# General Chemistry/Chemical Equilibria/Equilibrium

← Chemical Equilibria ·Chemical Equilibria/Le Chatelier's Principle →
Chemical equilibrium occurs when a reversible reaction is occurring backwards and forwards at the same time by the same amount. It is the balancing point of a chemical reaction, when it seems to stop happening. Although some reactions (like the combustion of propane) occur to completion (no backwards reaction), most reactions occur in both the forward and backward direction. One reaction will occur at a faster rate than the other, causing a net result. As the reaction progresses, the forward reaction slows down and the backward reaction speeds up. When the rates are equal, equilibrium has occurred. The reaction appears to have stopped, and the total amount of products and reactants remains stable.

## The Equilibrium Constant

The ratio of products to reactants, or $K_{eq}$ is known as the equilibrium constant. For the generic reaction mA + nB → xC + yD, the equilibrium constant is
$K_{eq}={\frac {[C]^{x}[D]^{y}}{[A]^{m}[B]^{n}}}$
where [X] detonates the activity of X. The activity of X is:
* its concentration if X is a gas or in a solution
* 1 if X is a pure solid or pure liquid
In other words, pure solids and liquids don't affect the equilibrium constant, as long as there is enough for the reaction to proceed. Their activity is 1, so they don't need to be written in the equilibrium constant.

### K$_p$ and K$_c$

There are two types of $K_{eq}$. One is $K_{c}$, and the activities are concentrations. You are already familiar with this expression; it is the normally-used $K_{eq}$. For gaseous reactions, you may use the concentration equilibrium. You may also use partial pressures instead of concentrations. This expression is denoted $K_{p}$. The activities are partial pressures instead of concentrations. This can only be used when all products and reactants are in the gaseous phase.
To convert between $K_{p}$ and $K_{c}$, there is an equation:
$K_{p}=K_{c}(RT)^{\Delta n}$
R is the Universal Gas Constant and T is the temperature at which the reaction is occurring. $\Delta n$ is the change (in moles) of gas molecules between products and reactants. It may be zero (in which case K$_p$ = K$_c$).

### Examples
#### Example 1

15.0 moles of X and 20.0 moles of Y are put into an empty 2.0 liter container. They react, and eventually reach equilibrium with 7.0 moles of Z according to the hypothetical equation:
${\hbox{X}}+2{\hbox{Y}}\to {\hbox{Z}}$
Calculate the equilibrium constant.
Solution: Because the container was originally empty and at equilibrium, all 7.0 moles of Z are produced by the reaction. Therefore, 7.0 moles of X and 14.0 moles of Y are consumed in the reaction, so at equilibrium there remain 8.0 moles of X, 6.0 moles of Y, and 7.0 moles of Z. Divide these values by 2.0 liters to determine the concentration, then substitute into the equilibrium expression:
$K_{eq}={\frac {[Z]}{[X][Y]^{2}}}={\frac {[3.5]}{[4.0][3.0]^{2}}}=0.097$

#### Example 2

5.00 moles of hydrogen are put into an empty 1.00-L container with 5.00 moles of iodine, where they react according to this equation:
${\hbox{H}}_{2}+{\hbox{I}}_{2}\to 2{\hbox{HI}}$
The equilibrium constant is 51.5 at the temperature of the reaction. Calculate the number of moles of HI that will form.
Solution: Suppose 2y moles of HI forms. Then, y moles of H$_2$ and I$_2$ are consumed. Therefore, at equilibrium, there are 2y moles of HI, and (5.00 - y) moles of both H$_2$ and I$_2$. Dividing by 1.00 liters (to find concentration) and writing an equilibrium expression gives:
$51.5={\frac {[2y]^{2}}{[5.00-y][5.00-y]}}$
Solving this equation gives y = 3.91 (reject the other solution, as it is greater than 5.00). At equilibrium, there are 7.82 mol of HI.

## Calculations With K

The exact value of K doesn't really matter. What is important is its magnitude.
| K > 1 | Reactions favors products |
| K < 1 | Reaction favors reactants |
If the reaction favors products, it will occur in the forward (left-to-right) direction. If K is very large, the reaction will occur mostly to completion, using up almost all the reactants.
If the reaction favors reactants, it will occur in the reverse (right-to-left) direction. If K is very small, the reaction will use up almost all the products and make them into reactants. The reverse reaction is favored.

### Reaction Arithmetic

If you reverse the reaction (changing the direction of the arrow, or flipping the left and right sides), the new value of K$_{eq}$ will be its reciprocal.
| A → B | (K = x) |
| B → A | (K = 1/x) |
If you add two reactions, their equilibrium constants are multiplied.
| A → B G → H | (K = x) (K = y) |
| A + G → B + H | (K = xy) |
These basic principles can help you determine an unknown equilibrium constant by combining known constants.

### Q

K$_{eq}$ is only used when a reaction is in equilibrium. To find it, write its concentration (or partial pressure) expression, then plug in all the measured values. The resulting K can be used to predict other equilibrium positions.
If a reaction is not at equilibrium, you can still plug in the measured concentrations. Instead of calling this value K$_{eq}$, it is called Q.
| Q > K | The products must decrease and the reactants will increase. |
| Q = K | There is equilibrium. |
| Q < K | The reactants must decrease and the products will increase. |
So, if you know the equilibrium constant for a reaction, and you know all the concentrations, you can predict what direction the reaction will proceed.

### I-C-E Charts

Let's say you know K for a reaction, and you know some concentrations. You can calculate the final (equilibrium) concentrations using an "Initial-Change-Equilibrium" chart. Here is an example.
${\hbox{CO}}_{(g)}+{\hbox{H}}_{2}{\hbox{O}}_{(g)}\to {\hbox{CO}}_{2(g)}+{\hbox{H}}_{2(g)}$
At 25°C, K$_c$ = 23.2 for this reaction.
First, write the equilibrium expression:
$K_{c}={\frac {[H_{2}][CO_{2}]}{[CO][H_{2}O]}}$
In a sealed container of 2.00 L: 3 molar H$_2$O, 4 molar CO, 1 molar H$_2$ have been added. (Write the ICE chart):
| | CO | H$_2$O | CO$_2$ | H$_2$ |
| Initial | 2 | 1.5 | 0 | 0.5 |
| Change | -x | -x | +x | +x |
| Equilibrium | 2-x | 1.5-x | x | 0.5+x |
Now, plug in the "equilibrium" values into the K expression. Substitute all known values and solve for x. Then, substitute x for the final values to determine the equilibrium concentrations.
By substituting everything into the K expression, we have:
$23.3={\frac {(0.5+x)\times (x)}{(2-x)\times (1.5-x)}}$
Solving this will require the use of the quadratic equation. It may be easier to use approximations or a graphing calculator. When solved, $x=1.34$ (disregard values of x that would give negative concentrations). By substituting x into the "equilibrium" values in the ICE chart, we can determine the concentrations of all substances when the reaction reaches equilibrium.
| CO | H$_2$O | CO$_2$ | H$_2$ |
| 0.66 M | 0.16 M | 1.34 M | 1.84 M |
To check your answer, you could substitute these values into the equilibrium expression and see if it equals 23.3 (the given K$_{eq}$ value).
The purpose of doing this is to see the final concentrations of the substances involved in a chemical reaction. Unless K$_{eq}$ is incredibly large, not all the reactants will be consumed.

## Equilibrium in Detail

The next few chapters in this book will provide more details on equilibrium.

### Shifts in Equilibrium

Le Chatelier's principle states that a system will adapt to minimize a change. If reactants are added, they will be consumed and products will form. Likewise, if products are added, they will be consumed (by the reverse reaction) and reactants will form.
The only change in a system that will affect the value of K$_{eq}$ is temperature. The value of K$_{eq}$ is never changed by the concentrations or pressures of the substances.

### Special Constants

| $K_{a}={\frac {[H^{+}][A^{-}]}{[HA]}}$ | Acids and bases have special ionization constants, which show how they react with water. The acid ionization constant, or K$_a$, tells how strong an acid is. |
| $K_{b}={\frac {[{\mbox{OH}}^{-}][{\mbox{BH}}]}{[{\mbox{B}}]}}$ | The base ionization constant, K$_b$, similarly tells how strong a base is. |
| $K_{sp}={\frac {[A^{+}][B^{-}]}{[AB]}}$ | The solubility product K$_{sp}$ shows how much an ionic substance will dissociate when dissolved. |

### Relation to E

There is a special relation between the standard potential of a redox reaction and the equilibrium constant $E={\frac {0.0592\log K}{n}}$, where n is the number of moles of electrons transferred. This is a significant equation, because this allows us to calculate the hard to measure K from the easy to measure E.