# The Equilibrium Constant, K

Chemical Equilibrium is the state in which the reactants and products have no net change over time. This is when the forward and reverse reactions occur at equal rates. This state of equilibrium can be described by the equilibrium constant, K.

### Introduction

The equilibrium constant, K, can be used to find the concentrations of each component of a reversible reaction at equilibrium. To effectively use this equilibrium constant, you must know a few simple equations first:

### The Equilibrium Constant: K

To determine the amount of each compound that will be present at equilibrium you must know the Equilibrium Constant. To determine the equilibrium constant you must consider the chemical reaction written in the form:

$aA + bB \rightleftharpoons cC + dD$

The equilibrium constant is defined as:

$K = \dfrac{(a_C)^{c}(a_D)^{d}}{(a_A)^{a}(a_B)^{b}}$

Where "a" stands for the activity of the respective species. The activity is defined as the dimensionless ratio [X]/co. In this ratio, [X] represents the concentration of a specific product or reactant while co reflects the concentration in the desired reference state. Generally, this is molarity, or mol/L.

#### The Equilibrium Constant: Kc

Using the chemical reaction given above, we can rewrite the Equilibrium constant expression to reflect concentration. The concentration is molarity, written as moles per liter (M=mol/L).

$$K_c = \dfrac{[C]^c [D]^d}{[A]^a [B]^b}$$

The products are in the numerator and those of the reactants are in the denominator. The equation is a function of the concentrations, the upper case letters are the molar concentrations of the reactants and products (A, B etc.). The lowercase letters are the stoichiometric coefficients that balance the equation.

#### The Equilibrium Constant: Kp

For equilibria in the gas phase, the equation is a function of the reactants and products Partial Pressures. The Equilibrium Constant is expressed as:

$K_p = \dfrac{(P_C)^c (P_D)^d}{(P_A)^a (P_B)^b}$

P represents partial pressure, usually in atmospheres.

### Relationship between Kp and Kc

The equilibrium constants Kpand Kc can easily be related by the following equation:

$$K_p = K_c(RT)^{\Delta{n}_{gas}}$$

The value of R is .08206 L atm mol-1 K-1. T represents the temperature in Kelvin (K) at which the equilibrium reaction takes place. To find $$\Delta{n}_{gas}$$ , simply subtract the sum of the coefficients of the reactants from the sum of the coefficients of the products.

Example 1

For the following balanced chemical reaction:

$CH_3OH (g) \rightleftharpoons CO_{(g)} + 2H_2 (g)$

The value $$\Delta{n}_{gas}$$ is (1+2)-1 = 2.

Knowing the Kc value for the above reaction is 14.5 and the temperature at equilibrium is 483K, we can find Kp.

$$K_p = (14.5)(.08206*483)^{2} = 22778.50877$$

Important Point Regarding Solids and Liquids

An important point you should know is that the chemical equilibrium constant expression includes only solutes and gases. It does not include solids or liquids. The concentrations of pure solids and pure liquids are treated as one for these substances, so they do not affect the equilibrium constant.

For example, the equilibrium expression for the reaction

$$H_2S_{(g)} + I_{2 \; (s)} \rightleftharpoons 2HI_{(g)} + S_{(s)}$$

is

$$Kc = \dfrac{[HI]^2}{[H_2S]}$$

Notice that the gas-phase species are included, yet the solid species are not.

#### Assumptions about the Reaction Based on the Value of K

When we know the numerical value of the equilibrium constant, we can make certain judgments about the extent of the chemical reaction.

K > 1

If K is larger than 1, the mixture contains mostly products.

K < 1

If K is less than 1, the mixture contains mostly reactants.

K = 1

If K is about equal to 1, the reaction will reach equilibrium as an intermediate mixture, meaning the amounts of products and reactants will be about the same.

#### Changing K when Changing the Balanced Equation

Since the value of K is dependent on the chemical reaction, when the balanced chemical equation is manipulated in any way, the value of K changes accordingly.

The Effect on K of Reversing a Balanced Equation

When an equation is reversed, the value of K must be inverted. Take, for example, the chemical equation listed below:

$2SO_2 (g)+O_2 (g) \rightleftharpoons 2SO_3 (g)$

where K = 2.8 x 102 at 1000 K. If this equation were to be reversed:

$2SO_3 (g) \rightleftharpoons 2SO_2 (g)+O_2 (g)$

K would now equal

$K = \dfrac{[SO_3]^2}{[SO_2]^2 [O_2]} = \dfrac{1}{2.8 \times 10^2}.$

#### The Effect on K of Multiplying the Coefficients in a Balanced Equation

When the coefficients of a balanced equation are multiplied by a common factor, the equilibrium constant must be raised to the respective factor.  Take again the reaction listed below:

$2SO_2 (g) + O_2 (g) \rightleftharpoons 2SO_3 (g)$

where K = 2.8 x 102 at 1000 K. If you decided to alter the equation so 4 moles of SO3 are produced instead of 2:

$4SO_2 (g) +2O_2 (g)\rightleftharpoons 4SO_3 (g)$

Then K would now equal:

$K = \dfrac{[SO_3]^4}{[SO_2]^4 [0_2]^2} = (2.8 \times 10^2)^2$

### The Effect on K of Dividing the Coefficients in a Balanced Equation

When the coefficients in a balanced equation are divided by a common factor, you must take the corresponding root of the equilibrium constant.

Once again, lets look at the following reaction:

$2SO_2 (g) +O_2 (g) \rightleftharpoons 2SO_3 (g)$

where K = 2.8 x 102 at 1000 K. If we were to divide the coefficients by 2, we would get the following equation:

$SO_2 (g)+ \dfrac{1}{2}O_2 (g) \rightleftharpoons SO_3 (g)$

So to get K, we would:

$K = \dfrac{[SO_3]}{[SO_2][O_2]^{\frac{1}{2}}} = \sqrt{(2.8 \times10^2)}$

#### The Effect of Combining Individual Equations on K

If we were to combine or add together individual equations, we must multiply the their equilibrium constants to get the K value for the resulting reaction.

Take for example the following balanced equation:

$2N_2O (g) +O_2 (g) \leftrightharpoons 4NO (g)$

where K is unknown.

To find the value of K, we can take other balanced equations with known K values that add up to the first equation and then multiply the equilibrium constants. To do this, we will use the following balanced chemical reactions:

$N_2 (g) + \dfrac{1}{2}O_2 (g) \leftrightharpoons N_2O (g) \tag{a}$

where K is 2.7 x 10-18

$N_2 (g) + O_2 (g) \leftrightharpoons 2NO (g) \tag{b}$

where K is 4.7 x 10-31

These types of problems generally require you to manipulate the given $$K$$ value in many different ways. This example is no exception.

a) $$2N_2O (g) \leftrightharpoons 2N_2 (g) + O_2 (g)$$

$K = \left(\dfrac{1}{2.7 \times 10^{-18}} \right)^{2} = 1.37 \times 10^{35}$

b) $$2N_2 (g) + 2O_2 (g) \leftrightharpoons 4NO (g)$$

$K = (4.7*10^{-31})^{2} = 2.21 \times 10^{-61}$

Overall: $$2N_2O (g)+O_2 (g) \leftrightharpoons 4NO (g)$$

$K = K(a)K(b) = (1.37 \times 10^{35})(2.21*10^{-61}) = 3.03 \times 10^{-26}$

### Problems

1. What is the general equation for $$K_c$$?

2. What is the equilibrium constant expression for

$(C (s) + H_2O (g) \leftrightharpoons CO (g) + H_2 (g)?$

3. Given the following reaction and the concentrations of .5M for H2S and .2M for HI, what is the value for Kc?

$$H_2S_{(g)} + I_{2 \; (s)} \rightleftharpoons 2HI_{(g)} + S_{(s)}$$

4. If you were to reverse the previous reaction, what would the $$K_c$$ value be?

5. What if you reversed the equation in problem 3 and multiplied the coefficients by 5?

6. You have a balanced reaction where K is more than 1. Which side of the equation does this equilibrium constant favor?

### Solutions

1. $$K_c = \dfrac{[C]^c [D]^d}{[A]^a [B]^b}$$
2. $$K = \dfrac{[CO][H_2]}{[H_2O]}$$
3. $$K_c = \dfrac{[.2M]^{2}}{.5M} = .08$$
4. $$K_c = \dfrac{1}{.08} = 12.5$$
5. $$K_c = (\dfrac{1}{.08})^{5} = 305175.7813$$
6. The products

### References

1. Petrucci, Harwood, Herring, Madura. General Chemistry: Principles & MOdern Application. Ninth Edition. Pages 636-638.

### Contributors

• Charlotte Hutton