The document provides calculations to determine if chemical reactions are spontaneous or nonspontaneous based on changes in enthalpy (ΔH), entropy (ΔS), and temperature (T). It analyzes three systems with different ΔH, ΔS and T values. For system I, ΔG is positive so the reaction is nonspontaneous. For system II, ΔG is negative, indicating a spontaneous reaction. For system III, ΔG is again positive, so the reaction is nonspontaneous.

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.