Simple Gas-Phase Equilibria

James Richard Fromm

When all of the substances which participate in a chemical equilibrium are gases, it is most convenient to express the amounts of substance present per unit volume in terms of the partial pressure of each of the gases. These gas-phase equilibria are described by equilibrium constants which use pressure units, and will generally carry a subscript "p" to indicate that pressure units are being used. A table of values of equilibrium constants for some of these equilibria is given below.

Table: Selected Gas-Phase Equilibrium Constants at 25oC
Equilibrium Kp (Pa) Kp (atm)
H2 + I2 larrow.GIF (55 bytes)rarrow.gif (63 bytes) 2HI 6.17 x 10+2 6.17 x 10+2
2HI larrow.GIF (55 bytes)rarrow.gif (63 bytes) H2 + I2 1.62 x 10-3 1.62 x 10-3
2HBr larrow.GIF (55 bytes)rarrow.gif (63 bytes) H2 + Br2 5.33 x 10-20 5.33 x 10-20
2HCl larrow.GIF (55 bytes)rarrow.gif (63 bytes) H2 + Cl2 4.06 x 10-34 4.06 x 10-34
2HF larrow.GIF (55 bytes)rarrow.gif (63 bytes) H2 + F2 1.88 x 10-96 1.88 x 10-96
N2 + 3H2 larrow.GIF (55 bytes)rarrow.gif (63 bytes) 2NH3 5.81 x 10-5 Pa-2 5.96 x 10+5 atm-2
N2O4 larrow.GIF (55 bytes)rarrow.gif (63 bytes) 2NO2 7.92 x 10+5 Pa 7.82 atm
2H2 + O2 larrow.GIF (55 bytes)rarrow.gif (63 bytes) 2H2O 1.23 x 10+75 Pa-1 1.24 x 10+80 atm-1
N2 + O2 larrow.GIF (55 bytes)rarrow.gif (63 bytes) 2NO 4.71 x 10-31 4.71 x 10-31
2NO + O2 larrow.GIF (55 bytes)rarrow.gif (63 bytes) 2NO2 2.23 x 10+7 Pa-1 2.26 x 10+12 atm-1
CO + 2H2 larrow.GIF (55 bytes)rarrow.gif (63 bytes) CH3OH 2.24 x 10-6 Pa-2 2.30 x 10+4 atm-2
CO + H2O larrow.GIF (55 bytes)rarrow.gif (63 bytes) CO2 + H2 1.03 x 10+5 1.03 x 10+5
CH4 + H2O larrow.GIF (55 bytes)rarrow.gif (63 bytes) CO + 3H2O 7.93 x 10+34 Pa+2 7.73 x 10+24 atm+2
2SO2 + O2 larrow.GIF (55 bytes)rarrow.gif (63 bytes) 2SO3 6.77 x 10+19 Pa-1 6.86 x 10+24 atm-1
2HCl + O2 larrow.GIF (55 bytes)rarrow.gif (63 bytes) H2O + Cl2 4.50 x 10+1 Pa-1 4.56 x 10+6 atm-1
CO + Cl2 larrow.GIF (55 bytes)rarrow.gif (63 bytes) COCl2 6.51 x 10+6 Pa-1 6.60 x 10+11 atm-1
3O2 larrow.GIF (55 bytes)rarrow.gif (63 bytes) 2O3 6.55 x 10-63 Pa-1 6.64 x 10-58 atm-1
PCl3 + Cl2 larrow.GIF (55 bytes)rarrow.gif (63 bytes) PCl5 3.29 x 10+1 Pa-1 3.33 x 10+6 atm-1
I2 + Cl2 larrow.GIF (55 bytes)rarrow.gif (63 bytes) 2ICl 8.19 x 10+1 8.19 x 10+1
CH4 + H2O larrow.GIF (55 bytes)rarrow.gif (63 bytes) CO + 3H2O 7.93 x 10+34 Pa+2 7.73 x 10+24 atm+2
2NO + Br2 larrow.GIF (55 bytes)rarrow.gif (63 bytes) 2NOBr 2.80 x 10-4 Pa-1 2.84 x 10+1 atm-1
SO2Cl2 larrow.GIF (55 bytes)rarrow.gif (63 bytes) SO2 + Cl2 2.95 x 10+8 Pa 2.91 x 10+3 atm

Notes to Table: The values in this table are calculated by the undersigned from table of thermodynamic properties of pure substances. The method of calculation and examples are given in another section. Constants have the nominal units given; those without units have no nominal units.

If the number of moles is the same for the products as it is for the reactants in the balanced stoichiometric equation, and the amount of substance per unit volume is measured in the same units for each of the products and reactants, then the nominal units in the numerator and denominator of the equilibrium constant expression will be identical and will divide out. The constant itself will have no nominal units. Moreover, its numeric value will then be independent of the units of the actual partial pressures inserted into the equilibrium constant expression as long as those units are themselves consistent.

If the number of moles is not the same for the products as it is for the reactants, or if all reactants and products are not measured in the same units, then the equilibrium constant cannot have identical units in the numerator and denominator. The constant itself then has nominal units, and its numeric value is not independent of those units. Care must be taken to ensure that the units of amount of substance per unit volume, or pressure, are consistent with the nominal units of the equilibrium constant whose value is being used.

Example. The equilibrium partial pressure of chlorine in a container which has a partial pressure of 100 kPa of HCl and 1.00 Pa of hydrogen can be calculated. The equilibrium is:

2HCl &lt rarrow.gif (63 bytes) H2 + Cl2.

Kp = 4.06 x 10-34 = p(H2)p(Cl2)/p2(HCl)

4.06 x 10-34 = (1.00 Pa)p(Cl2)/(1.00 x 10+5 Pa)2

4.06 x 10-34 = (1.00 Pa)p(Cl2)/1.00 x 10+10 Pa2

4.06 x 10-24 Pa2 = (1.00 Pa)p(Cl2)

4.06 x 10-24 Pa = p(Cl2)

Example. We can calculate the value of the equilibrium constant for the formation of the interhalogen compound ICl(g) from the gaseous elements, given that at 25oC a partial pressure of 50.0 kPa ICl(g) is found when the partial pressure of chlorine is 2.34 kPa and the partial pressure of I2(g) is 13.0 kPa. The equilibrium is:

I2 + Cl2 &lt rarrow.gif (63 bytes) 2ICl and therefore

Kp = p2(ICl)/p(I2)p(Cl2)

Kp = (50.0 x 10+3)2 Pa2/(2.34 x 10+3)(13.0 x 10+3) Pa2

Kp = 82.2; it has no nominal units. The difference between this value and the calculated thermodynamic value, 81.9, is well within the precision to which the thermodynamic value is known.

In many studies of chemical equilibria, it is either inconvenient or impractical to measure the equilibrium concentrations, partial pressures, or activities of all of the substances involved in the equilibrium. When this happens, chemists make use of stoichiometric reactions to supply additional chemical information.

Example. We can calculate the equilibrium partial pressures of PCl3 and Cl2 when an equilibrium partial pressure of PCl5 of 86.43 kPa is found after injection of PCl5 into a sample container at 25oC. The equilibrium is:

PCl3 + Cl2 &lt rarrow.gif (63 bytes) PCl5 and therefore

Kp = 3.29 x 10+1 Pa-1 = p(PCl5)/p(PCl3)p(Cl2)

Since the only source of either PCl3 or Cl2 is the original PCl5, and each mole of PCl5 which dissociates gives one mole of PCl3 and one mole of Cl2, we know that p(Cl2) = p(PCl3).

3.29 x 10+1 Pa-1 = 86.43 x 10+3 Pa/p2(PCl3)

p2(PCl3) = 86.43 x 10+3 Pa/3.29 x 10+1 Pa-1 = 2.627 x 10+3 Pa2

p(PCl3) = 51.3 Pa = p(Cl2)

Example. At 25oC, the gas-phase equilibrium constant for the formation of water vapor from hydrogen and oxygen has the value 1.262 x 10+85 Pa. Let us compute the partial pressure of hydrogen and of oxygen found at equilibrium when the pressure of water vapor is 101325 Pa, which is one standard atmosphere. The equilibrium is:

2H2 + O2 &lt rarrow.gif (63 bytes) 2H2O and therefore

1.262 x 10+85 Pa = p2(H2O)/p2(H2)p(O2)

If the only source of hydrogen and oxygen is water then from the stoichiometry of the equilibrium reaction p(H2) must be twice p(O2), p2(H2) = 4p2(O2), and

4p3(O2) = (101325 Pa)2/(1.262 x 10+85 Pa-1)

p3(O2) = 2.0333 x 10-76 Pa3, p(O2) = 5.880 x 10-26 Pa

p(H2) = 11.76 x 10-26 Pa, p(O2) = 5.88 x 10-26 Pa

The partial pressures of hydrogen and of oxygen are so low as to be totally negligible under these conditions. Effectively, water vapor does not dissociate under these conditions.

Stoichiometric reactions are not the only additional information that chemists use in studying gas-phase equilibria. The total pressure of a gas mixture is readily measurable and can be used together with Dalton's law of partial pressures in equilibrium calculations.

Example. We can calculate the equilibrium partial pressure of N2O4(g) and of NO2(g) at 25oC in a mixture of the two gases whose total pressure is 1.30 atm. The equilibrium is:

N2O4(g) &lt rarrow.gif (63 bytes) 2NO2(g),

for which Kp = 0.1484 atm = p2(NO2)/p(N2O4). By Dalton's law of partial pressures, at a total pressure of 1.30 atm the sum of the partial pressures of the two gases must total 1.30 atm, so 1.30 = p(NO2) + p(N2O4).

0.1484 = p2(NO2)/(1.30 - p(NO2))

0.19291 - 0.1484p(NO2) - p2(NO2) = 0

This equation can be solved by use of the quadratic formula to give p(NO2) = 0.371 atm, which makes p(N2O4) = 0.929 atm.

Copyright 1997 James R. Fromm