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)?

DGo = -RT ln K => large, negative DGo favors products

DGo = DHo - TDSo

At 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)