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. `-650.3 kJ mol ^(-1)`
C. `-554 . 2 kJ mol ^(-1)`
D. `-435.5 kJ mol ^(-1)`

Answer

A

Explanation

Give data are `S (s) + O _(2) (g) to SO _(2) (g) , Delta H =- 29 . 2 kJ mol ^(-1)`
`SO _(2) g + ((1)/(2)) O _(2) (g) to SO _(3) (g)," "Delta H =85 98.7 kJ mol ^(-1)`
`H _(2) (g) + ((1)/(2)) O _(2) (g) to H _(2) O (l)," " Delta H = 287. 3 kJ mol ^(-1)`
`SO _(3) (g) + H _(2) O (1) to H _(2) SO _(4) (l) ," " Delta H =- 130 .2 kJ mol ^(-1)`
The formation of `H _(2) SO _(4) (l)` impiles the reaction `H _(2) (g) +S (s) + 2 O _(2) (g) t H _(2) OS _(4) (1)`
This reacton is obained by adding the aboe four ractions. Hence
`Delta H =- (298.2 + 98.7 + 287.3+ 130.2) kJ mol ^(-1) =- 814. 4 kJ mol ^(-1)`