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If, S+O2 rarr SO2, Delta H =-298.2 kj ...

If, `S+O_2 rarr SO_2, Delta H =-298.2 kj`
`SO_2 + 1/2 O_2 rarr SO_3, Delta H=-98.7 kJ `
`SO_3 + H_2 O rarrr H_2SO_4 , Delta H =-130.2 kJ`
`H_2 + 1/2 O_2 rarr H_2 O, Delta H =-227.3 kJ`
Find `Delta H` for formation of `H_2SO_4` at 298 K.

A

`-814.4 kJ`

B

`+320.5 kJ`

C

`-650.3 kJ`

D

`-933.7 kJ`

Text Solution

Verified by Experts

The correct Answer is:
A
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`.
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