1. GRAPHICAL DETERMINATION OF ΔHº AND ΔSº
ΔGº = ΔHº - TΔSº
ΔHº and ΔSº are constant over a wide temperature range, but not ΔGº
ΔGº = ΔHº - TΔSº
______ _____ _______
T T T
ΔGº = ΔHº ΔSº
______ ___
T T y = m x + b
For a plot of ΔGº /T vs. 1/T
: the slope is ΔHº and the y-intercept is –ΔSº
ΔGº can be determined from Keq’s at various temperatures:
ΔGº = -RT ln Keq
E15-1 (of 3)

2. The Ksp for a chemical reaction was determined at 2 different temperatures:
298 K Ksp = 3.0 x 102
348 K Ksp = 8.0 x 104
ΔGº1 = -14,132 J
ΔGº1 = -32,664 J
Calculate ΔHº and ΔSº for the reaction.
ΔGº1 = -RT1 ln K1
= -(8.314 J/K)(298 K) ln 3.0 x 102
= -14,132 J
ΔGº2 = -RT2 ln K2
= -(8.314 J/K)(348 K) ln 8.0 x 104
= -32,664 J
E15-2 (of 3)

3. Calculate ΔHº and ΔSº for the reaction.
The Ksp for a chemical reaction was determined at 2 different temperatures:
298 K Ksp = 3.0 x 102
348 K Ksp = 8.0 x 104
ΔGº1 = -14,132 J
ΔGº1 = -32,664 J
1/T
ΔG/T
1/298
1/348
Calculate ΔHº and ΔSº for the reaction.
ΔGº = ΔHº ΔSº
______ ___
T T
ΔHº = 96,381 J
ΔSº = J/K
E15-3 (of 3)