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)toHCl(g)`, `DeltaH_(f)=-90 kJ mol^(-1)`
`DeltaH_(f)=(1)\(2)BE(H_(2))+(1)\(2)BE(Cl_(2))-BE(HCl)`
`-90=(1)/(2)(430)+(1)/(2)(240)-Be(HCl)`
`:. BE(HCl)=215+120+90`
`=425 kJ mol^(-1)`