If `S+O_(2)toSO_(2),DeltaH=-298.2 " kJ" " mole"^(-1)`
`SO_(2)+(1)/(2)O_(2)toSO_(3)DeltaH=-98.7 " kJ" " mole"^(-1)`
`SO_(3)+H_(2)OtoH_(2)SO_(4),DeltaH=-130.2 " kJ" " mole"^(-1)`
`H_(2)+(1)/(2)O_(2)toH_(2)O,DeltaH=-287.3 " kJ" " mole"^(-1)`
the enthlapy of formation of `H_(2)SO_(4)` at 298 K will be

Options

A. `-814.4 kJ`
B. `+320.5 kJ`
C. `-650.3 kJ`
D. `-933.7 kJ`

Answer

A

Explanation

Enthalpy of formation of `H_2SO_4` can be represented by the following equation:
`H_2 + S + 2O_2 to H_2SO_4`
Given, `S + O_(2) to SO_(2)," "Delta H = -298.2 kJ`
`SO_(2) + 1/2O_(2) to SO_(3), " "Delta H = -98.7 kJ`
`SO_(3) + H_(2)O to H_(2)SO_(4), " "Delta H = -130.2 kJ`
`H_2 + 1/2 O_(2 ) to H_(2)O , " "Delta H = -287.3 kJ`.
On adding all the four equations, we get the equation (i),
Hence, `Delta H_(f) = (-298.2) + (-98.7) + (-130.2) + (-287.3) = -814.4 kJ`.