1.

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

Answer»

`- 433.7 kJ`
`- 650.3 kJ`
`+320.5 kJ`
`- 813.9 kJ`

Solution :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`


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