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

In problem 1, you determined that step 1 in the dry process is reactant-favored at 25 ºC. Now, use DGfºs to calculate the free energy changes of all four steps.

What is the DGº of the overall process? If all four steps could be coupled together, would the overall process be product- or reactant-favored at 25 ºC?

Step 1: 2 Ca3(PO4)2 (s) + 6 SiO2 (s) + 10 C (s) 10 CO (g) + P4 (g) + 6 CaSiO3 (s)

Step 2: P4 (g) P4 (s, red)

Step 3: P4 (s, red) + 5 O2 (g) P4O10 (s)

Step 4: P4O10 (s) + 6 H2O () 4 H3PO4 (aq)

Use the table of thermodynamic values by clicking on the thermodynamics icon at top right.

DGº= Product-favored
Reactant-favored

Correct! If all four steps could somehow be coupled together, the overall process would indeed be spontaneous at 25 ºC. In reality, however, this would be difficult, if not impossible. For example, while the condensation of phosphorus in step 2 is exergonic, can you think of a way to harness the free energy liberated to drive step 1? It may be fun for you to try to think of some other ways the steps could be coupled to make the overall process more efficient. Click on the Next button below to return to the Problem-Solving Page or on the Close button to return to close this window.

    

The solution is shown below. If all four steps could somehow be coupled together, the overall process would indeed be spontaneous at 25 ºC. In reality, however, this would be difficult, if not impossible. For example, while the condensation of phosphorus in step 2 is exergonic, can you think of a way to harness the free energy liberated to drive step 1? It may be fun for you to try to think of some other ways the steps could be coupled to make the overall process more efficient. Click on the Next button below to return to the Problem-Solving Page or on the Close button to close this window.

DGº1=10 mol • DGfº (CO(g)) + 1 mol • DGfº (P4(g)) + 6 mol • DGfº (CaSiO3(s)) - 2 mol • DGfº (Ca3(PO4)2(s)) - 6 mol • DGfº (SiO2(s)) - 10 mol • DGfº (C(s))
     = 10 mol (-137.168 kJ/mol) + 1 mol (24.48 kJ/mol) + 6 mol (-1549.7 kJ/mol) - 2 mol (-3885 kJ/mol) - 6 mol (-856.64 kJ/mol) - 10 mol (0)
     = 2264.44 kJ

DGº2=1 mol • DGº (P4(s, red)) - 1 mol • DGº (P4(g))
     = 1 mol (-48.4 kJ/mol) - 1 mol (24.48 kJ/mol)
     = -72.88 kJ

DGº3=1 mol • DGº (P4O10(s)) - 1 mol • DGº (P4(s, red)) - 1 mol • DGº (O2(g))
     = 1 mol (-2697.7 kJ/mol) - 1 mol (-48.4 kJ/mol) - 5 mol (0)
     = -2649.3 kJ

DGº4=4 mol • DGº (H3PO4(aq)) - 1 mol • DGº (P4O10(s)) - 6 mol • DGº (H2O())
     = 4 mol (-1019 kJ/mol) - 1 mol (-2697.7 kJ/mol) - 6 mol (-237.129 kJ/mol)
     = 44.474 kJ

DGº=DGº1 + DGº2 + DGº3 + DGº4
     = 2264.44 kJ - 72.88 kJ - 2649.3 kJ + 44.474 kJ
     = -413.3 kJ

    

While numerically correct, your value of DGº has the wrong number of significant digits. Try again.
Remember if DG < 0, the reaction is product-favored.
Did you remember to multiply each DGfº by the number of moles of the compound in the target equation?
Did you remember to change the signs of DGfºs of compounds that are reactants?
That answer is incorrect. Try again.
How do you know the process is reactant- or product-favored without calculating the DGº?