The bond dissociation energy of gaseous H2,Cl2 and HCl are 104, 58 and 103 kcalmol−1 respectively. The enthalpy of formation for HCl gas will be
A
−44.0kcal
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B
−22.0kcal
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C
22.0kcal
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D
44.0kcal
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Solution
The correct option is B−22.0kcal H2(g)→2H(g);△H=104kcal....(1)Cl2(g)→2Cl(g)△H=58kcal....(2)HCl(g)→H(g)+Cl(g)△H=103kcal....(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=81kcal....(4)
HCl(g)→H(g)+Cl(g);△H=103kcal Subtracting equation (3) from equation (4) gives enthalpy of formation of HCl ∴ Enthalpy of formation of HCl gas = −22.0kcal