Heat of Formation

The bond diss...

Question

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

- 44.0 k c a l

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- 22.0 k c a l

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22.0 k c a l

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44.0 k c a l

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Solution

The correct option is B $- 22.0 k c a l$
$\begin{matrix} H_{2} (g) \to 2 H (g); & △ H = 104 k c a l . . . . (1) C l_{2} (g) \to 2 C l (g) & △ H = 58 k c a l . . . . (2) H C l (g) \to H (g) + C l (g) & △ H = 103 k c a l . . . . (3) \end{matrix}$
Heat of formation for $H C l$
$\frac{1}{2} H_{2} (g) + \frac{1}{2} C l_{2} (g) \to H C l (g) △ H = ?$
Divide equations (1) and (2) by 2, and then add
$\frac{1}{2} H_{2} (g) + \frac{1}{2} C l_{2} (g) \to H (g) + C l (g); △ H = 81 k c a l . . . . (4)$

$H C l (g) \to H (g) + C l (g); △ H = 103 k c a l$
Subtracting equation (3) from equation (4) gives enthalpy of formation of HCl
$∴$ Enthalpy of formation of HCl gas = $- 22.0 k c a l$

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Heat of Formation

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