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

Given, `DeltaH_(H-H)=434` KJ/mol
`DeltaH_(Cl-Cl)=242` KJ/mol
`DeltaH_(H-Cl) = 431` KJ/mol
`(1)/(2)H_(2) + (1)/(2) Cl_(2) to HCl, DeltaH_(r)=?`
`DeltaH_(r)=(1)/(2)xxDeltaH_(H-H)+(1)/(2)xxDeltaH_(Cl-Cl)-DeltaH_(H-Cl)`
`=(1)/(2)xx434 + (1)/(2)xx 242 - 431`
`=217xx121 - 431 =- 93` KJ/mol