Equilibrium Constant For Gaseous Reactions Example Problem

This example problem demonstrates how to find the equilibrium constant and pressure equilibrium constant of a gaseous reaction from the pressures of the gases.

Problem:

What is the equilibrium constant and pressure equilibrium constant for the reaction

CH₃OH(g) ↔ CO(g) + 2 H₂(g)

if the equilbrium pressures at 298.15 K are found to be

P_CH3OH = 6.10 x 10^-4 atm
P_CO = 0.387 atm
P_H2 = 1.34 atm?

Solution

The equilibrium constant expression takes the same form for concentration or pressure. In other words, the equilibrium constant K is a ratio of conditions between products and reactants. For this reaction, the concentration equilibrium constant is

K = [CO]²[H₂]/[CH₃OH]

The same is true for the pressure equilibrium constant, K_P.

K_P = (P_CO)²·P_H2/P_CH3OH

Step 1 - Find K_P

Substitute values into the above equation

K_P = (0.387 atm)²·(1.34 atm)/6.10 x 10^-4 atm
K_P = 1.14 x 10³ atm²

Step 2 - Find K

To find K, we need to know the equilibrium concentrations of the products and reactants. Concentration is a measure of the number of moles per volume. The ideal gas law says

PV = nRT

where
P = Pressure
V = Volume
n = number of moles
R = ideal gas constant = 0.08206 L·atm/K·mol
T = absolute temperature

Solve for n/V

n/V = concentration = P/RT

Substitute into the equilibrium constant formula:

K = (P_CO/RT)²[P_H2/RT]/[P_CH3OH/RT]

K = (P_CO²·P_H2/P_CH3OH)(1/RT²)

or

K = K_P/RT²

Substitute K_P from Step 1:

K = 1.14 x 10³ atm²/(0.08206 L·atm/K·mol·298.15 K)²
K = 1.14 x 10³ atm²/(24.47 L·atm/mol)²
K = 1.14 x 10³ atm²/598.59 L²·atm²/mol²
K = 1.90 mol²/L²

Answer:

The pressure equilibrium constant, K_P of the reaction at 298.15 K is 1.14 x 10³ atm² and the concentration equilibrium contant, K is 1.90 mol²/L².