We mentioned in the previous post that the equilibrium-constant expression can be formulated in terms of partial pressures when the reactants and products in a chemical reaction are gases.

For example, when the equilibrium constant of the reaction between sulfur dioxide and oxygen, can be expressed based on the molar concentrations (*K*_{c}) or the partial pressures (*K*_{p}):

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

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

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

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 K_{p} **is defined using partial pressures.

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:**

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

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

**The Relationship Between ***K*_{p} and *K*_{c}

*K*

_{p}and

*K*

_{c}

Although the values of *K*_{p} and *K*_{c} are generally different, it is possible to calculate one from the other using the ideal gas law equation. The idea is to rearrange the ideal gas law equation to get an expression for the partial pressure of each gas and use in the equation of *K*_{p}.

*PV = nRT*, so for P, we can write:

\[P\; = \;\frac{n}{V}RT\]

Remember that n/V is the molarity, so the partial pressure of gas “A” is equal to:

\[{P_A}\; = \;\frac{{{n_A}}}{V}RT\; = \,\left[ {\rm{A}} \right]RT\]

The equivalent expression for the partial pressure of each gas is then entered in the equation of *K*_{p}:

This expression is then simplified by separating the terms of concentration and RT:

And now the cool part; Notice that the term in brackets on the left side is equal to *Kc*, so we can further simplify this expression:

Let’s put this as separate formula correlating Kp and Kc since you are going to use it quite often when solving equilibrium problems:

*a*A *+ b*B ⇆ *c*C *+ d*D

* ** *

The quantity Δ*n *is the number of moles of gaseous products minus the number of moles of gaseous reactants.

Δ*n *= (*c *+ *d*) – (*a *+ *b)*

*R *is the gas constant, 0.08206 (L · atm)/ (K · mol) , and *T *is the absolute temperature in Kelvins.

**For example,** the decomposition of 1 mol of N_{2}O_{4} produces 2 mol of NO_{2}, soΔ*n *= 2 – 1 = 1, and *K*_{p }= *K*_{c}(*RT*):

* *

N_{2}O_{4}*(g)* ⇆ 2NO_{2}*(g)*

* *

The equilibrium constant of this reaction is* K*_{c} = 4.6 x 10^{-4 }, and therefore, *K*_{p} = *K(RT) = *4.6 x 10^{-4}(0.08206 L · atm/K · mol × 298 K) = 1.1 x 10^{-2}_{.}

^{ }

**Another example**: The *K*_{c }for the synthesis reaction of ammonia at a certain temperature is 6.7 x 10^{9}. Calculate the Kp of the reaction.

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

There are more reactant than product molecules in the equation, so the is going to be a negative number:

Δn = 2- (3+ 1) = -2

*K*_{p} = *K(RT) = *6.7 x 10^{9 }(0.08206 L · atm/K · mol × 298 K)^{-2} = 1.6 x 10^{11}

When the number of products and reactant molecules is equal, then *K*_{c} = *K*_{p} because *K*_{p} = *K(RT)*^{0 }= **K**.

**For example**, suppose the* K _{c} of the reaction between hydrogen and bromine gases is *5.20 x 10

^{18. }

* *

H_{2}(*g*) + B_{2}(*g*) ⇆ 2HBr(*g*) *K*_{c} = 5.20 x 10^{18}

^{ }

Determine the *K*_{p} of this reaction.

Solution: *K*_{p} = *K(RT)*^{0 }= *K =* 5.20 x 10^{18}

*K***p Changes with Chemical Equation**

*K*

The expression of *K*p changes with the chemical equation just like we saw it for *K*c.

*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 Kp 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 reactions:

1) H_{2}(*g*) + I_{2}(*g*) ⇆ 2HI(*g*)

2) H_{2}(*g*) + I_{2}(*s*) ⇆ 2HI(*g*)

In which reaction are the *K *and *K*p equal?

At 300 K, the equilibrium concentrations for the following reaction are [CH_{3}OH] = 0.240 *M*, [CO] = 0.350 *M*, and [H_{2}] = 1.65 *M *for the reaction

CH_{3}OH(*g*) ⇆ CO(*g*) + 2H_{2}(*g*)

Calculate *K*p at this temperature.

Given the equilibrium constant, calculate *K*p for each of the following reactions at 298 K.

a) N_{2}O_{4}(*g*) ⇆ 2NO_{2}(*g*) *K*c = 4.6 x 10^{-4}

b) 3H_{2}(*g*) + N_{2}(*g*) ⇆ 2NH_{3}(*g*) *K*c = 6.7 x 10^{9}

c) H_{2}(*g*) + B_{2}(*g*) ⇆ 2HBr(*g*) *K*c = 5.20 x 10^{18}

Calculate *K*c for each reaction.

a) CO(*g*) + Cl_{2}(*g*) ⇆ COCl_{2}(*g*) *K*p = 5.3 x 10^{6}

b) CH_{4}(*g*) + H_{2}O(*g*) ⇆ CO(*g*) + 3H_{2}(*g*) *K*p = 7.7 x 10^{8}

c) 2SO_{2}(*g*) + O_{2}(*g*) ⇆ 2SO_{3}(*g*) *K*p = 6.26 x 10^{5}

**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**