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Equilibrium partial pressures in reaction (4)
Initial n(N 2 O 4 ) P(N 2 O 4 ) (atm) P(NO 2 ) (atm) K p
0.1 0.00764 0.0336 0.148
0.5 0.0710 0.102 0.148
1 0.166 0.156 0.148
1.5 0.267 0.198 0.148
2 0.371 0.234 0.148
2.5 0.478 0.266 0.148
3 0.586 0.294 0.148
3.5 0.695 0.320 0.148
4 0.805 0.345 0.148
4.5 0.916 0.368 0.148
5 1.027 0.389 0.148

Recall that we have seen a reaction equilibrium constant before in our study of solubility. The solubility product K sp is one example of a reaction equilibrium constant. Remember that, in the expression for K sp , each concentration was raised to a power equal to its stoichiometric coefficient in the balanced chemical equation.

It is very interesting to compare this to the functional form of the equilibrium constant in Equation (9). The product NO 2 pressure appears in the numerator, and the exponent 2 on the pressure is the stoichiometric coefficient on NO 2 in the balanced chemical equation. The reactant N 2 O 4 pressure appears in the denominator, and the exponent 1 on the pressure is the stoichiometric coefficient on N 2 O 4 in the chemical equation. This is very similar to what we observed in solubility equilibrium, except that we now have the added factor of the reactant appearing in the denominator.

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, reaction (2):

N 2 (g) + 3 H 2 (g) → 2 NH 3 (g)

In section II, 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 Equation 10:

K p = P N H 3 2 P N 2 · P H 2 3

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, K p in Equation (10) is essentially a constant for all of the initial conditions examined and is thus the “reaction equilibrium constant” for Reaction (2).

Equilibrium partial pressures of reaction (2)
V(L) n(N 2 ) n(H 2 ) P(N 2 ) P(H 2 ) P(NH 3 ) K p
10 1 3 0.0342 0.1027 4.82 6.2·10 5
10 0.1 0.3 0.0107 0.0322 0.467 6.0·10 5
100 0.1 0.3 0.00323 0.00968 0.0425 6.1·10 5
100 3 3 0.492 0.00880 0.483 6.1·10 5
100 1 3 0.0107 0.0322 0.467 6.0·10 5
1000 1.5 1.5 0.0255 0.00315 0.0223 6.2·10 5

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.”

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Source:  OpenStax, Concept development studies in chemistry 2013. OpenStax CNX. Oct 07, 2013 Download for free at http://legacy.cnx.org/content/col11579/1.1
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