Bond dissociation enthalpy of `H_(2),Cl_(2)` and `HCl` are `434,242` and `431kJ mol^(-1)` respectively. Enthalpy of formation of `HCl` is `:`

Options

A. `-93kJ mol^(-1)`
B. `245kJ mol^(-1)`
C. `93kJ mol^(-1)`
D. `-245kJ mol^(-1)`

Answer

A

Explanation

`H_(2) rarr 2H`,
`:. E_(H-H)=434kJ mol^(-1)`
`Cl_(2) rarr 2Cl`
`:. E_(Cl-Cl)=242kJ mol^(-1)`
`HCl rarr H+Cl,`
`e_(H-Cl)431kJ mol^(-1)`
we know
`(1)/(2)H_(2)+(1)/(2)Cl_(2) rarr HCl`
`Delta_(r)H=` Bond dissociation enthalpy of `H-Cl` in `HCl+(1)/(2)[e_(H-H)+e_(Cl-Cl)]`
`=-431+(1)/(2)[434+242]`
`=-93kJ mol^(-1)`