0.3 Reaction equilibrium in the gas phase  (Page 3/6)

This reasoning reveals that the amounts of reactant and product present at equilibrium are determined by therates of the forward and reverse reactions. If the rate of theforward reaction ( e.g. decomposition of $N_2{O}_4$ ) is faster than the rate of the reverse reaction, then atequilibrium we have more product than reactant. If that difference in rates is very large, at equilibrium there will be much moreproduct than reactant. Of course, the converse of these conclusions is also true. It must also be the case that the rates of theseprocesses depends on, amongst other factors, the volume of the reaction flask, since the amounts of each gas present atequilibrium change when the volume is changed.

Observation 2: equilibrium constants

It was noted above that the equilibrium partial pressures of the gases in a reaction vary depending upon avariety of conditions. These include changes in the initial numbers of moles of reactants and products, changes in the volume of thereaction 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 $N_2{O}_4\left(g\right)$ into a 100L reaction flask at 298K. Now, however, we vary the initial number of moles of $N_2{O}_4\left(g\right)$ in the flask and measure the equilibrium pressures of both the reactant and product gases. The results of a number of such studiesare given here .

We might have expected that the amount of $N{O}_2$ produced at equilibrium would increase in direct proportion to increases in the amount of $N_2{O}_4$ we begin with. shows that this is not the case. Note that when we increase the initial amountof $N_2{O}_4$ by a factor of 10 from 0.5 moles to 5.0 moles, the pressure of $N{O}_2$ at equilibrium increases by a factor of less than 4.

The relationship between the pressures at equilibrium and the initial amount of $N_2{O}_4$ is perhaps more easily seen in a graph of the data in , as shown in . There are some interesting features here. Note that, when the initial amount of $N_2{O}_4$ is less than 1 mol, the equilibrium pressure of $N{O}_2$ is greater than that of $N_2{O}_4$ . These relative pressures reverse as the initial amount increases,as the $N_2{O}_4$ equilibrium pressure keeps track with the initial amount but the $N{O}_2$ pressure falls short. Clearly, the equilibrium pressure of $N{O}_2$ does not increase proportionally with the initial amount of $N_2{O}_4$ . In fact, the increase is slower than proportionality, suggestingperhaps a square root relationship between the pressure of $N{O}_2$ and the initial amount of $N_2{O}_4$ .

We test this in by plotting $P_{N{O}_2}$ at equilibrium versus the square root of the initial number of moles of $N_2{O}_4$ . makes it clear that this is not a simple proportional relationship, but it is closer. Note in that the equilibrium pressure $P_{N_2{O}_4}$ increases close to proportionally with the initial amount of $N_2{O}_4$ . This suggests plotting $P_{N{O}_2}$ versus the square root of $P_{N_2{O}_4}$ . This is done in , where we discover that there is a very simple proportional relationshipbetween the variables plotted in this way. We have thus observed that