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` is

Options

A. 245 kJ `mol^(-1)`
B. 290 kJ `mol^(-1)`
C. 380 kJ `mol^(-1)`
D. 425 kJ `mol^(-1)`

Answer

D

Explanation

`(1)/(2)H_(2(g))+(1)/(2)Cl_(2(g))rarrHCl_((g)) , Delta H_(f) =-90 kJ mol^(-1)`
`Delta H = Sigma (B.E.)_("Reactants")- Sigma (B.E.)_("Products")`
`-90=((1)/(2)xx430+(1)/(2)xx240)-(B.E.)_(HCl)`
`thereforer (B.E.)_(HCl)=425 kJ mol^(-1)`