0.20 Reaction equilibrium in the gas phase  (Page 3/8)

We begin with a dynamic equilibrium description. We know from our studies of phase transitions that equilibrium occurs when the rate of the forward process (e.g. evaporation) is matched by the rate of reverse process (e.g. condensation). Since we have now observed that gas reactions also come to equilibrium, we should imagine that at equilibrium the forward reaction rate is equal to the reverse reaction rate. For example, in Reaction (4) above, the rate of decomposition of N ₂ O ₄ molecules at equilibrium must be exactly matched by the rate of recombination (or “dimerization”) of NO ₂ molecules.

How can we show that this model is correct and the forward and reverse reactions continue to happen at equilibrium? We can do as we did in the case of phase equilibrium by varying the volume. With the NO ₂ /N ₂ O ₄ mixture at equilibrium, we can vary the volume of the flask containing the mixture and determine what happens to the equilibrium. If we increase the volume and the reaction is allowed to come to equilibrium, we observe that the amount of NO ₂ at equilibrium is larger and the amount of N ₂ O ₄ is smaller. This shows that the amounts of the gases at equilibrium depend on the reaction conditions, including the volume. However, imagine that the forward and reverse reactions stop once the equilibrium amounts of material are achieved. If this were the case, the molecules would not “know” that the volume of the container had increased and there would be no change in the amounts of gas at equilibrium. Since the reaction equilibrium can and does respond to a change in volume, it must be that the change in volume affects the rates of both the forward and reverse processes. This means that both reactions must be occurring at equilibrium, and that their rates must exactly match at equilibrium.

This reasoning reveals that the amounts of reactant and product present at equilibrium are determined by the rates of the forward and reverse reactions. If the rate of the forward reaction (e.g. decomposition of N ₂ O ₄ ) is faster than the rate of the reverse reaction, then at equilibrium we have more product than reactant. If that difference in rates is very large, at equilibrium there will be much more product than reactant. Of course, the converse of these conclusions is also true. It must also be the case that the rates of these processes depends on, amongst other factors, the volume of the reaction flask, since the amounts of each gas present at equilibrium change when the volume is changed.

Since we have studied how the rates of reactions depend on concentrations, we should be able to quantify these effects at equilibrium. First, however, we quantify the equilibrium itself.

Observation 2: equilibrium constants

We observed 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.