CHM 1046
General Chemistry II
Dr. Michael Blaber


Chemical Equilibrium

Applications of Equilibrium Constants


This section of our study of equilibrium constants deals with various types of calculations, including:

  1. How to figure out in which direction a reaction will go (i.e. towards making product, or more reactant)
  2. Calculating equilibrium concentrations. This may involve knowing equilibrium values for some of the reactants and products and determining the concentration of an unknown.
  3. Alternatively, we may be provided with the starting concentrations of reactants and products and may be asked to find the equilibrium concentrations

The Magnitude of K

The magnitude of the equilibrium constant, K, indicates the extent to which a reaction will proceed:

Knowing the value of the equilibrium constant, K, will allow us to determine:

Predicting the Direction of a Reaction

The value of Kc for the Haber reaction at 472C is 0.105. If we place the following amounts of H2(g), N2(g) and NH3(g) in a 3.0L container at 472C will the N2(g) and H2(g) react to form more NH3(g)?

N2(g) + 2H2(g) ó 2NH3(g)

First of all, we need to convert the amounts of the components into concentrations (mol/L or M). Thus, we will divide the mol value by the volume of the container (gases expand to fill their container)

We need to state the equilibrium expression for this reaction, based upon the balanced equation:

N2(g) + 3H2(g) <=> 2NH3(g)

Inserting the initial values of reactants and products into the equilibrium expression we get the following:

= (0.600)2/((2.77)*(0.167)3)

= 27.9

How does this number compare to the value of the equilibrium constant at this temperature?

When we substituted the initial values for the concentrations of the reactants and products into the equilibrium expression the number we came up with is called the Reaction Quotient (Q) (i.e. "the initial value of the equilibrium constant")

Calculation of the Equilibrium Concentration of a Reactant or Product

Many types of equilibrium problems deal with determining how much of a product (or reactant) we will have once a reaction reaches equilibrium.

The following example involves Kp and partial pressures for our friend, the Haber reaction:

At 500 Kp = 1.45 x 10-5 for the Haber reaction:

N2(g) + 3H2(g) <=> 2NH3(g)

In a sample of N2(g), H2(g) and NH3(g) at equilibrium in a Haber reaction the partial pressure of the H2(g) is 1.32 atm, and the partial pressure of the N2(g) is 0.648 atm. What is the partial pressure of the NH3(g) in the equilibrium mixture?

In this problem we have a homogeneous equilibrium of gases and the equilibrium constant, Kp, is given in terms of partial pressures of the component gases.

1.45 x 10-5 = (PNH3)2 / ((0.648)*(1.32)3)

PNH3 = (2.16 x 10-5)1/2

PNH3 = 4.64 x 10-3 atm

If there is any doubt that we have done the calculations correctly, we can substitute this value (along with the equilibrium partial pressures for H2(g) and N2(g)) into the equilibrium expression and make sure that we get the correct value for Kp


Solving equilibrium concentrations of all components in a reaction

Sometimes an equilibrium problem will provide the value for the equilibrium constant and the initial concentration of all species. The question is then to calculate the concentration of all species at equilibrium. This is a more difficult type of problem.

The following example is one for the reaction between nitrogen and oxygen.

Nitrogen and oxygen can react in the following equation:

N2(g) + O2(g) ó 2NO(g)

Kc = 2.1 x 10-3 (2500K)

A reaction is setup with 5.75M N2(g), 4.89M O2(g) and 0M NO(g). What will be the equilibrium concentrations of all components?

XM = conc of N2(g) that reacts as reaction goes to equilibrium. Therefore:

Equilibrium conc of N2(g) = (5.75M - XM)

From the stoichiometry of the balanced equation, if XM of N2(g) reacts, then an equivalent concentration of O2(g) must react. Therefore:

Equilibrium conc of O2(g) = (4.89M - XM)

Furthermore, from the stoichiometry of the balanced equation, XM of N2(g) reacting will result in 2XM of NO(g) being produced. Therefore:

Equilibrium concentration of NO(g) = 2XM

Setting up the equilibrium constant equation:

2.1 x 10-3 = [NO]2 / ([N2][O2]) = (2X)2 / {(5.75 - X)(4.89 - X)}

2.1 x 10-3 = 4X2 / (28.1 - 10.6X + X2)

(2.1 x 10-3)*(28.1 - 10.6X + X2) = 4X2

0.0590 - 0.0223X + 0.0021X2 = 4X2

4.00X2 + 0.0223X - 0.059 = 0

Even for simple reaction we are often faced with a quadratic solution for x:

 

Quadratic solutions: X = 0.119 or -0.124 (can't have a negative value). Therefore, the equilibrium concentrations are:

N2(g) = (5.75 - 0.119) = 5.63M

O2(g) = (4.89 - 0.119) = 4.77M

NO(g) = (2 * 0.119) = 0.238M

Note: plugging these values back into the equilibrium constant expression will allow us to check our answer:

Kc = [NO]2 / ([N2][O2]) = (0.238)2 / (5.63)(4.77)

Kc = 2.11 x 10-3

(this is pretty close considering round off errors)


Here is a more complex problem relating to the Haber reaction.

The value of Kc for the Haber reaction at 472C is 0.105. A 2.0L flask is filled with 0.2 mol of H2(g), 1.8 mol of N2(g) and 0.5 mol of NH3(g). What are the concentrations of H2(g), N2(g) and NH3(g) at equilibrium?

The equilibrium constant Kc defines the amount of reactant and product in terms of concentration (moles/L). So the first thing to do is convert all amounts into molar concentrations:

Kc defines the equilibrium expression for this reaction:

Next, calculate the reaction quotient, Q, to figure out whether the reaction will go to the right or left to attain equilibrium:

Q = (0.25)2 / ((0.9)*(0.1)3)

Q = 69.4

When the reaction does goes to the left, as predicted by the reaction quotient, the breakdown of NH3(g) product to produce N2(g) and H2(g) reactants will follow the stoichiometry of the balanced equation:

N2(g) + 3H2(g) <=> 2NH3(g)

Thus, the concentrations of H2(g), N2(g) and NH3(g) at equilibrium will equal the following:

The above amounts describe the equilibrium concentrations of the reactants and products and can therefore be inserted into the description of the equilibrium expression, and are set equal to the value for Kc:

0.105 = (0.25 - x)2 / ((0.9 + 0.5x)*(0.1 + 1.5x)3)

Although there is only one variable, this is a fairly complicated expression. It turns out that there are several possible mathematical solutions for x in this equation:

 


© 2000 Dr. Michael Blaber