What is the equilibrium constant `K_(eq)` for the following reaction at 400 K.
`2NOCl_((g)) hArr 2NO_((g)) +Cl_(2(g))` , given that `DeltaH^0=77.2 "kJ mol"^(-1)` and `DeltaS^0=122 "JK"^(-1) "mol"^(-1)` .

Explanation

Given :T=400 K , `DeltaH^0=77.2 "kJ mol"^(-1) =77200 "J mol"^(-1) ,DeltaS^0 =122 "JK"^(-1) "mol"^(-1)`
`DeltaG^0=-2.303 RT log K_(eq)`
`log K_(eq)=(DeltaG^0)/(2.303 RT)`
`log K_(eq)=-((DeltaH^0-TDeltaS^0))/(2.303RT)`
`log K_(eq)=-((77200-400xx122)/(2.303xx8.314xx400))`
`log K_(eq)=-(28400/7659)`
`log K_(eq)=-3.7080`
`K_(eq)`=antilog (-3.7080)
`K_(eq)=1.95xx10^(-4)`