1.

given H_(2)O(l) to H^(+)(aq) + OH^(-)(aq),DeltaH= 57.32 kJ H_(2)(g) + 1/2O_(2)(g) to H_(2)O(l),DeltaH =-286.02 kJ then , calculate the enthalpy of formation of OH^(-) at 25^(@)C

Answer»

`-228.8kJ`
`-343.52 kJ`
`+228.8 kJ`
`+ 343. 52 kJ`

SOLUTION :consider the formation of `H_(2)O`
`H_(2)(g)+1/2O_(2)(g) to H_(2)O(l),DELTAH=-286.20 kJ`
`DeltaH_(r)= DeltaH_(t)[H_(2)O(l)]-DeltaH_(t)[H_(2)(g))]-1/2 DeltaH_(t)[O_(2)(g)] `
` DeltaH_(t)(H_(2)O(l))=-286.20`
now consdier the ionisation of `H_(2)O`
`H_(2)O(l) to H^(+)(aq)+ OH^(-)(aq), DeltaH = 57.32KJ H_(2)O(l) to H^(+)(aq)+ OH^(-)(aq), DeltaH = 57.32KJ`


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