Calculate the standard free energy change for the reaction,
`4NH_(3(g))+5O_(2(g))rarr4NO_((g))+6H_(2)O_((l))`. Given that the standard free energies of formation `(Delta_(f)G^(0))` for `NH_(3(g)),NO_((g))andH_(2)O_((l))` are - 16.8, + 86.7 and - 237.2 KJ `mol^(-1)` respectively. Predict the feasiblity of the above reaction at the standard state.

Explanation

Here, we are given
`Delta_(r) G^(@) (NH_(3))= - 16.8 kJ mol^(-1)`
`Delta_(f)G^(@) (NO) = + 86.7 kJ mol^(-1)`
`Delta_(f) G^(@) (H_(2)O) = -237.2 kJ mol^(-1)`
`:. Delta_(r) G^(@) = Sigma Delta_(f) G^(@) ` ( Products) `- Sigma Delta_(f) G^(@) `( Reactants ) `= [ 4 xx Delta_(f) G^(@) (NO) + 6 xx Delta_(f) G^(@) (H_(2)O) ]-[4 xx Delta _(f) G^(@) ( NH_(3))+ 5 xx Delta G^(@) (O_(2))]`
`= [ 4 xx (86.7) + 6 xx ( -237.2) ]- [ 4 xx ( -16.8) + 5 xx0]= -1009.2 kJ `
Since` Delta_(r) G^(@)` is negative, the process is feasible.