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Heat of Reaction

Bond dissocia...

Question

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

- 44 kcal

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- 88 kcal

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- 22 kcal

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- 11 kcal

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Solution

The correct option is A - 44 kcal
$H_{2 (g)} + C l_{2 (g)} \to 2 H C l_{(g)}$
$\sum (B E o f r e a c t a n t s) - \sum (B E o f p r o d u c t s) = Δ H_{r}$
$= 104 + 58 - 2 \times 103$

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Similar questions

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

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Bond dissociation enthalpies of H2(g),Cl2 and HCl(g) are 104, 58 and 103 kcal respectively. The enthalpy of formation of HCl is

The correct option is A - 44 kcalH2(g)+Cl2(g)→2HCl(g) ∑(BE of reactants)−∑(BE of products)=ΔHr =104+58−2×103

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

The correct option is A - 44 kcal
$H_{2 (g)} + C l_{2 (g)} \to 2 H C l_{(g)}$
$\sum (B E o f r e a c t a n t s) - \sum (B E o f p r o d u c t s) = Δ H_{r}$
$= 104 + 58 - 2 \times 103$