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Equilibrium Constants

Author: John Hutchinson

It was noted above that the equilibrium partial pressures of the gases in a reaction vary depending upon a variety of conditions. These include changes in the initial numbers of moles of reactants and products, changes in the volume of the reaction flask, and changes in the temperature. We now study these variations quantitatively.

Consider first the reaction here. Following on our previous study of this reaction, we inject an initial amount of N2O4(g) into a 100L reaction flask at 298K. Now, however, we vary the initial number of moles of N2O4(g) in the flask and measure the equilibrium pressures of both the reactant and product gases. The results of a number of such studies are given here.

Table 1: Equilibrium Partial Pressures in Decomposition Reaction
Initial n(N2O4) p(N2O4) [atm] p(NO2) [atm]
0.10.007640.033627
0.50.0710110.102517
10.1661360.156806
1.50.267350.198917
20.3717910.234574
2.50.4783150.266065
30.5863270.294578
3.50.6954720.320827
40.8055170.345277
4.50.9162970.368255
51.0276950.389998

We might have expected that the amount of NO2 produced at equilibrium would increase in direct proportion to increases in the amount of N2O4 we begin with. table 1 shows that this is not the case. Note that when we increase the initial amount of N2O4 by a factor of 10 from 0.5 moles to 5.0 moles, the pressure of NO2 at equilibrium increases by a factor of less than 4.

The relationship between the pressures at equilibrium and the initial amount of N2O4 is perhaps more easily seen in a graph of the data in table 1, as shown in figure 1. There are some interesting features here. Note that, when the initial amount of N2O4 is less than 1 mol, the equilibrium pressure of NO2 is greater than that of N2O4. These relative pressures reverse as the initial amount increases, as the N2O4 equilibrium pressure keeps track with the initial amount but the NO2 pressure falls short. Clearly, the equilibrium pressure of NO2 does not increase proportionally with the initial amount of N2O4. In fact, the increase is slower than proportionality, suggesting perhaps a square root relationship between the pressure of NO2 and the initial amount of N2O4.

Figure 1: Equilibrium Partial Pressures in Decomposition Reaction

We test this in figure 2 by plotting PNO2 at equilibrium versus the square root of the initial number of moles of N2O4. figure 2 makes it clear that this is not a simple proportional relationship, but it is closer. Note in figure 1 that the equilibrium pressure PN2O4 increases close to proportionally with the initial amount of N2O4. This suggests plotting PNO2 versus the square root of PN2O4. This is done in figure 3, where we discover that there is a very simple proportional relationship between the variables plotted in this way. We have thus observed that

[6]

where c is the slope of the graph. equation 6 can be rewritten in a standard form

[7]

To test the accuracy of this equation and to find the value of Kp, we return to table 1 and add another column in which we calculate the value of Kp for each of the data points. Table 2 makes it clear that the "constant" in equation 7 truly is independent of both the initial conditions and the equilibrium partial pressure of either one of the reactant or product. We thus refer to the constant Kp in equation 7 as the reaction equilibrium constant.

Figure 2: Relationship of Pressure of Product to Initial Amount of Reactant.
Figure 3: Equilibrium Partial Pressures.

Table 2: Equilibrium Partial Pressures in Decomposition Reaction
Initial n(N2O4) p(N2O4) [atm] p(NO2) [atm] Kp
0.10.007640.03360.148
0.50.07100.1020.148
10.1660.1560.148
1.50.2670.1980.148
20.3710.2340.148
2.50.4780.2660.148
30.5860.2940.148
3.50.6950.3200.148
40.8050.3450.148
4.50.9160.3680.148
51.0270.3890.148

It is very interesting to note the functional form of the equilibrium constant. The product NO2 pressure appears in the numerator, and the exponent 2 on the pressure is the stoichiometric coefficient on NO2 in the balanced chemical equation. The reactant N2O4 pressure appears in the denominator, and the exponent 1 on the pressure is the stoichiometric coefficient on N2O4 in the chemical equation.

We now investigate whether other reactions have equilibrium constants and whether the form of this equilibrium constant is a happy coincidence or a general observation. We return to the reaction for the synthesis of ammonia.

In a previous section, we considered only the equilibrium produced when 1 mole of N2 is reacted with 3 moles of H2. We now consider a range of possible initial values of these amounts, with the resultant equilibrium partial pressures given in table 3. In addition, anticipating the possibility of an equilibrium constant, we have calculated the ratio of partial pressures given by:

Kp = PNH32PN2PH23 [8]

In table 3, the equilibrium partial pressures of the gases are in a very wide variety, including whether the final pressures are greater for reactants or products. However, from the data in table 3, it is clear that, despite these variations, Kp in equation 8 is essentially a constant for all of the initial conditions examined and is thus the reaction equilibrium constant for this reaction.

Table 3: Equilibrium Partial Pressures of the Synthesis of Ammonia
V [L] n(N2) n(H2) p(N2) p(H2) p(NH3) Kp
10130.03420.10274.826.2105
100.10.30.01070.03220.4676.0105
1000.10.30.003230.009680.04256.1105
100330.4920.008800.4836.1105
100130.01070.03220.4676.0105
10001.51.50.02550.003150.02236.2105

Studies of many chemical reactions of gases result in the same observations. Each reaction equilibrium can be described by an equilibrium constant in which the partial pressures of the products, each raised to their corresponding stoichiometric coefficient, are multiplied together in the numerator, and the partial pressures of the reactants, each raised to their corresponding stoichiometric coefficient, are multiplied together in the denominator. For historical reasons, this general observation is sometimes referred to as the Law of Mass Action.




Last Update: 2011-04-07