The bond dissociation enthalpies of H2(g),Cl2(g) and HCl(g) are 435, 243 and 431 kJ mol−1 respectively. The enthalpy of formation of HCl(g) will be

The bond dissociation enthalpies of H2(g),Cl2(g) and HCl(g) are 435, 243 and 431 kJ mol1 respectively. The enthalpy of formation of HCl(g) will be

  1. A

    247 kJ mol1

  2. B

    770 kJ mol1

  3. C

    1109 kJ mol1

  4. D

    92 kJ mol1

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    Solution:

    H2(g)+Cl2(g)2HCl(g)ΔfH(HCl,g)=12[ε(HH)+ε(ClCl)2ε(HCl)]=12(435+2432×431)kJ mol1=92 kJ mol1

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