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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)
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