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) .

What is the equilibrium constant K_(eq) for the following reaction at

1.	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) .
Answer» Solution :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)`