Gas-Phase Equilibria: Effects of Pressure and Temperature

James Richard Fromm


Effect of Total Pressure

The effect of changing the total pressure on a gas-phase equilibrium will depend upon the nature of the equilibrium reaction. If the number of moles of gas is not changed by the equilibrium reaction then no significant effect will be observed. If, however, the number of moles of gas is increased by the equilibrium reaction then the equilibrium will be driven backwards by the increased pressure as it reacts to minimize the effect of the pressure increase. If the number of moles of gas is decreased by the equilibrium reaction then the equilibrium will be driven forwards by the increased pressure; it will again react to minimize the effect of the pressure increase.

When the equilibrium reaction involves no change in the number of moles of gas, the relative partial pressures of the gases do not change when the total pressure changes.


Example. We calculate the equilibrium partial pressures of hydrogen, iodine, and hydrogen iodide at a total pressure of 20 kPa and compare them with the equilibrium partial pressures at 200 kPa.

The equilibrium constant is Kp = 617 = p(H2p(I2/p2(HI). Assuming stoichiometric ratios, p(H2 = p(I2). Substitution of this into the above equation simplifies it to 617 = p2(H2)/p2(HI). Taking the square root of both sides of this equation gives 24.84 = p(H2)/p(HI).

At 20 kPa, 20000 = p(HI) + 2p(H2). Solution of this equation for either partial pressure and substituting it into the equation above will give an equation whose solution gives p(H2) = p(I2 = 9803 Pa, so p(HI) is 395 Pa. The partial pressure of hydrogen (or iodine) is 4.0% of the partial pressure of hydrogen iodide.

At 200 kPa, 200000 = p(HI) + 2p(H2). The procedure of the above paragraph now gives p(H2 = p(I2) = 98.03 kPa, so p(HI) is 3.95 kPa. The partial pressure of hydrogen (or iodine) is again 4.0% of the partial pressure of hydrogen iodide.


When the equilibrium reaction does involve a change in the number of moles of gas, then the relative partial pressures of the gases do change when the total pressure changes. An increase in the total pressure will favor the formation of fewer moles of gas.


Example. The equilibrium constant for the decomposition of the dimer of nitrogen dioxide is 1.48 x 10+4 Pa for the equilibrium N2O4(g) rarrow.gif (63 bytes) 2NO2(g) at 25oC.

At a total pressure of 20 kPa, using Dalton's law of partial pressures,

14800 Pa = p2(NO2)/p(N2O4) and p(N2O4) = 20000 - p(NO2), so

p2(NO2) + 14800 p(NO2) - 2.96 x 10+8 = 0

Solution of this equation by means of the quadratic formula gives p(NO2) = 11.33 kPa, so p(N2O4) = 20.00 - 11.33 = 8.67 kPa.

At a total pressure of 200 kPa, using Dalton's law of partial pressures,

14800 Pa = p2(NO2)/p(N2O4) and p(N2O4) = 200000 - p(NO2),so

p2(NO2) + 14800 p(NO2) - 2.96 x 10+9 = 0

Solution of this equation by means of the quadratic formula gives p(NO2) = 47.5 kPa, so p(N2O4) = 200.0 - 47.5 = 152.5 kPa.

At a total pressure of 2000 kPa, using Dalton's law of partial pressures,

14800 Pa = p2(NO2)/p(N2O4) and p(N2O4) = 2000000 - p(NO2), so

p2(NO2) + 14800 p(NO2) - 2.96 x 1010 = 0

Solution of this equation by means of the quadratic formula gives p(NO2) = 164.6 kPa, so p(N2O4) = 2000 - 165 = 1835 kPa. The results of these calculations are summarized in the Table below. The value in square brackets was calculated as an example in a different section.


Effect of Increasing Pressure on N2O4 Dissociation
p (total) (kPa) p (NO2) (kPa) p (N2O4) (kPa) NO2 (% N2O4) NO2 (% total)
20 11.33 8.67 132 57
[1.30 atm] [0.37 atm] [0.93 atm] 40 28
200 47.5 152.5 31 24
2000 165 1835 9 8

These calculations, a quantitative result of the use of an equilibrium constant, explain the qualitative predictions of the Principle of Le Chatelier which were presented qualitatively in our earlier discussion of how the equilibrium constant explains Le Chatelier's Principle. These sections should be at least reviewed now if they have not previously been studied.

Effect of Temperature

The effect of temperature on the position of any equilibrium, including a gas-phase equilibrium, occurs through a change in the value of the equilibrium constant. The effect of a temperature change is usually significant, or in other words the value of an equilibrium constant is unlikely to remain constant as the temperature is changed, and changes of several orders of magnitude can be expected over reasonable temperature ranges. The magnitude of the change in equilibrium constant with temperature can more easily be determined by calculations involving the free energy change of a reaction, but if only the standard enthalpy change of the reaction is available an estimate is still possible as shown below.

To a first approximation the value of the equilibrium constant is given by the van't Hoff equation:

log(K2/K1) = DH0(T2 - T1)/2.303...RT1T2

In this equation the value of the equilibrium constant at temperature T1 is K1 and the value of the same equilibrium constant at temperature T2 is K2. We will defer the derivation and discussion of the van't Hoff equation to another section. In general, the van't Hoff equation should be used only when no other data are available.


Copyright 1997 James R. Fromm