Calculate `DeltaG_(298)^(@)` and `DeltaS_(298)^(@)` for the reaction :
`2H_(2(g))+O_(2(g)) rarr 2H_(2)O_((l)): DeltaH_(298)^(@) = -136.64kcal`.
Given that
`O_(2(g))+4H^(+)+4e rarr 2H_(2)O_((l)), E^(@) = 1.23V`
`2H_(2(g)) rarr 4H^(+)+4e, E^(@) = 0.00V`

Explanation

Given `{:(O_(2(g))+4H^(+)+4erarr2H_(2)O_(4), E^(@)=1.23V),(2H_(2(g))rarr4H^(+)+e^(-), E^(@)=0.00V):} /("On adding" 2H_(2(g))+O_(2(g))rarr2H_(2)O_((l)))` , `E^(@)=1.23+0 = 1.23V)`
`because DeltaG_(298)^(@) = -nE^(@)F`
`= 4 xx 1.23 xx 96500 = -474780J`
`= -474.78J = -113.58 kcal`
Also `DeltaG_(298)^(@) = DeltaH^(@)-TDeltaS^(@)`
`DeltaS_(298)^(@) = [(DeltaG^(@)-DeltaH^(@))/(T)]= (DeltaH^(@)-DeltaG^(@))/(T)`
`= (-136.64-(-113.58))/(298)`
`= -0.77 kcal`