If `DeltaH_(f)^(@)` for `H_(2)O_(2)` and `H_(2)O` are `-188` kJ/mole and `-286` kJ/mole, what will be the enthalpy change of the reaction `H_(2)O_(2) rarr H_(2)O+1/2 O_(2)`

Options

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

Answer

A

Explanation

`H_(2)(g) + O_(2)(g) rarr H_(2)O_(2)(l) " " Delta H= -188`
`H_(2)(g) + (1)/(2) O_(2)(g) rarr H_(2)O(l) Delta H= -286`
By Lavoisier and Laplace Law:
`2H_(2)O_(2)(l) rarr 2H_(2)(g) + 2O_(2)(g), Delta H= +376`
`2H_(2)(g) + O_(2)(g) rarr 2H_(2)O(l), Delta H= -572`
Adding the above two equations by using Hess. Law :
`2H_(2)O_(2)(l) rarr 2H_(2)O(l) + O_(2) (g) Delta H= -196`