1. Question 1
Using the following values for ΔH, ΔS, and T, determine the change in free energy and if the
reaction is spontaneous or nonspontaneous.
I) ΔH = 40 kJ, ΔS = 300 J/K, T = 130 K
II) II) ΔH = 40 kJ, ΔS = 300 J/K, T = 150 K
III) III) ΔH = 40 kJ, ΔS = -300 J/K, T = 150 K
Question 2
i. Careful measurements show that the entropy of water is +0.0320Kcal/moleK at 25
degree Celsius and the enthalpy is +31.39Kcal. Calculate the free energy.
ii. Water-gas reactin occurs at the temperature of white hot carbon, which is
approximately 900 degree Celsius. The values of ∆H and ∆S are different at
different temperature. However, if we assume they remain abot the same at the
reaction temperature of 900 degree Celsius, Calculate the approximate value of
∆G. retaining the value of ∆H as +31.39Kcal/mol and entropy as +0.0320Kcal
System I
ΔG = ΔH - TΔS
ΔG = 40 kJ - 130 K x (300 J/K x 1 kJ/1000 J)
ΔG = 40 kJ - 130 K x 0.300 kJ/K
ΔG = 40 kJ - 39 kJ
ΔG = +1 kJ
ΔG is positive, therefore the reaction will not be spontaneous.
System II
ΔG = ΔH - TΔS
ΔG = 40 kJ - 150 K x (300 J/K x 1 kJ/1000 J)
ΔG = 40 kJ - 150 K x 0.300 kJ/K

2. ΔG = 40 kJ - 45 kJ
ΔG = -5 kJ
ΔG is negative, therefore the reaction will be spontaneous.
System III
ΔG = ΔH - TΔS
ΔG = 40 kJ - 150 K x (-300 J/K x 1 kJ/1000 J)
ΔG = 40 kJ - 150 K x -0.300 kJ/K
ΔG = 40 kJ + 45 kJ
ΔG = +85 kJ
ΔG is positive, therefore the reaction will not be spontaneous.
Answer:
A reaction in system I would be nonspontaneous.
A reaction in system II would be spontaneous.
A reaction in system III would be nonspontaneous.