Question

The bond dissociation energies of gaseous `H_(2),C1_(2)`, and `HC1` are `100, 50`, and `100 kcal mol^(-1)`, respectively. Calculate the enthalpy of formation of `HC1(g)`.

Answer 1

The required equation is
`(1)/(2)H_(2)(g) +(1)/(2)C1_(2)(g) rarr HC1(g), DeltaH^(Theta) = ?`
`DeltaH^(Theta) =[(1)/(2)DeltaH_(H-H) +(1)/(2)DeltaH_(C1-C1)] =[DeltaH_(H-CI)^(Theta)]`
`= (1)/(2) xx 100 +(1)/(2) xx 50 - 100`
`=- 25 kcal mol^(-1)`