2+12Cl2→HClΔfH(HCl)=ΔH(Reactant)−ΔH(Product) =(12(104)+12(58))−103 =−22 kcal/mol

Bond dissociation energies of $H_{2}, C l_{2}$ and $H C I^{(g)}$ are 104, 58 and 103 kcal $m o l^{- 1}$ respectively, Calculate the enthalpy of formation of $H C I$ gas.

The correct option is A -22 kcal
Bond enthalpies are given;
$\frac{1}{2} H_{2} + \frac{1}{2} C l_{2} \to H C l$
$Δ_{f} H_{(H C l)} = Δ H (R e a c t a n t) - Δ H (P r o d u c t)$
$= (\frac{1}{2} (104) + \frac{1}{2} (58)) - 103$
$= - 22 k c a l / m o l$