A word about equilibrium

We are often in the habit of writing chemical reactions with a single arrow, like this:

CH_{4 (g)} + 2 O_{2 (g)} → CO_{2 (g)} + 2 H₂O _(g)

That's actually OK for that particular reaction, the combustion of methane, which runs to completion in the forward direction, and is very difficult to make happen in reverse (combining CO₂ and water to make CH₄).

The thermal decomposition of ammonium chloride, however,

NH₄Cl _(s) ⇌ NH_{3 (g)} + HCl _(g),

is quite reversible, and in fact at room temperature there is always some of both reactants and products around (you can smell the ammonia in a jar of ammonium chloride). That's why we write the double arrow, ⇌ .

In this section we'll explore the nature of that double arrow, something very important in many aspects of chemistry.

Examples of equilibrium

You are already familiar with many examples of equilibrium. A book sitting on a table is in equilibrium with the table. The force of gravity pulling down on the book is exactly matched by the repulsive force between the electrons in the outer atoms of the book and those of the outer atoms of the table.

A mass suspended on a spring is another example. Stretch or compress the spring and the mass will move up and down for a while, but with diminished amplitude (travel range) until it achieves equilibrium, the point at which the upward pull of the spring matches the downward pull of gravity on the mass and spring.

The law of mass action

The basis for making a mathematical model of chemical equilibrium is the law of mass action. It says that the rate of a chemical reaction is directly proportional to the product of the concentrations of the reactants, considering each instance (each mole) of a substance as a separate concentration (see below – it's not difficult).

Consider the model reaction:

$$aA + bB \; \rightleftharpoons \; cC + dD$$

In this reaction a moles of A and b moles of B react to form c moles of C, and d moles of D. C and D could be viewed as the "reactants" in the reverse reaction. If we let k_f and k_r be constants of proportionality (f = forward, r = reverse), then we can define the rates of the forward and reverse reactions like this:

forward rate $= k_f[A]^a[B]^b$

reverse rate $= k_r[C]^c[D]^d$

Now at equilibrium, the key point is that the rates of the forward and backward reactions are the same – i.e. there's no further accumulation of products on either side. Therefore we can make this equality:

$$k_f [A]^a [B]^b = k_r [C]^c [D]^d$$

Rate constants and equilibrium

Here are a few examples of ratios of k_f to k_r and how they affect the equilibrium states of a reaction. Here we'll just consider a simple transformation, A ⇌ B, where [A] and [B] are the molar concentrations, as usual. In looking at these graphs, remember that K_eq = k_f/k_r.

For many important reactions, the rate constant of the reverse reaction, k_r, is very close to the rate constant of the forward reaction. In this case, the ratio is 0.5, or K_eq = 0.5. Once equilibrium is achieved, the ratio of the concentrations of B to A will be ½.

When the forward rate constant is twice the reverse rate constant, K_eq = 2, and more A will be converted to B than B converted back to A. Thus the concentration of A in the reaction mixture will be ½[B].

When the reverse rate constant far exceeds the forward rate constant (K_eq = 2 is small), the forward reaction is said to be kinetically limited. It just won't go (or is just too slow to observe) because any B formed would be immediately reverted to A. Such reactions may require a catalyst to proceed at all.

The concentrations of liquids and solids

When working with chemical equilibria, we're mostly going to be working with aqueous equilibria – reactions run in aqueous solution, and gas-phase equilibria – reactions between gases that produce other gases. In the latter, we'll replace molar concentrations with partial pressures,

$$K_{eq} = \frac{[C]^c [D]^d}{[A]^a [B]^b} = \frac{P_C^c P_D^d}{P_A^a P_B^b}$$

more on that later. Certain components of all reaction mixtures, however, have essentially an infinite concentration; these are solids and liquids.

For example, in an aqueous reaction mixture, if there is a precipitate, then the concentration of that precipitate, a solid, is taken to be one.

Here's why: Recall that K_eq can be expressed in terms of the forward and backward rate constants:

Many reactions go only one-way

Many reactions are, for all practical purposes, one-way reactions. One reason for this is that some simple-looking reactions are actually quite complex, consisting of many steps.

Consider the combustion of hydrogen to form water:

On the surface, that's a pretty simple reaction. It's the one that some rockets use for propulsion because it's so exothermic. But the details (some of which are still not fully worked out) are more complicated.

We know that combustion reactions require a "spark" of some kind to get them started. That spark results in the creation of free-radicals – non-ionic free atoms of H and O. These are referred to as "chain initiation steps":

Those radicals then participate in "chain branching steps," in which new radical O and H atoms are formed, each of which can go on to form more:

In those steps the OH is a radical, too. Those radicals go on to participate in the "chain propagation steps," in which water is formed.

Example 1

0.025 mol of H₂ and 0.025 mol of Br₂ are placed in a 4.0 L reaction vessel and heated to 700K. Calculate the concentration of each reactant and the product (HBr) at equilibrium

H_{2 (g)} + Br_{2 (g)} ⇌ 2 HBr _(g)       K_eq = 64.0 at 700K

Solution: Although these are gases, we can use the volume to calculate concentrations just as though it were an aqueous solution.

The first thing to do is to write out an equilibrium constant expression (make sure you have a balanced reaction, too, of course):

Now we calculate the concentrations. The H₂ concentration is:

and the Br₂ concentration will obviously be the same. These are initial concentrations, before the mixture comes to equilibrium. The initial concentration of HBr is [HBr] = 0.

Now the trick for most of these equilibrium problems is to build what we call an ICE table:

ICE: Initial - Change - Equilibrium

It's a way of organizing our information. For this problem it starts like this:

Now once equilibrium is established, we will have lost some H₂ and Br₂ to the formation of HBr. If we form x moles of HBr, we use x/2 moles of H₂ and Br₂. It might be easier, though, to make that 2x moles of HBr and x moles each of H₂ and Br₂. We put that into the table like this:

Example 2

Determine whether the reaction

2 SO_{2 (g)} + O_{2 (g)} ⇌ 2 SO_{3 (g)}     K_eq = 4.36

is at equilibrium if [SO₂] = 2.00 M, [O₂] = 1.50 M and [SO₃] = 1.25 M.

Solution: In this type of problem, we'll calculate a quantity, Q:

which is just the equilibrium constant expression for this reaction. We compare Q to K_eq = 4.36. If Q < K_eq, the reaction needs to proceed to the right to reach equilibrium, and if Q > K_eq, it has proceeded "too far," and needs to move to the left in order to achieve equilibria.

Example 3

Phosgene, COCl₂, is a toxic gas that was used as a nerve agent (a poison) in World War II. It decomposes at high temperature into two other toxic gases, CO and Cl₂:

COCl_{2 (g)} ⇌ CO _(g) + Cl_{2 (g)}       K_eq = 0.0041 at 600K

Calculate the equilibrium composition of the mixture of gases after 0.11 atm of COCl₂ is allowed to reach equilibrium at 600K.

Solution: In this example, we'll work with the partial pressures of the components. If you haven't worked with the gas laws and partial pressures before, just know that the equilibrium calculation is the same, just with different units (atmospheres [atm.] or Pascals [Pa] instead of molarity).

The equilibrium constant expression is:

where P_CO is the partial pressure of CO, and so on. We'll work with an ICE table again. It starts with what is known about the initial concentrations – before equilibrium is established.

The balanced decomposition equation tells us that for every mole of COCl₂ that decomposes, we gain 1 mole of Cl₂ and 1 mole of CO, so the rest of the table is:

Now we can plug those equilibrium pressures into the equilibrium expression to get

Practice problems

Calculate the concentrations of all reaction-mixture components given the balanced equation, initial concentrations*, and K_eq

* If no initial concentration is given, assume it is zero.

Equilibrium: macroscopic vs. microscopic

There is an important difference between the equilibrium you see when you look at a beaker or measure the temperature of a gaseous reaction mixture, and what's going on atom-by-atom, molecule-by-molecule at the microscopic level.

Macroscopically, a chemical system at equilibrium is static – unchanging. The concentrations of all components of the system are constant, and other measurable properties, such as pressure and temperature are constant, too. But that's not necessarily so at the microscopic level. On the level of atoms and molecules, we expect tiny fluctuations to be occurring always.

For the general reaction