If `S + O_(2) rarr SO_(2), Delta H = -298.2 kJ`
`SO_(2) + (1)/(2) O_(2) rarr SO_(3), Delta H = -98.2 kJ`
`SO_(3) + H_(2)O rarr H_(2)SO_(4), Delta H = -130.2 kJ`
`H_(2) + (1)/(2) O_(2) rarr H_(2)O, Delta H = -287.3 kJ`
the enthalpy of formation of `H_(2)SO_(4)` at 298 K will be

Options

A. `- 433.7 kJ`
B. `- 650.3 kJ`
C. `+320.5 kJ`
D. `- 813.9 kJ`

Answer

D

Explanation

The required equation is
`H_(2) (g) + S + 2O_(2) rarr H_(2)SO_(4) Delta H = ?`
(i) `(S + O_(2) rarr SO_(2) Delta H = - 298.2 kJ mol^(-1)`
(ii) `SO_(2) + (1)/(2) O_(2) rarr SO_(3) Delta H = - 98.2 kJ mol^(-1)`
(iii) `SO_(2) + (1)/(2) O_(2) rarr SO_(3) Delta H = - 130.2 kJ mol^(-1)`
(iv) `H_(2) + (1)/(2) O_(2) rarr H_(2)O Delta H = - 287.3 kJ mol^(-1)`
Add eqns. (i), (ii), (iii) and (iv),
`H_(2) + S + 2O_(2) rarr H_(2)SO_(4) Delta H = -813.9 kJ`