1.

Delta H for the reaction, SO_(2) (g) + (1)/(2) O_(2)(g) hArr SO_(3)(g) " " Delta H = -98.3kJ If the enthalpy of formation of SO_(3)(g) is -395.4kJ, then enthalpy of formation of SO_(2)(g) is:

Answer»

`-297.1kJ`
`493.7kJ`
`-493.7kJ`
`297.1kJ`

Solution :`Delta H = Delta_(f) H(SO_(3)) - [Delta_(f) H(SO_(2)) + (1)/(2) Delta_(f) H(O_(2))]`
`-98.3 = -395.4 - [Delta_(f) H(SO_(2)) + 0]`
or `Delta_(f)H(SO_(2)) = -395.4 + 98.3 = -297.1 kJ`


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