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 H_{2}O_{ (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 CO_{2} and water to make CH_{4}).

The thermal decomposition of ammonium chloride, however,

NH_{4}Cl_{ (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.

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.

Another example comes from the flow of heat from hotter to cooler objects. Absent any other places for the heat to go, a 50˚C block of metal placed in contact with a 0˚C block will lose heat to the cooler block until both are at 25˚C, the equilibrium temperature.

You can probably come up with many more examples of these kinds of physical equilibria.

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:

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

Now at equilibrium, the key point is that the rates of the forward and backward reactions are the same. Therefore we can make this equality:

Now if we define the ratio of forward and backward rate constants to be the **equilibrium constant**, **K _{eq}**, we have

And finally, the equilibrium constant expression in terms of the concentrations of A, B, C and D is

Notice that in the equilibrium constant expression, it's moles of product(s) divided by moles of reactant(s).

At a given temperature, the equilibrium constant will give us the ratio of the components of a reaction mixture.

If **K _{eq}** is large, then there are more products present than reactants at equilibrium.

If **K _{eq}** is small then there are more reactants present than products at equilibrium.

We'll work through some examples of how to use **K _{eq}** below.

For a generic chemical reaction

the rate of the reaction in the forward direction (left to right) is

and the rate for the reverse reaction is

where **k _{f}** and

**chemical equilibrium**, the forward and backward rates of reaction are the same. The *rate constants* are a property of each reaction; they and the concentrations of reactant(s) and products(s) may differ.

For the general reaction

The equilibrium constant expression is

where **[A]** is the molar concentration of **A**, and so on.

Here are a few examples of ratios of **k _{f}** to

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

When the rate constants of the forward and reverse reactions are equal, the equilibrium constant, **K _{eq} = 1**. In this situation, half of

When the forward rate constant is twice the reverse rate constant, **K _{eq} = 2**, and more

When the forward rate constant is much larger (that's what >> means) than the backward rate constant (that is, **K _{eq} = 2** is large) there will be no appreciable reverse reaction. This is the case for many reactions we perceive to be "one-way" in the forward direction. Any

When the reverse rate constant far exceeds the forward rate constant (**K _{eq} = 2** is small), the forward reaction is said to be

We've kind of glossed over rate constants as simple constants of proportionality in the law of mass action. Mathematically, that's true, but there is so much more to them. Rate constants, rate "laws," reaction rates and the Gibbs energy are all intertwined. These relationships will be covered in later sections:

- Le Chatelier's Principle
- Equilibrium and free energy
- Temperature dependence of K
_{eq} - Kinetics

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 of gases that produce gases. In the latter, we'll replace molar concentrations with **partial pressures**,

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:

These rate constants are not functions of concentration, so neither is **K _{eq} **. In a reaction mixture at equilibrium,

is true no matter what the concentration of any one component. If the mixture is at equilibrium, the mass-action quotient will yield the same **K _{eq} **. Now some concentrations, like those of liquids or solids, never change (or we might agree that if you need some, it's always there, there's so much of it in that condensed state). We might as well just absorb those things that don't change right into

The bottom line is:

*ignored*.

Many reactions are, for all practical purposes, one-way reactions. One reason for that 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.

Somehow it all has to stop, and it does when (1) the reactants are gone, (2) two H radicals collide together at the wall to reform H_{2} (the wall is necessary to absorb excess energy present from radical formation), or (3) through the formation of the short-lived HO_{2} radical. In the last equation M represents any third atom or molecule that can serve to carry away excess energy from the formation of the H radical.

Now I don't mean for you to understand all of that just now, only to realize that some reaction mechanisms can be quite complex.

Now imagine reversing all of those steps and you get an idea of why the forward rate of this combustion reaction is so high compared to the reverse.

A few more examples of one-way reactions are:

- The cooking of food, in which heat, through a complex series of steps, breaks and rearranges bonds.
- The rusting of metal, which is another oxidation process, just much slower than burning. The iron oxides that form in the rusting process are more stable than iron.
- The neutralization of a strong acid with a base to form a salt and water.

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 H_{2} concentration is:

and the Br_{2} 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: I**nitial - **C**hange - **E**quilibrium

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 H_{2} and Br_{2} to the formation of HBr. If we form **x** moles of HBr, we use **x/2** moles of H_{2} and Br_{2}. It might be easier, though, to make that **2x** moles of HBr and **x** moles each of H_{2} and Br_{2}. We put that into the table like this:

Notice that we subtract **x** from the initial concentrations of reactants, and add the **2x** to the initial concentration of the product. Now at equilibrium, we simply sum the first two columns:

Now we can plug those equilibrium concentrations into the K_{eq} expression and solve for **x**:

The left side is a perfect square,

so we can just take the square root of both sides

Multiplication on both sides by **[0.00625 - x]**, and some grouping and rearrangement gives:

So the final concentrations (.00625 - x and 2x) are:

The ICE table is a nice way to organize and solve these problems; it's worth using. It's also worth checking an answer like this. If you plug these concentrations back into the K_{eq} expression, you'll indeed get 64.

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.

Now Q < K_{eq}, so the reaction has not yet reached equilibrium. The concentrations of the reactants are too large and the concentration of the product is too small for Q to equal K_{eq}, so we conclude that the reaction has to move to the *right*.

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 COCl_{2} that decomposes, we gain 1 mole of Cl_{2} 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

The resulting quadratic equation can be solved by completing the square or using the quadratic formula

If x = 0.193 atm, then the equilibrium partial pressures of all components are.

We could have made a handy approximation in the equilibrium constant expression above, by noting that Keq is small, K_{eq} = 0.0041, so the equilibrium amounts of CO and Cl_{2} should be small, much smaller than the initial pressure of COCl_{2}, 0.11 atm, thus we might just ignore it.

If we'd done that, the resulting equation and solution would have been

which is only about a 10% error. With smaller equilibrium constants, it's a better approximation. Very often it's a useful approximation to make. On the other hand, it's not that difficult to solve the quadratic formula on a calculator.

When the equilibrium constant is measured and used in relation to molar concentrations, it's often referred to as **K _{c}**,

1. | H_{2} + I_{2} ⇌ 2 HI |
[H_{2}]_{o} =[I _{2}]_{o} = 0.25 M |
K_{eq} = 64.0 |

2. | PCl_{3} + Cl_{2} ⇌ PCl_{5} |
[PCl_{5}]_{o} = 0.90 M |
K_{eq} = 16.0 |

3. | COCl_{2} ⇌ CO + Cl_{2} |
[CO]_{o} = 0.60 M[Cl _{2}]_{o} = 1.1 M |
K_{eq} = 0.678 |

4. | 2 NOCl_{ (g)} ⇌ 2 NO_{2 (g)} + Cl_{2 (g)} |
[NOCl]_{o} = 0.50 M |
K_{eq} = 1.60 x 10^{-5} |

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 *micro*scopic 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

at the microscopic level there can be small fluctuations of the concentrations of each component at equilibrium, but those fluctuations always average to the equilibrium concentrations given enough time – and we're talking about very short times, here.

Many factors contribute to the microscopic behavior of reactants and products in a mixture, including

- Intermolecular forces
- Orientation of particles during a collision
- Relative size and speed of particles
- Overall concentration of the solution or total pressure of the gas

This site is a one-person operation. If you can manage it, I'd appreciate anything you could give to help defray the cost of keeping up the domain name, server, search service and my time.

**xaktly.com** by Dr. Jeff Cruzan is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. © 2016, Jeff Cruzan. All text and images on this website not specifically attributed to another source were created by me and I reserve all rights as to their use. Any opinions expressed on this website are entirely mine, and do not necessarily reflect the views of any of my employers. Please feel free to send any questions or comments to jeff.cruzan@verizon.net.