Calculate (a) `DeltaG^(Θ)` and (b) the equilibrium constant for the formation of `NO` and `O_(2)` at `298 K`
`NO(g)+1//2 O_(2)(g) hArr NO_(2)(g)`
where
`Delta_(f)G^(Θ)(NO_(2))=52.0 kJ mol^(-1)`
`Delta_(f)G^(Θ)(NO)=87.0 kJ mol^(-1)`
`Delta_(f)G^(Θ)(O_(2))=0 kJ mol^(-1)`

Answer

a) `-35.0 kJ`, b) `1.365 xx 10^(6)`

Explanation

(a) For the given reaction,
`DeltaG^(@) = DeltaG^(@)("Products")-DeltaG^(@)("Reactants")`
`DeltaG^(@) = 52.0 - {87.0 + 0}`
`= - 35.0 kJ "mol"^(-1)`
(b) We know that
`DeltaG^(@) = RT "log" K_(c)`
`DeltaG^(@) = 2.303 RT "log" K_(c)`
`K_(c) = (-35.0 xx 10^(-3))/(-2.303 xx 8.314 xx 298)`
`= 6.134`
`:. K_(c) = "antilog" (6.134)`
`= 1.36 xx 10^(6)`
Hence, the equilibrium constant for the given reaction `K_(c)` is `1.36 xx 10^(6)`