The bond dissociation energies of gaseous H2, Cl2 and HCl are 104, 58 and 103 K.Cal / mole respectively. If magnitude of heat of formation of HCl = 11x, x =?

12H2(g)+12Cl(g)→HCl(g)ΔH=∑B⋅E of reactions −∑B.E of products=12(104)+12(58)−103=−22 11x=22x=2

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The bond dissociation energies of gaseous H_2,Cl₂ and HCl are 104, 58 and 103 K.Cal / mole respectively. If magnitude of heat of formation of HCl = 11x, x =?

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answer is 2.

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Detailed Solution

$\frac{1}{2} H_{2} (g) + \frac{1}{2} Cl (g) \to HCl (g) ΔH = \sum B \cdot E$ of reactions $- \sum B . E$ of products
$= \frac{1}{2} (104) + \frac{1}{2} (58) - 103 = - 22$

11x=22

x=2

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