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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 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)`

Text Solution

Verified by Experts

The correct Answer is:
A
`H_(2)+Cl_(2)rarr 2HCl` ltBrgt `DeltaH_("reaction")=Sigma(BE)_("reactant")-Sigma(BE)_("product")`
`=[(BE)_(H-H)+(BE)_(Cl-Cl)]-[2(BE)_(H-Cl)]`
`=434+242-(431)xx2`
`=-186kJ`
As `DeltaH_("reaction")=-186kJ`
So enthalpy of formation of HCl
`=(-186kJ)/(2)`
`=-93" kJ mol"^(-1)`
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