If `DeltaH_(f)^(@)` for `H_(2)O_(2)(l) and H_(2)O(l)` are -188kJ and `mol^(-1)` and -286 kJ `mol^(-1)`, what will be the enthalpy change of the reaction `2H_(2)O_(2)(l)to2H_(2)O(l)+O_(2)(g)`?

Options

A. `-196` kJ/mole
B. `146 kJ//"mole"`
C. `-494 kJ//"mole"`
D. `-98 kJ//"mole"`

Answer

A

Explanation

`H_(2)+O_(2)rarrH_(2)O_(2) " "DeltaH_(f)^(@)=-188 kJ//"mole ...(i)"`
`H_(2)+(1)/(2)O_(2)rarrH_(2)O_(2)" " DeltaH_(f)^(@)=-286 kJ//"mole ...(ii)"`
eq. (i) - eq. (ii) `xx2` gives the required result.