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ChemWiki: The Dynamic Chemistry E-textbook > Physical Chemistry > Equilibria > Chemical Equilibria > Calculating an Equilibrium Concentration > Calculating an Equilibrium Constant, Kp, with Partial Pressures

Calculating an Equilibrium Constant, Kp, with Partial Pressures

Kp is the equilibrium constant calculated 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.

Introduction

In calculating Kp, the partial pressures of gases are used. The partial pressures of pure solids and liquids are not included. To use this equation, it is beneficial 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

\[P_{total} = P_A + P_B + ...\]

Individual gases maintain their respective pressures when combined. However, their individual pressures add up to the total pressure in the system

Mole Fractions: Although mole fractions can be used for non-gases, this description is referring specifically to those involving gases. This fraction expresses the moles of an individual gas compared to the total 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

\[x_A = \dfrac{mol_A}{mol_{total}}\]

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

\[P_A = X_A \times P_{total}\]

Calculations

Partial pressures are used for calculating the pressure constant. Consider this general equation for a gas-phase reaction:

 \[aA + bB \rightleftharpoons cC + dD\]

The formula uses the partial pressures, PA, PB, PC, and PD, raised to exponents equal to their respective coefficients in the chemical equation. Kp is calculated as follows:

\(K_p = \frac{P_C^cP_D^d}{P_A^aP_B^b}\)

Warning: This expression is similar to Kc; however, Kc is calculated from molar concentrations. To prevent confusion, do not use brackets around partial pressures.

Kp can also be obtained from Kc (the concentration equilibrium constant), temperature, change in moles, and the gas constant, as shown:

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

Calculating Kc is not discussed in this article; this equation simply relates the two. For more information see the page about Kc

References

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

Practice Problems

1.

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

Give the equilibrium constant expression for the above equation.

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

\[K_p = \frac{P_{H_2O}}{P_H^2P_O}\]

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 Kp?

Answer: set up an ICE table

  2SO2 O2 2SO3
Initial 0 0
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. At first this equation may look identical to Problem 1, but look at the equation again.This time the H2O is a liquid this time and not a gas. Therefore, it is not included in the equation.

Answer: The equation is

\[K_p = \dfrac{1}{0.9968^2 \times 1.105} = 0.9108\]

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 Kp?

Answer: first find the partial pressures

\(P_{(Cl_2)} = 0.243 \times 3 atm = 0.729 atm\)

\(P_{(O_2)} = 0.274 \times 3 atm = 0.822 atm\)

\(P_{(Cl_2O_5)} 0.483 \times 3 atm = 1.449 atm\)

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

5.

\[2N_2O_5 (g) \rightleftharpoons O_2 + 4NO_2 (g)\]

N: 14.01 g/mol, O: 16 g/mol, total pressure: 4. Find Kp.

Answer: First find the mole fractions:

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

\(O_2 = 16 \times 2 = 32 mol\)

\(NO_2 = 4(14.01 + 16 \times 2) = 184.04 mol\)

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

Next find partial pressures:

\(N_2O_5 = 0.5 \times 4 atm = 2 atm\)

\(O_2 = 0.0741 \times 4 atm = 0.2964 atm\)

\(NO_2 = 0.425 \times 4 atm = 1.7 atm\)

Last, find Kp:

\[K_p = \dfrac{(0.2964)(1.7)^4}{(2)^2} = 0.619\]

Contributors

  • Sarah Reno (UC Davis)

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Last Modified
13:38, 10 Jun 2014

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