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Bond dissociation enthalpy of H(2),Cl(2)...

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

A

`- 93 kJ mol^(-1)`

B

`245kJ mol^(-1)`

C

`93kJmol^(-1)`

D

`- 245kJmol^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
A
`(1)/(2) H_(2)+ (1)/(2) Cl_(2) rarr HCl`
`Delta H= (1)/(2) xx 434 + (1)/(2) xx 242 - 431`
`=217 +121 - 431 = - 93`kJ/mole
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