# Chemistry: Equilibrium Constants

## Equilibrium Constants

All equilibria have different ratios of products to reactants. As an example, let's consider a reaction in which the forward reaction takes place quickly and the reverse reaction is slow. When the system reaches equilibrium, a vast majority of the chemicals in the beaker will be products (we usually say that the "equilibrium favors the products"). On the other hand, if the forward reaction is slow and the reverse reaction is fast, the equilibrium favors the reactants.

It probably won't be a big surprise to you to find that scientists have figured out a way to quantify the position of an equilibrium. This was done by the "law of mass action."

What the law of mass action says is simple. Let's say that we have a reaction taking place in solution with the following equation:

- aA + bB ⇔ cC + dD

The equilibrium condition can be expressed using the following equation:

*K*_{eq}=^{[C]c [D]d}⁄_{[A]a[B]b}

##### The Mole Says

We can relate the equilibrium constant of a chemical equilibrium to the rates of the forward and reverse reactions. For example, consider the process A ⇔ B. The rate of the forward reaction, A → B, is k_{f}[A], where the little "f" after the k denotes the rate constant for the forward reaction. Likewise, the rate of the reverse reaction, B → A, is k_{r}[B], where "r" denotes the rate constant for the reverse process. Since at equilibrium, the rates for the forward and backward reactions are the same (k_{f}[A] = k_{r}[B]), we can say that the equilibrium constant, K_{eq}, is equal to [B]/[A] and k_{f}/k_{r}.

where K_{eq} is a constant for the process called the "equilibrium constant," each of the letters in brackets stands for the concentration of that chemical in mol/L (M) when the reaction has reached equilibrium, and the superscripts stand for the coefficient of each chemical. For gases, the equilibrium constant is determined in almost the same way, except that partial pressures are used in place of concentrations.

The equilibrium constant is important because it gives us an idea of where the equilibrium lies. The larger the equilibrium constant, the further the equilibrium lies toward the products. For example, an equilibrium constant of 1.0 × 10^{-6} suggests that very little of the reactants have formed products at equilibrium, while an equilibrium constant of 1.0 × 10^{6} suggests that at equilibrium, most of the chemical species present are products.

K_{eq} is a neat constant because it allows us to determine the ratios of the concentrations of each chemical species in the equilibrium, which is pretty handy for reasons we'll see in this example:

**Example**: Acetic acid breaks up in water via the following process:

- C
_{2}H_{3}O_{2}H_{(aq)}+ C_{2}H_{3}O_{2}^{-}_{(aq)}

Write the expression for the equilibrium constant, and determine what the value of K_{eq} is given these concentrations of each of the chemical species at equilibrium:

Chemical Species | Concentration at Equilibrium (M) |
---|---|

C_{2}H_{3}O_{2}H | 5.6 × 10^{-4} |

H^{+} | 1.0 × 10^{-5} |

C_{2}H_{3}O_{}^{-} | 1.0 × 10^{-5} |

To figure out the expression for the chemical equilibrium, use the equation we learned for finding K_{c}.

*K*_{eq}=^{[C2H3O2-][H+]}⁄_{[C2H3O2H]}

##### Bad Reactions

Be careful not to mistake the equilibrium constant K_{eq} for the rate constant k!

To determine the value of K_{eq}, all we need to do is substitute the values given to us for the concentrations in the equilibrium expression. As a result, we get:

*K*_{eq}=^{(1.0×10-4M)(1.0×10-4M)}⁄_{5.6×10-4M)}- K
_{eq}= 1.8 × 10^{-5}

Once we have an equilibrium constant, we can use it to figure out what the equilibrium concentrations of the products will be given an initial concentration of the reactants. Let's see another example:

**Example**: Given the reaction A ⇔ B + C, what will be the equilibrium concentrations of B and C if the initial concentration of A is 1.00 M and the equilibrium constant is 1.86 × 10^{-6}?

**Solution**: Let's walk through this problem step by step. The first step is to write the equilibrium expression for this process:

*K*=_{eq}^{[B][C]}⁄_{[A]}

Our next step is to figure out what the concentrations of each species will be at equilibrium. We do this by setting up a chart that shows the initial concentrations of all species, how the concentrations of each will change, and what the final concentrations of each species will be. For this process, the chart is given next (don't panic, we'll explain how we got all these values in a minute):

Species | Initial Concentration | Change | Final Concentration |
---|---|---|---|

A | 1.00 M | -x | (1.00 - x) M |

B | 0 M | x | x M |

C | 0 M | x | x M |

Okay, let's talk about where these came from:

- The initial concentration of A is defined by the problem as 1.00 M. Between the time that the reaction starts and the system reaches equilibrium, some of it will turn into the products. How much? We have no idea, so we'll just say that the change was "-x." As a result, our final concentration will be the initial concentration minus the amount of change, or (1.00 - x) M.
- The initial concentrations of both B and C are zero because neither was initially present. However, from the equation, you can see that every time a molecule of A breaks apart, one molecule each of B and C are formed. As a result, if the concentration of A decreases by x, the concentration of both products must increase by x. At equilibrium, the concentration of both B and C is x.

To figure out what the concentrations of each species are, we plug these values into the expression for finding the equilibrium constant. Since we know that the equilibrium constant is 1.86 × 10^{-6}, all we have to do is solve for x in the following expression:

- 1.86×10
^{-6}=^{[x][x]}⁄_{[1.00 - x]}

Now, solving for x won't be a lot of fun because we'll need to use the quadratic equation. Let's face it, we don't really want to memorize the quadratic equation, much less use it.

##### You've Got Problems

Problem 1: Given the reaction H_{2} + I_{2} ⇔ 2 HI, find the following:

a) The general expression for the equilibrium constant.

b) The equilibrium concentration of HI if I start with 2.00 M H_{2} and 2.00 M I_{2} and the equilibrium constant is 5.00.

Fortunately, there's a shortcut we can use to get around using the quadratic equation when the K_{eq} values are very small. If K_{eq} is small, then we can safely guess that the amount of product formed (x) will be very, very small when compared to the initial quantity of the reactant. This is because a very small K_{eq} value means that very little product has been formed. As a result, we can omit the "x" in the denominator to simplify the [1.00 - x] term because [1.00 - x] will be roughly equal to 1.00. Our new (and easier) expression to solve is transformed into:

- 1.86×10
^{-6}=^{[x][x]}⁄_{[1.00]} - x = 1.36 × 10
^{-3}M

Let's see if this was a good assumption. If the amount of product formed was 1.36 × 10^{-3} M, then the final concentration of acetic acid will be (1.00 - 0.00136) M, which, within the constraints of significant figure rules, is still 1.00 M. As a result, our assumption was valid and we saved ourselves a lot of mathematical heartache and toil. I love it when I save myself toil!

Our final concentrations of B and C are 1.36 × 10^{-3} M.

Excerpted from The Complete Idiot's Guide to Chemistry © 2003 by Ian Guch. All rights reserved including the right of reproduction in whole or in part in any form. Used by arrangement with **Alpha Books**, a member of Penguin Group (USA) Inc.

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**See also:**