Question

# The bond dissociation energy of gaseous H2,Cl2 and HCl are 104, 58 and 103 kcal mol−1 respectively. The enthalpy of formation for HCl gas will be

A
44.0 kcal
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B
22.0 kcal
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C
22.0 kcal
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D
44.0 kcal
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Solution

## The correct option is B −22.0 kcalH2 (g)→2H (g);△H=104 kcal ....(1)Cl2 (g)→2Cl (g)△H=58 kcal ....(2)HCl (g)→H (g)+Cl (g)△H=103 kcal ....(3) Heat of formation for HCl 12H2 (g)+12Cl2 (g)→HCl (g) △H=? Divide equations (1) and (2) by 2, and then add 12H2(g)+12Cl2(g)→H(g)+Cl(g); △H=81 kcal....(4) HCl (g)→H (g)+Cl (g); △H=103 kcal Subtracting equation (3) from equation (4) gives enthalpy of formation of HCl ∴ Enthalpy of formation of HCl gas = −22.0 kcal

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