Thermodynamics Solutions: #28
28.* (1993 3 5) A. What must the signs of DHo and DSo for a reaction at equilibrium that proceeds mostly to products for low T but not at high T (where T is the temperature)? D Go = -RT ln K => large, negative DGo favors productsD Go = DHo - TDSoAt low T, DGo depends mostly on DHo.At high T, DSo becomes important.Here, at low T, products favored => DHo = (-) (negative) (DGo < 0 )At high T, products not favored => DGo = (+) (positive)=> DSo = (-) (negative)B. What must be the signs of DHo and DSo for a reaction at equilibrium that proceeds mostly to products for both low and high T?By the same reasoning: At low T, products favored => DHo = (-) (negative)At high T, products still favored => DGo < 0=> DSo > 0 (positive) |