The first studies and the observations of patterns in the ratio of the products and reactants were made in 1864 by Norwegian mathematician Cato Guldberg and a chemist Peter Waage when working on reactions of gases.

They noticed that in a reaction between gas molecules, the ratio of the partial pressures of the products over the product of the partial pressures of reactants raised to the power of their coefficients stays constant no matter what the initial amounts of were. This was how the expression for the equilibrium constant was derived.

For example, when sulfur dioxide was mixed with oxygen, it was found the following ratio of the partial pressures is always the same number:

2SO_{2}(*g*) + O_{2}(*g*) ⇆ 2SO_{3}(*g*)

* *

\[{K_p} = \frac{{{P^2}_{{\rm{S}}{{\rm{O}}_{\rm{3}}}}}}{{{P^{\rm{2}}}{{_{{\rm{SO}}}}_{_2}}{P_{{{\rm{O}}_{\rm{2}}}}}}}\]

Where *P*SO_{3}, *P*SO_{2}, and *P*O_{2} are the partial pressure of the reaction components.

The subscript p on *K* is used to point out that the **equilibrium constant Kp **is defined using partial pressures.

The equilibrium constant can also be expressed in terms of the molar concentrations:

\[{K_c} = \frac{{{{\left[ {{\rm{S}}{{\rm{O}}_{\rm{3}}}} \right]}^2}}}{{{{\left[ {{\rm{S}}{{\rm{O}}_{\rm{2}}}} \right]}^2}\left[ {{{\rm{O}}_{\rm{2}}}} \right]}}\]

The values of *K*_{p} and *K*_{c} are often different, however, we can calculate one from the other based on the coefficients of the balanced chemical equation. We will cover this in the next article.

For a general reaction of gases, the equilibrium constant *K*_{p} can be represented as:

*a*A *+ b*B ⇆ *c*C *+ d*D

Let’s see an example of calculating *K*_{p}.

**Example:**

The following equilibrium pressures were observed at a certain temperature for the Haber process:

3H_{2}(*g*) + N_{2}(*g*) ⇆ 2NH_{3}(*g*)

*P(*NH_{3}) = 5.2 x 10^{9 }atm

*P(*N_{2}) = 6.1 x 10^{2} atm

*P(*H_{2}) = 4.7 x 10^{3} atm

Calculate the value for the equilibrium constant *K*_{p} at this temperature.

**Solution:**

\[{K_p} = \frac{{{P^2}_{{\rm{N}}{{\rm{H}}_{\rm{3}}}}}}{{{P^3}_{{{\rm{H}}_{\rm{2}}}}{P_{{{\rm{N}}_{\rm{2}}}}}}}\]

\[{K_p} = \frac{{{{(5.2 \times {{10}^9})}^2}}}{{{{(4.7 \times {{10}^3})}^3}(6.1 \times {{10}^2})}} = 4.3 \times {10^5}\]

As for the *K*_{c}, the expression and value of *K*_{p} depend on the chemical equation, mainly the coefficients and the direction of the reaction.

Let’s see how these are related.

*K***p for the reverse reaction**

*K*

The following reaction has an equilibrium constant of *K*_{p} = 4.42 x 10^{-5 }at 298 K:

CH_{3}OH(*g*) ⇆ CO(*g*) + 2H_{2}(*g*)

Calculate the equilibrium constant for this process if the reaction is represented as follows:

CO(*g*) + 2H_{2}(*g*) ⇆ CH_{3}OH(*g*)

Remember, when a reaction is **reversed**, then *K*_{new }**= 1/ K_{original}**.

Therefore,

*K*_{new} = 1/*K*_{original}= 1/4.42 x 10^{-5}= 22,624.43 = 2.26 x 10^{4}

*K***p When the Coefficients are Changed**

*K*

Based on the *K*_{p} value given above, what is the *K*_{p} for the following reaction?

2CH_{3}OH(*g*) ⇆ 2CO(*g*) + 4H_{2}(*g*)

Comparing this equation to the original, we see that it is **multiplied by 2**. Remember when the coefficients in the equation are multiplied by any factor, we raise the equilibrium constant to the same factor. Therefore,

*K*_{new} = (*K*_{original})^{n }= (4.42 x 10^{-5})^{2}= 1.96 x 10^{-9}

#### Practice

Consider the following reaction at 298 K:

2NO(*g*) + Cl_{2}(*g*) ⇆ 2NOCl(*g*) *K*p = 31.6

In a reaction mixture at equilibrium, the partial pressure of NO is 128 torr and that of Cl_{2} is 176 torr. Calculate the partial pressure of NOCl at equilibrium.

*P*_{(NOCl)} = 0.455 atm, or 346 torr

Remember, when converting *K*c and *K*p, the value for R is most often used as 0.08206 L *atm*/ mol K, and therefore, the units for pressure should also be in atm.

\[{P_{({\rm{NO}})}}{\rm{ = }}\;\;{\rm{128torr}}\;\;{\rm{x}}\;\frac{{{\rm{1}}\;{\rm{atm}}}}{{{\rm{760}}\;{\rm{torr}}}}\;{\rm{ = }}\;\;{\rm{0}}.{\rm{168}}\;{\rm{atm}}\]

\[{P_{({\rm{C}}{{\rm{l}}_{\rm{2}}})}}\; = \;{\rm{176}}\;{\rm{torr\;}}\; \times \;{\rm{\;}}\frac{{{\rm{1}}\;{\rm{atm}}}}{{{\rm{760}}\;{\rm{tor}}}}\,{\rm{\; = \;}}\,{\rm{0}}.{\rm{232}}\,{\rm{atm}}\]

Next, use the *K*p expression to assign the numbers and an unknown for the *P*_{(NOCl)}:

\[{K_p} = \frac{{{P^2}_{{\rm{NOCl}}}}}{{{P^2}_{{\rm{NO}}}{P_{{\rm{C}}{{\rm{l}}_{\rm{2}}}}}}}\]

\[{K_p} = \frac{{{{\rm{X}}^{\rm{2}}}}}{{{{\left( {{\rm{0}}{\rm{.168}}} \right)}^{\rm{2}}}{\rm{0}}{\rm{.232}}}}\;{\rm{ = }}\;{\rm{31}}{\rm{.6}}\]

**X = 0.455 atm = 346 torr**

Check:

\[{K_p} = \frac{{{{\rm{0.455}}^{\rm{2}}}}}{{{{\left( {{\rm{0}}{\rm{.168}}} \right)}^{\rm{2}}}{\rm{0}}{\rm{.232}}}}\;{\rm{ = }}\;{\rm{31}}{\rm{.6}}\]

**✔**

**Check Also**

- Chemical Equilibrium
- Equilibrium Constant
*K*_{p}and Partial Pressure*K*and_{p }*K*Relationship_{c}*K*Changes with Chemical Equation- Equilibrium Constant K from Two Reactions
- Reaction Quotient –
*Q* - ICE Table – Calculating Equilibrium Concentrations
- ICE Table Practice Problems
- Le Châtelier’s principle
- Le Châtelier’s principle Practice Problems
**Chemical Equilibrium Practice Problems**