For a reaction, X2(g) =2X(g), ∆U = 40 kJmol^-1, ∆S = 70 JK^-1mol^-1 at 600 K then ∆G will be2.3 kJ mol^-11.95 kJ mol^-12.998 kJ mol^-1-0.5 kJ mol^-1

For a reaction, X2(g) =2X(g), ∆U = 40 kJmol^-1, ∆S = 70 JK^-1mol

1.	For a reaction, X2(g) =2X(g), ∆U = 40 kJmol^-1, ∆S = 70 JK^-1mol^-1 at 600 K then ∆G will be2.3 kJ mol^-11.95 kJ mol^-12.998 kJ mol^-1-0.5 kJ mol^-1
Answer» The change in GIBBS Free energy is GIVEN byΔH=ΔU+Δn g RTwhereΔH is the enthalpy of the reaction ΔS is the entropy of the reactionand ΔU is the change in INTERNAL energy Δn g is the (NUMBER of gaseous moles in product) - (number of gaseous moles in reactant)=2-0=2R is the gas constant =2 calBut,ΔH=(2.1×10 3 )+(2×2×300)=3300calHence,ΔG=ΔH=TΔSΔG=3300−(300×20)ΔG=−2700cal=−2.7cal