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. 290 `KJ "mol"^(-1)`
B. `380 KJ "mol"^(-1)`
C. `425 KJ "mol"^(-1)`
D. `245 KJ "mol"^(-1)`

Answer

B

Explanation

`DeltaH_("reaction")=Delta_(H-H) + DeltaH_(Cl-Cl)-2DeltaH_(HCl)`
or `DeltaH_(Cl-Cl) =(430+ 240 -(-90))/(2)`
`=(760)/(2)=380 KJ "mol"^(-1)`