The enthalpy change for first proton neutralisation of `H_(2)S` is `-37.1kJ mol^(-1)`. What is the enthalpy change for first ionisation of `H_(2)S`.

Options

A. `94.2`
B. `-20.0`
C. `20.0`
D. `-94.2`

Answer

C

Explanation

`H_(2)S + overset(Theta)OH rarr H_(2)O + HS^(Theta), DeltaH_(1) =- 37.1 kJ`
`H^(o+) + overset(Theta)OH rarr H_(2)O, DeltaH_(2) =- 57.1 kJ`
`(DeltaH_(2)` is the heat of neutralisation of `S_(A)` and `S_(B))`.
`:. H_(2)S hArr HS^(Theta) + H^(o+), (DeltaH = ?)`
`DeltaH = DeltaH_(1) - DeltaH_(2)`
`= -37.1 - (-57.1) = 20.0kJ`