Thermodynamics:
Problem-Solving Example
Thermodynamics Gateway Page
In this module:
Problem 1
Problem 2
Problem 3

Is step 1 product-favored or reactant-favored?

Click on the thermodynamics icon at right to see a partial table of thermodynamic values.

2 Ca3(PO4)2 (s) + 6 SiO2 (s) + 10 C (s) 10 CO (g) + P4 (g) + 6 CaSiO3 (s)

Product-favored
Reactant-favored
The solution is shown below. Since the DGº > 0, the reaction is reactant-favored at 25 ºC. Click on the next button to see what must be done to make the reaction product-favored.

At 25 ºC:

DHº=10 mol • DHfº (CO(g)) + 1 mol • DHfº (P4(g)) + 6 mol • DHfº (CaSiO3(s)) - 2 mol • DHfº (Ca3(PO4)2(s)) - 6 mol • DHfº (SiO2(s)) - 10 mol • DHfº (C(s))
    = 10 mol (-110.525 kJ/mol) + 1 mol (58.91 kJ/mol) + 6 mol (-1634.9 kJ/mol) - 2 mol (-4121 kJ/mol) - 6 mol (41.84 kJ/mol) - 10 mol (0)
    = 2851.9 kJ

DSº=10 mol • Sº (CO(g)) + 1 mol • Sº (P4(g)) + 6 mol • Sº (CaSiO3(s)) - 2 mol • Sº (Ca3(PO4)2(s)) - 6 mol • Sº (SiO2(s)) - 10 mol • Sº (C(s))
    = 10 mol (197.674 J/K•mol) + 1 mol (279.9 J/K•mol) + 6 mol (81.92 J/K•mol) - 2 mol (236 J/K•mol) - 6 mol (41.84 J/K•mol) - 10 mol (5.740 J/K•mol)
    = 1967.72 J/K

DGº=DHº - TDSº
    = 2851.9 kJ - 298.15 K • 1967.72 J/K • 1kJ/1000J
    = 2262.5 kJ

Therefore, the reaction is reactant-favored at 25 ºC.

Correct! Because the DGº > 0, the reaction is reactant-favored at 25 ºC. Click on the next button to see what must be done to make the reaction product-favored.