Calculating an Equilibrium Constant, Kp, with Partial Pressures

$$K_p$$ is the constant which is derived from the partial pressures of a reaction equation. It is used to express the relationship between product pressures and reactant pressures. It is a unitless number, although it relates the pressures. Although this may seem somewhat strange, just think of it as the the fraction that is able to relate the two sides of a reaction.

Introduction

For calculating $$K_p$$, the partial pressures of gasses are used. The partial pressures of pure solids and liquids are not used. This is very important! Don't forget this! To be able to successfully use this equation, it is helpful to have an understanding of partial pressures and mole fractions.

Partial Pressures: All of the partial pressures add up to the total pressure, as shown in the equation. Dalton's law

$P_{total} = P_A + P_B + ...$

This is due to the individual gases maintaining their pressures, even when combined. However, once they are combined, their individual pressures are combined for a total, overall pressure.

Mole Fractions: Although mole fractions can be used for elements other than gases, this description is referring specifically to those involving gases. This fraction expresses the moles of an individual gas compaired to the total amount of moles. It is used for reactions in which more than one gas is involved. In general the equation for finding the mole fraction of a gas (represented by A) is

$x_A = \dfrac{molesA}{total moles}$

Note: If the mole fraction is multiplied by 100, it becomes a percent expressing the mole amount for the individual gas.

When combined with the total pressure, this will equal the partial pressure (for gas A). The equation is thus

$P_A = X_A * P_{total}$

Calculations

The partial pressures are used for calculating the pressure constant. Using this general equation in the gas phase:

$aA + bB \rightleftharpoons cC + dD$

The formula uses the partial pressures, PA, PB, PC, and PD, which are to the power of their coefficient amounts. The equation for this is:

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

Warning! This looks similar to $$K_c$$, except $$K_c$$ uses concentrations. To prevent confusion, do not use brackets around the partial pressures.

The $$K_p$$ can also be found using $$K_c$$ (the concentration constant at equilibrium), temperature, change in moles, and the gas constant.

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

Because this ($$K_c$$) is a concept which would need a page of it's own, this page will not go into how to find it. This equation is simply for how to relate the two. For more information see the page about $$K_c$$.

References

1. Petrucci, et al. General Chemistry Principles & Modern Applications. 9th ed. Upper Saddle River, NJ: Pearson Prentice Hall, 2007

Practice Problems

1.

$2H_{(g)} + O_{(g)} \rightleftharpoons H_2O_{(g)}$

For this one just set up what the equation would be.

Answer: For this basic equation, all of the products and reactants are gases. The equation will be:

$K_p = \dfrac{P_{H_2O}}{P_H^2 * P_O}$

2.

$$2SO_2 + O_2 \rightleftharpoons 2SO_3$$

The initial pressure of SO2 is 3 atm. If x is the final pressure of O2, what is $$K_p$$?

Answer: set up an ICE table

 2SO2 O2 2SO3 Initial 0 0 3 Change +x +x -x Equilibrium x x 3-x

so it will be

$K_p = \frac{(3-x)^2}{x^2 * x}$

3.

$2H_{(g)} + O {(g)} \rightleftharpoons H_2O{(l)}$

The partial pressures are PH= 0.9968 atm, and PO= 1.105 atm

Stop. If you are thinking, "This is the same equation as number one", look at the equation again. Hopefully this time you noticed that the H2O is a liquid this time and not a gas. Therefore, it is not included in the equation.

$K_p = \dfrac{1}{0.9968^2 * 1.105} = 0.9108$

4.

$2Cl_2O_5 (g) \rightleftharpoons 2Cl_2 (g) + 5O_2 (g)$

If the mole fractions are- Cl2:0.243, O2:0.274, Cl2O5:0.483. The total pressure is:3 atm.What is the $$K_p$$?

Answer: first find the partial pressures

$$P_{(Cl_2)} = 0.243 * 3 atm = 0.729 atm$$

$$P_{(O_2)} = 0.274 * 3 atm = 0.822 atm$$

$$P_{(Cl_2O_5)} 0.483 * 3 atm = 1.449 atm$$

$$K_p = \dfrac{(0.729)^2 * (0.274)^5}{(0.483)^2} = 0.00352$$

5.

$2N_2O_5 (g) \rightleftharpoons O_2 + 4NO_2 (g)$

N: 14.01 g/mol, O: 16 g/mol, total pressure: 4. What is the $$K_p$$?

Answer: first find the mole fractions

$$N_2O_5 = 2(14.01 * 2 + 16 * 5) = 216.04 mol$$

$$O_2 = 16 * 2 = 32 mol$$

$$NO_2 = 4(14.01 + 16 * 2) = 184.04 mol$$

mole fractions- N2O5: 0.5, O2: 0.0741, NO2: 0.425

Next find Partial Pressures.

$$N_2O_5 = 0.5 * 4 atm = 2 atm$$

$$O_2 = 0.0741 * 4 atm = 0.2964 atm$$

$$NO_2 = 0.425 * 4 atm = 1.7 atm$$

Last, find the

$K_p = \dfrac{(0.2964) * (1.7)^4}{(2)^2} = 0.619$

Contributors

• Sarah Reno (UC Davis)

20:18, 21 Jan 2014