The enthalpies of combustion of S, `SO_(2) and H_(2)` are`-298.2, -98.7 and -287.3.kJ mol^(-1)` respectively. If enthalpy of the reaction `SO _(x (g)) + H _(2) O _((f)) to H _(2 ) SO _(4 (f)) is - 130.2 kJ mol ^(-1),` the enthalpy of formation of `H_(2)SO_(4)` is

Options

A. `-814.4 kJ mol ^(-1)`
B. `-650.4 kJ mol ^(-1)`
C. `- 554.2 kJ mol ^(-1)`
D. `-435. 5 kJ mol ^(-1)`

Explanation

Given data are: `S (s) + O _(2) (g) to SO _(2) (g)," " Delta H =- 298 . 2 kJ mol ^(-1)`
`SO _(2) (g) + ((1)/(2)) O _(2) (g) toSO _(3) " " Delta H =- 98.7 kJ mol ^(-1)`
`H_(2) (g )+ ((1)/(2)) O _(2) (g) to H _(2) O (1)" " Delta H =- 287. 3 kJ mol ^(-1)`
`SO _(3) (g) + H _(2) O (1) to H _(2) SO _(4) (g) " " Delta H =- 130. 2kJ mol 6(-1)`
The formation of `H _(2) SO _(4) `(l) implies the reaction `H _(2) (g) + S (s) + 2 O _(2) (g) to H _(2) SO _(4) (1)`
This reaction is obtained by adding the above four reactions. Hence.
`Delta H =- ( 298. 2 + 98. + 287. 3 130.2) kJ mol ^(-1) =- 814 . 4 kJ mol ^(-1)`