Question

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

Answer 1

Correct Answer - A
`(1)/(2)H_(2)+(1)/(2)Cl_(2)rarrHCl`
`DeltaH=(1)/(2)xx434+(1)/(2)xx242-431`
`=217+121-431=-93 kJ//"mole"`.