1.

Given the bond energies of `H - H` and `Cl - Cl` are `430 kJ mol^(-1)` and `240 kJ mol^(-1)`, respectively, and `Delta_(f) H^(@)` for `HCl` is `- 90 kJ mol^(-1)`. Bond enthalpy of `HCl` isA. `245 kJ mol^(-1)`B. `290 kJ mol^(-1)`C. `380 kJ mol^(-1)`D. `425 kJ mol^(-1)`

Answer» Correct Answer - D
The enthalpy of formation of `HCl` is the enthalpy change when 1 mol of `HCl`is the enthalpy from its elements in their stantard states:
`(1)/(2) H_(2) (g) + (1)/(2) Cl_(2) (g) rarr HCl (g) , Delta_(g) H^(@) = - 90 kJ mol^(-1)`
According to Eq. , we have
`Delta_(r) H^(@) = sum "Bond enthalpies"_("reactants") - sum "Bond enthalpies"_("products")`
`((1)/(2) Delta_(H - H) H^(@) + (1)/(2) Delta_(Cl - Cl) H^(@)) - Delta_(H - Cl) H^(@)`
`- 90 = (1)/(2) (430) + (1)/(2) (240) - Delta_(H - Cl) H^(@)`
`:. Delta_(H - Cl) H^(@) = (215) + (120) + (90) = 425 kJ mol^(-1)`


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