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Question

Bond dissociation energies of $$H_{2}, Cl_{2}$$ and $$HCl_{(g)}$$ are $$104, 58$$ and $$103$$ kcal mol$$^{-1}$$ respectively. Calculate the enthalpy of formation of HCl gas.

A
$$- 184 kcal$$
B
$$- 22 kcal$$
C
$$+ 22 kcal$$
D
$$+ 184 kcal$$
Solution
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Correct option is A. $$- 22 kcal$$
$$ \dfrac{1}{2} H_{2} + \dfrac{1}{2}Cl_{2} \rightarrow HCl $$

$$\Delta H =\dfrac12 B.E._{(H-H)} + \dfrac12 B.E._{(Cl-Cl)} -B.E._{(HCl)}$$

$$ \Delta H = \dfrac{1}{2} \times 104 + \dfrac{1}{2} \times 58 - 103 = - 22 \ kcal $$

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