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+//Chapter 3: Thermodynamic and Chemical Equilibrium
+//Problem: 3
+clc;
+
+//Declaration of Variables
+delta_g = -16.0 // Kelvin cal
+delta_h = -10.0 // Kelvin cal
+T = 300 // Kelvin
+
+// Solution
+delta_s = (delta_h - delta_g) * 10 ** 3 / T // cal/deg
+new_t = 330 // Kelvin
+new_delta_g = (delta_h * 10 ** 3) - new_t * delta_s
+
+mprintf("The free energy at 330K is: %.2e K cal",new_delta_g)