1.

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

Answer»

`- 93 KJ mol^(-1)`
245 kJ `mol^(-1)`
93 kJ `mol^(-1)`
`-245` kJ `mol^(-1)`

Solution :`(1)/(2)H_(2)+(1)/(2)Cl_(2)rarrHCl`
`DeltaH=(1)/(2)xx434+(1)/(2)xx242-431`
`=217+121-431=-93 kJ//"MOLE"`.


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