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The Equilibrium Constant

The equilibrium constant, \(K\), is an expression expressing the relationship between products and reactants of a reaction at equilibrium with respect to a specific unit. There are two different types of equilibria reactions: (1) heterogeneous and (2) homogeneous. Below we illustrate the difference between the two, explain how to write equilibrium constants for both, and introduce calculations involved with both the concentration and the pressure equilibrium constant.


As mentioned above, there are two types of equilibrium reactions. It is important to understand the difference between the two, because we express their equilibrium constants differently. 


The simpler one, a homogeneous reaction, is one where the states of matter of the products and reactions are all the same (homo=same). In most cases, the solvent determines the state of matter for the overall reaction. For example, the synthesis of methanol from a carbon monoxide-hydrogen mixture is a gaseous homogeneous mixture, which contains two or more substances:

\[ CO (g)+ 2H_2 (g) \rightleftharpoons CH_3OH (g)\]

At equilibrium, the rate of the forward and reverse reaction are equal, which is demonstrated by the arrows. The K however tells you the ratio of the units (pressure or concentration) when the reaction is at equilibrium. 

The synthesis of ammonia is another example of a gaseous homogeneous mixture:

\[ N_2(g) + 3H_2(g)  \rightleftharpoons 2NH_3(g) \]


A heterogeneous reaction is one where one or more states within the reaction differ (heteros=different). For example, the formation of an aqueous solution of lead(II) iodide is a heterogeneous mixture dealing with molecules in both the solid and aqueous states:

\[PbI_2 (s)  \rightleftharpoons Pb^{+2}(aq) + 2I^-(aq)\]

The decomposition of sodium hydrogen carbonate (baking soda) at high elevations is another example of a heterogeneous mixture, this reaction deals with molecules in both the solid and gaseous states:

\[ 2NaHCO_3 (s)  \rightleftharpoons Na_2CO_3 (s) + H_2O (g) + CO_2 (g) \]

\[ C(s) + O_2 (g) \rightleftharpoons CO_2 (g) \]

Once again, this difference is emphasized so that students remember that equilibrium constant calculations are different from heterogeneous mixtures. 

Writing Equilibrium Constants

Before we write an equilibrium constant, we should become familiar with the concept of an equilibrium constant. An equilibrium constant is obtained by letting a single reaction proceed to equilibrium and then measuring the concentrations of each molecule involved in that reaction and creating a ratio of products to reactants. Because we must allow the reaction to proceed to equilibrium, the equilibrium constant will remain the same for each reaction independant of initial concentrations which determine the speed of a reaction (for an ideal reaction). This knowledge allowed scientists to derive a model expression that can serve as a "template" for any reaction. We will now look at the basic "template" form of a homogeneous equilibrium constant.

If we have a hypothetical reaction:

\[aA (g) + bB (g)  \rightleftharpoons gG (g) + hH (g) \]

*The lower case letters represent the number of moles of each molecule, the upper case letters represent the molecule itself, and the letters in the parenthesis always represents the state of matter of the molecule.

Equilibrium Constant of Concentration

The Equilibrium Constant of Concentration gives the ratio of concentrations of products over reactants for a reaction that is at equilibrium. This is usually used when the state of matter for the reaction is (aq). The equilibrium constant expression is written as \(K_c\), which would be as follows:

\[ K_c = \dfrac{[G]^g[H]^h}{[A]^a[B]^b}\]

  • If K > 1 then it Favors Product
  • If K < 1 then it favors the Reactant

Here, the letters inside the brackets represent the concentration of each molecule. Notice the sum of the products is the numerator of the ratio and the sum of the reactants is the denominator. This will be the case for every equilibrium constant. Also note that each concentration is raised to a power of its coefficient; this will also be seen in every equilibrium constant problem. Keep in mind that this expression was obtained by a homogeneous equilibrium reaction. \(K\) represent equilibrium constants and \(c\) represents concentration (\(K_c\)). This means that every molecule shows up in the expression, as long as it is a solution or a gas.

Equilibrium Constant of Pressure

Gaseous reactions are not expressed by concentration, but instead, by partial pressures. The Equilibrium Constant of Pressure gives the ratio of pressure of products over reactants for a reaction that is at equilibrium. The equilibrium constant is written as \(K_p\),  which would be as follows:

\[ K_p= \dfrac{P^g_G \times P^h_H}{ P^a_A \times P^b_B} \]

  • Where \(p\) can have units of atm or bar. 

The procedure for this is the same as the procedure for the concentration constant above.

Conversion of Kc to Kp

If you have \(K_c\) and want it converted into \(K_p\) use the following Equation:

Kp = Kc(RT)^Δngas 

R= 0.8206 L*atm*K-1*Mol-1

T= Temperature in Kelvin

Δngas= Mole of gas (product) - Mole of Gas (Reactant)

Quotient Equilibrium Constant

There is also \(Q_c\), which is the constant at a particular concentration that is not at equilibrium. You would calculate it the same as \(K_c\) but this is a comparison with \(K_c\). It is used to determine which way the reaction will proceed at any one given time.

Qc = [G]g[H]h/[A]a[B]b

  • If Qc > KcShift to the left
  • Qc < KcShift to the Right
  • Qc = Kc Nothing has changed (still at equilibrium) 

*Same process for Qp

Heterogeneous Mixture

So far, we have dealt with reaction is which all the states of matter were the same. So how do we write heterogeneous reaction constants? Most importantly, solids and pure liquids will always be excluded from the equilibrium constant.  In a mathematical perspective, concentration of solids and liquids  equal one, which do not affect the overall K value. Solvents also equal one. The molarity of solids, liquids, and solvents remain constant throughout the reaction, which means their value can be denoted as 1. This rule is extremely important to remember, especially in dealing with heterogeneous solutions.

Example 1

If we have a hypothetical reaction:

\[ aA (s) + bB (l) \rightleftharpoons gG (aq) + hH (aq) \]

  • \([A] (s)= [B] (l)=1 \)

The equilibrium constant expression would be as follows:

Kc = [G]g[H]h/1*1 = [G]g[H]h

*In this case, since solids and liquids do not affect the equilibrium constant expression, the expression is independent from the concentration of the reactants. Thus, we must leave A and B out of the final expression. When the product of the reaction is a solvent, the numerator equals one, which is illustrated in the following reaction:

\[ H^+(aq) + OH^–(aq) \rightarrow H_2O (l)\]

The equilibrium constant expression would be:

\[ K_c= \dfrac{1}{ [H^+][OH^-]}\]

Manipulation of Constants

So, what do we do if a reaction is reversed, or a reaction is split into its elementary steps. In this section, you will be taught these two manipulations.

1. When the reaction is reversed, the equilibrium constant expression is inverted. The expression would be written as:

K'= 1/[G]g[Hh]/ [A]a[B]b

2. When there are multiple steps in the reaction, each with its own K (similar to Hess' problems), then you would multiply the different K's of the different steps in the same reaction to get the final K.

Activity Series

Since the concentration of reactants and products are not dimensionless (numbers that do not have units) in a reaction, the units of the equilibrium constants is represented by activity. Activity is expressed by the dimensionless ratio [X]/c0 where X signifies the molarity of the reaction and c is the chosen reference state, which is illustrated here:

\[ a_b=\dfrac{[B]}{c^0} \]

For gases that do not follow the ideal gas laws, using activity will accurately determine the equilibrium constant that changes when concentration or pressure varies. Thus, the units are cancelled and \(k\) becomes unitless.

Practice Problems

Write the equilibrium constant expression for each reaction.

  1. \(2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g) \)
  2. \(N_2O (g) + \frac{1}{2} O_2(g) \rightleftharpoons 2NO(g) \)
  3. \(Cu (s) + 2Ag^+(aq) \rightleftharpoons Cu^{+2}(aq) + 2Ag(s) \)
  4. \(CaCO_3 (g) \rightleftharpoons CaCO(s) + CO_2 (g) \)
  5. \(2NaHCO_3 (s) \rightleftharpoons Na_2CO_3 (s) + CO_2 (g) + H_2O (g) \)

What is the \(K_c\) of the following reaction?

\[ 2SO_2 (g) + O_2 (g) \rightleftharpoons 2SO_3 (g) \]

with concentration \(SO_2(g) = 0.2 M O_2 (g) = 0.5 M SO_3 (g) = 0.7 M\)

Also, What is the \(K_p\) of this reaction? At room temperature

2. For the same reaction, the differing concentrations: \(SO_2 (g) = 0.1 M O_2(g) = 0.3 M SO_3 (g) = 0.5 M\)

 Would this go towards to product or reactant?

Write the Partial Pressure Equilibrium.

1. \[ C (s) + O_2 (g) \rightarrow CO_2 (g)\]

Write the reaction for this Partial Pressure:

2. \(K_p= \dfrac{P^2_{HI}}{P_{H_2} \times P_{I_2}}\)

*Answers may be found at bottom of page (under contributors).


  1. Petrucci, Ralph H. General Chemistry: Principles and Modern Applications 9th Ed. New Jersey: Pearson Education Inc. 2007.

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Answers to Practice Problems

  1. \(K_c = \dfrac{ [SO_3]^2}{[O_2][SO_2]^2}\)
  2. \(Kc = \dfrac{ [NO]^2}{[O_2]^{0.5}[N_2O]}\)
  3. \(Kc = \dfrac{ [Cu^{+2}]}{[Ag^+]^2 }\)
  4. \(Kc = \dfrac{ [CO_2]}{[CaCO_3]} \)
  5. \(K_c = [H_2O][CO_2] \)

What is \(K_c\) for the Reaction

1) Kc: 24.5

Kp: 1.002 Atm

2) Qc= 83.33 > Kc therefore the reaction shifts to the left

  1. \( K_p= \dfrac{P_{CO_2}} {P_{O_2}} \)
  2. \( H_2 (g)+ I_2 (g) \rightarrow 2HI(g) \)


  • Heather Voigt

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Last Modified
20:46, 19 Dec 2013

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