#### xaktly | Chemistry | Thermodynamics

Chemical equilibrium

### A word about equilibrium

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

CH4 (g) + 2 O2 (g) → CO2 (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) ⇌ NH3 (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.

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 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:

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 kf and kr 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:

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

Now if we define the ratio of forward and backward rate constants to be the equilibrium constant, Keq, we have

$$K_{eq} = \frac{k_f}{k_r}$$

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

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

Notice that in the equilibrium constant expression, it's moles of product(s) divided by moles of reactant(s), each raised to the power of its numerical coefficient in the balanced reaction.

At a given temperature, the equilibrium constant will give us a ratio (though it might be a little complicated) of the components of a reaction mixture.

If Keq is large, then there are more products present than reactants at equilibrium.

If Keq is small then there are more reactants present than products at equilibrium.

We'll work through some examples of how to use Keq below.

#### The law of mass action

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 kf and kr are the rate constants of the reaction.

At chemical equilibrium, the forward and backward rates of reaction are the same. The rate constants are a property of each reaction at a given temperature (all reaction rates speed up with increasing temperature).

#### Equilibrium constant expression

For the general reaction

The equilibrium constant expression is

where [A] is the molar concentration of A, and a is the coefficient of A in the balanced reaction, and so on.

### Rate constants and equilibrium

Here are a few examples of ratios of kf to kr 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 Keq = kf/kr.

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

When the rate constants of the forward and reverse reactions are equal, the equilibrium constant, Keq = 1. In this situation, half of A is converted to B and a dynamic equilibrium is established keeping the concentrations of A and B equal, [A] = [B].

When the forward rate constant is twice the reverse rate constant, Keq = 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 forward rate constant is much larger (that's what >> means) than the backward rate constant (that is, Keq = 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 B that does become transformed into A is rapidly re-converted to B

When the reverse rate constant far exceeds the forward rate constant (Keq = 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.

#### About rate constants

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:

### 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 Keq can be expressed in terms of the forward and backward rate constants:

$$K_{eq} = \frac{k_f}{k_r}$$

These rate constants are not functions of concentration, so neither is Keq . In a reaction mixture at equilibrium,

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

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 Keq . 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 Keq , and that's just what we do.

The bottom line is:

The concentrations of pure liquids and solids are set to 1 in equilibrium constant expressions, or better yet, they are ignored.

### 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.

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 H2 (the wall is necessary to absorb excess energy present from radical formation), or (3) through the formation of the short-lived HO2 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.

### 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

H2 (g) + Br2 (g) ⇌ 2 HBr (g)       Keq = 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:

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 Keq 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 Keq expression, you'll indeed get 64.

### Example 2

Determine whether the reaction

2 SO2 (g) + O2 (g) ⇌ 2 SO3 (g)     Keq = 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 Keq = 4.36. If Q < Keq, the reaction needs to proceed to the right to reach equilibrium, and if Q > Keq, it has proceeded "too far," and needs to move to the left in order to achieve equilibria.

Now Q < Keq, 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 Keq, so we conclude that the reaction has to move to the right.

### 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:

COCl2 (g) ⇌ CO (g) + Cl2 (g)       Keq = 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 PCO 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

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.

#### An approximation

We could have made a handy approximation in the equilibrium constant expression above, by noting that Keq is small, Keq = 0.0041, so the equilibrium amounts of CO and Cl2 should be small, much smaller than the initial pressure of COCl2, 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.

#### A note on nomenclature: Keq, Kc & Kp

When the equilibrium constant is measured and used in relation to molar concentrations, it's often referred to as Kc, c for "concentration." When partial pressures are used, Kp is frequently used. There is no real difference between the three notations; only the context is different.

### Practice problems

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

##### * If no initial concentration is given, assume it is zero.
 1 H2 + I2 ⇌ 2 HI [H2]o =[I2]o = 0.25 M Keq = 64.0 2 PCl3 + Cl2 ⇌ PCl5 [PCl5]o = 0.90 M Keq = 16.0 3 COCl2 ⇌ CO + Cl2 [CO]o = 0.60 M[Cl2]o = 1.1 M Keq = 0.678 4 2 NOCl (g) ⇌ 2 NO2 (g) + Cl2 (g) [NOCl]o = 0.50 M Keq = 1.60 x 10-5

### 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

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
• Temperature

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### equilibria

Equilibria is the plural of "equilibrium," just like "data" is the plural of "datum."

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### aqueous

An aqueous solution is one in which the solvent is water (root = "aqua"). Typically, but not always, aqueous solutions are ionic salts dissolved in water.

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