Question

(ii) 2 NO2
(g) + N2O4
(g), 298K ij
∆H = –57.2 KJ, ∆S = –175.6JK–1

Answer 1

DG=DX−TDS

DX=DH

form

(products)−DH

reactants

⇒9.7KJ∣mo∣−2×90.5K3∣mo∣

⇒−171.3K3∣mo∣

DS=S

product

−S

reactant

⇒304−205−2×210

⇒304−205−420⇒−321JK

−1

∣mo∣

D4=−171.3kJ∣mo∣+29+321JK

−1

∣mo∣

⇒−75.64kJ∣mo∣