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Q.

Bond dissociation enthalpy of  H2,Cl2  and HCl respectively are 104 k cal/mole, 58 kcal/mole and 103 kcal/mole. Then standard enthalpy of formation of HCl will be equal to

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a

+ 32 kcals

b

- 64 kcals

c

+ 54 kcals

d

- 22 kcals

answer is B.

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Detailed Solution

H2+Cl2⟶2HCl

ΔH=(+52+29)−(103)=−22kcals

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Bond dissociation enthalpy of  H2,Cl2  and HCl respectively are 104 k cal/mole, 58 kcal/mole and 103 kcal/mole. Then standard enthalpy of formation of HCl will be equal to