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

Options

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

Answer

C

Explanation

`(1)/(2)H_(2)(g)+(1)/(2)Cl_(2)(g)toHCl(g)`, `DeltaH^(@)=?`
`Delta_(r )H^(@)=(1)/(2)BE(H_(2))+(1)/(2)BE(Cl_(2))-BE(HCl)`
`=(1)\(2)(434)+(1)/(2)(242)-431`
`=217+121-431=-93 kJ mol^(-1)`