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Calculating Equilibrium Pressures


Concentrations or gas pressures of reactants and products at equilibrium can be calculated from the initial concentrations or gas pressures of reactants and products and equilibrium constants. There are many variations of this procedure, but because our goal is only to introduce you to this category of calculations, we will restrict our examples to gas-phase reactions, for which you are given the initial partial pressures of reactants, the equation for the reaction, and the KP that corresponds to the equation. The equations for the reactions of this kind that you will see will have one of two forms:

A(g)  +  B(g)    Double arrow    C(g)  +  D(g)

or      A(g)  +  B(g)    Double arrow    2C(g)

Sample Study Sheet: Calculating Equilibrium Pressures   

Tip-off – You are asked for the equilibrium partial pressures of reactants and products, and you are given the equation for the reaction, its KP (or a table of KP values), and the initial partial pressures of reactants.

General Steps

  • Write the equilibrium constant expression that corresponds to the chemical equation.

  • Set up a table for displaying the initial pressures, the changes in pressure, and the equilibrium pressures.

For our examples, assign x to the decrease in pressure of each reactant. The equilibrium partial pressure of each reactant will be the given initial partial pressure minus x. The equilibrium partial pressure of product will be x for reactions that have the first general form below or 2x for reactions that have the second form.

A(g)  +  B(g)    Double arrow    C(g)  +  D(g)

or      A(g)  +  B(g)    Double arrow    2C(g)

  • Plug the equilibrium concentration terms into the equilibrium constant expression.

  • Solve for x.

  • Use x to calculate the equilibrium concentrations of reactants and products.  

 

EXAMPLE – Calculating Equilibrium Pressures: 

Chloromethane, CH3Cl, which has been used as a refrigerant and a local anesthetic, can be made from the following reaction.

CH3OH(g)  +  HCl(g)    Double arrow    CH3Cl(g)  +  H2O(g)

KP = 5.9 × 103 at 120 °C

If enough methanol and hydrogen chloride are added to a container at 120 °C to yield an initial pressure of 0.75 atm for each, what will the equilibrium pressures of all of the reactants and products be?:

Solution

We start by writing the equilibrium constant expression that corresponds to the given chemical equation:

Image of equation for the eqilibrium constant equal to the pressure of CH3Cl times pressure of H2O over presure of CH2OH times pressure of HCl

Next, we set up a table describing the initial gas pressures, the changes in pressures, and the equilibrium pressures in terms of x. For our problem, we assign "x" to the decrease in pressure of CH3OH and HCl.

CH3OH

HCl

CH3Cl

H2O

Initial

0.75

0.75

0

0

Change

-x

-x

x

x

Equilibrium

0.75 - x

0.75 - x

x

x

 

We now plug in the terms for the equilibrium concentrations of each reactant and product into the equilibrium constant expression.

Image of equation for the eqilibrium constant equal to the pressure of CH3Cl times pressure of H2O over presure of CH2OH times pressure of HCl, which is equal to x squared over 0.75 minus x times 0.75 minus x equals 5.9 times 10 to the third

Next, we solve for x. (We can simplify our equation by taking the square root of both sides.)

Image showing the algebraic steps to solve for x, which is 0.74

We complete the problem by calculating the equilibrium concentrations of each reactant and product from the calculated x.

Image that shows that the pressures of CH3Cl and H2O are equal to 0.74 atmospheres and the pressures of CH3OH and HCl are 0.01 atmospheres