The dry process is a four-step process used to produce phosphoric acid industrially. We will take
a look at the dry process to see how Thermodynamics can be used to answer practical problems
encountered in real life.
In the first step, a common mineral containing phosphorus and contaminated with silica is heated
with coal in an electrical oven. Gaseous carbon monoxide, tetraphosphorus, and solid calcium
silicate is produced:
1. 2 Ca3(PO4)2 (s) + 6 SiO2 (s) + 10 C (s)  10 CO (g) + P4 (g) + 6 CaSiO3 (s)
In the second step, the gaseous phosphorus is condensed into the solid form. The allotrope that is
formed is the red phosphorus variety:
2. P4 (g)  P4 (s, red)
In the third step, the red phosphorus is burned in air to yield solid tetraphosphorus decaoxide:
3. P4 (s, red) + 5 O2 (g)  P4O10 (s)
Finally, the tetraphosphorus decaoxide is dissolved in water to form phosphoric acid:
4. P4O10 (s) + 6 H2O (l)  4 H3PO4 (aq)
The table of Thermodynamic values on the last page will be useful as you work through this
problem.
Part 1: Is step 1 product-favored or reactant-favored at 25 C? If it is reactant-favored, at what
temperature will it become product-favored (assuming there are no phase changes in reactants or
products or that the H or S’s do not otherwise change)?
Solution:
H = 10 (-110.525 kJ) + 1 (58.91 kJ) + 6 (-1634.9 kJ) – 2 (-4121 kJ) – 6 (-910.94 kJ)
= 2851.9 kJ
S = 10 (197.674 J/K) + (279.9 J/K) + 6 (81.92 J/K) – 2 (236 J/K) – 6 (41.84 J/K) – 10 (5.740
J/K)
= 1967.72 J/K
G = H - TS
At 25 C (298.15 K) G = 2851.9 kJ – 298.15 K ·1967.72 J/K · 1 kJ/1000 J = 2262.5 kJ
 reactant-favored at 25 C
G = H - TS < 0
2851.9 kJ – T·(1967.72 J/K ·1 kJ/1000 J) < 0
-1.96772T kJ/K < -2851.9 kJ
T > 2851.9 kJ/(1.96772 kJ/K)
T > 1450 K = 1180 C
Part 2: How much water is required to dissipate the heat released when the phosphorus is
condensed from the gas phase at 1180 C to the solid state at 25 C? At 43 atm, red phosphorus
sublimes to gaseous tetraphosphorus at 590 C. In your calculation, neglect the heat of
sublimation since it is relatively small compared to the heat needed to change the temperature of
the phosphorus. Assume that the water comes into the plant at 25 C and is discharged at 30 C.
Express your answer in gallons of water per mole of P4 condensed
c (P4(g)) = 67.15 J/mol·K
c (P4(s,red)) = 28.84 J/mol·K
c (H2O(l)) = 4.184 J/g · K
1 g H2O = 1 mL H2O
1 L = 1.056710 qt
4 qt = 1 gal
Part a: Calculate the heat released by the phosphorus:
q = mcT (gas) + mcT (solid)
= (67.15 J/mol·K (590 – 1180) K + 28.84 J/mol·K (25 – 590) K) · 1 kJ/1000 J
= -52 kJ/mol
Part b: Calculate the amount of water required to absorb and dissipate that much heat:
52 kJ/mol P4 · 1mol/123.8952 g · 1000 g/1 kg · 1/(4.184 J/g H2O K · 5 K) · 1000 J/kJ · 1 mL/1 g
· 1L/1000 mL · 1.056710 qt/1L · 1 gal/4 qt = 5.30 gal H2O/kg P4 condensed
Part 3: Use the Gf  values listed in the table to find the G  for the overall process, the sum of
steps 2-4. If these steps could be coupled to the reactant-favored first step, would the overall
process be product-favored?
Solution:
Step 2: G = Gf  (P4(s, red)) - Gf  (P4(g))
= -48.4 kJ – 24.48 kJ = -72.88 kJ
Step 3: G = Gf  (P4O10(s)) - Gf  (P4(s, red)) – 5 Gf  (O2(g))
= -2697.7 kJ – (-48.4 kJ) – 5 (0) = -2649.3 kJ
Step 4: G = 4 Gf  (H3PO4(aq)) - Gf  (P4O10(s)) – 6 Gf  (H2O(l))
= 4 (-1019 kJ) – (-2697.7 kJ) – 6 (-237.129 kJ) = 44.474 kJ
Overall (2-4) G = -72.88 kJ + -2649.3 kJ + 44.474 kJ = -2677.706 kJ
Overall G = 2262.5 kJ – 2677.706 kJ = -415.2 kJ
Therefore, the overall dry process is product-favored, and would proceed without the input of
energy if reactions 2-4 could be coupled to reaction 1. In reality, this would be difficult to do. I
can’t think of any way to couple the condensation of tetraphosphorus to reaction 1. It may be
possible to generate some electricity from the burning of tetraphosphorus in reaction 3 by boiling
water to turn a turbine. Step 4 is reactant-favored according to these calculations (which doesn’t
make much sense to me). Burning the carbon monoxide produced in step 1 may be another way
to make the overall process product-favored, but that free energy change is not accounted for in
these calculations.
Some Thermodynamic Values (298.15 K)
Species
Ca3(PO4)2 (s)
CaSiO3 (s)
C (s, graphite)
CO (g)
H2O (l)
O2 (g)
P4 (s, red)
P4 (g)
P4O10 (s)
H3PO4 (aq)
SiO2 (s)
Hfº (kJ/mol)
-4121
-1634.9
0
-110.525
-285.830
0
-70.4
58.91
-2984.0
-1277
-910.94
Sº (J/mol·K)
236
81.92
5.740
197.674
69.91
205.138
91.2
279.9
228.86
-220
41.84
Heat capacity of P4 (g) = 67.15 J/mol·K
Heat capacity of P4 (s, red) = 28.84 J/ J/mol·K
Sublimation point of P4 (s, red) = 590 °C
Heat capacity of H2O (l) = 4.184 J/mol·K
1 g H2O = 1 mL H2O
1 L = 1.056710 qt
4 qt = 1 gal
Gfº (kJ/mol)
-3885
-1549.7
0
-137.168
-237.129
0
-48.4
24.48
-2697.7
-1019
-856.64