The standard heat of formation of `H_(2)O_((l))` from its elements `H_(2) and O_(2)` is `-290.83 kJmol^(-1)` and the standard entropy change for the same reaction is `-330 JK^(-1)" at " 25^(@)C.` Will the reaction be spontaneous at `25^(@)C`.
Given: `DeltaH^(@)=-290.83 kJmol^(-1)`
`=-290830Jmol^(-1)`
`DeltaS^(@)=-330JK^(-1)`
`T=25^(@)C=298K`

Explanation

`DeltaG^(@)=DeltaH^(@)-TDeltaS^(@)`
`=-290830-298(-330)`
`=-290830+98340=-192490`
`DeltaG^(@)=-192490 J mol^(-1)`
Since `DeltaG^(@)` is negative , the reaction is spontaneous.