Reaction Equilibrium

Author: John Hutchinson

We begin by analyzing a significant industrial chemical process, the synthesis of ammonia gas, NH₃, from nitrogen and hydrogen:

N2(g)+3H2(g) 2NH3(g)

[2]

If we start with 1 mole of N₂ and 3 moles of H₂, the balanced equation predicts that we will produce 2 moles of NH₃. In fact, if we carry out this reaction starting with these quantities of nitrogen and hydrogen at 298K in a 100.0L reaction vessel, we observe that the number of moles of NH₃ produced is 1.91 mol. This "yield" is less than predicted by the balanced equation, but the difference is not due to a limiting reagent factor. Recall that, in stoichiometry, the limiting reagent is the one that is present in less than the ratio of moles given by the balanced equation. In this case, neither N₂ nor H₂ is limiting because they are present initially in a 1:3 ratio, exactly matching the stoichiometry. Note also that this seeming deficit in the yield is not due to any experimental error or imperfection, nor is it due to poor measurements or preparation. Rather, the observation that, at 298K, 1.91 moles rather than 2 moles are produced is completely reproducible: every measurement of this reaction at this temperature in this volume starting with 1 mole of N₂ and 3 moles of H₂ gives this result. We conclude that the reaction achieves reaction equilibrium in which all three gases are present in the gas mixture. We can determine the amounts of each gas at equilibrium from the stoichiometry of the reaction. When nNH₃ = 1.91 mol are created, the number of moles of N₂ remaining at equilibrium is nN₂ = 0.045 mol and nH₂ = 0.135 mol.

It is important to note that we can vary the relative amount of NH₃ produced by varying the temperature of the reaction, the volume of the vessel in which the reaction occurs, or the relative starting amounts of N₂ and H₂. We shall study and analyze this observation in detail in later sections. For now, though, we demonstrate that the concept of reaction equilibrium is general to all reactions.

Consider the reaction

H2(g)+I2(g) 2HI(g)

[3]

If we begin with 1.00 mole of H₂ and 1.00 mole of I₂ at 500K in a reaction vessel of fixed volume, we observe that, at equilibrium, nHI=1.72mol, leaving in the equilibrium mixture nH₂ = 0.14 mol and nI₂ = 0.14 mol.

Similarly, consider the decomposition reaction

N2O4(g) 2NO2(g)

[4]

At 298K in a 100.0L reaction flask, 1.00 mol of N₂O₄ partially decomposes to produce, at equilibrium, nNO₂=0.64mol and nN₂O₄=0.68mol.

Some chemical reactions achieve an equilibrium that appears to be very nearly complete reaction. For example,

H2(g)+Cl2(g) 2HCl(g)

[5]

If we begin with 1.00 mole of H₂ and 1.00 mole of Cl₂ at 298K in a reaction vessel of fixed volume, we observe that, at equilibrium, nHCl is almost exactly 2.00 mol, leaving virtually no H₂ or Cl₂. This does not mean that the reaction has not come to equilibrium. It means instead that, at equilibrium, there are essentially no reactants remaining.

In each of these cases, the amounts of reactants and products present at equilibrium vary as the conditions are varied but are completely reproducible for fixed conditions. Before making further observations that will lead to a quantitative description of the reaction equilibrium, we consider a qualitative description of equilibrium.

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 postulate that at equilibrium the forward reaction rate is equal to the reverse reaction rate. For example, in the reaction here, the rate of decomposition of N₂O₄ molecules at equilibrium must be exactly matched by the rate of recombination (or dimerization) of NO₂ molecules.

To show that the forward and reverse reactions continue to happen at equilibrium, we start with the NO₂ and N₂O₄ mixture at equilibrium and we vary the volume of the flask containing the mixture. We observe that, if we increase the volume and the reaction is allowed to come to equilibrium, the amount of NO₂ at equilibrium is larger at the expense of a smaller amount of N₂O₄. We can certainly conclude that the amounts of the gases at equilibrium depend on the reaction conditions. However, if the forward and reverse reactions stop once the equilibrium amounts of material are achieved, the molecules would not "know" that the volume of the container had increased. 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 N₂O₄) 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.