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:

• If K is a large number, it means that the equilibrium concentration of the products is large. In this case, the reaction as written will proceed to the right (resulting in an increase in the concentration of products)
• If K is a small number, it means that the equilibrium concentration of the reactants is large. In this case, the reaction as written will proceed to the left (resulting in an increase in the concentration of reactants)

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

• The direction a reaction will proceed to achieve equilibrium
• The ratios of the concentrations of reactants and products when equilibrium is reached

Predicting the Direction of a Reaction

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

N2(g) + 2H2(g) ó 2NH3(g)

• H2(g) = 0.5 mol
• N2(g) = 8.3 mol
• NH3(g) = 1.8 mol

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)

• Concentration of H2(g) = 0.5mol/3.0L = 0.167M
• Concentration of N2(g) = 8.3mol/3.0L = 2.77M
• Concentration of NH3(g) = 1.8mol/3.0L = 0.600M

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?

• The "initial value of the equilibrium constant" is 27.9, at equilibrium we know that this must decrease to 0.105
• For the value of Kc to decrease to the experimentally expected value at equilibrium, the concentration of NH3(g) must decrease, and/or the concentrations of N2(g) and H2(g) must increase
• Therefore, the reaction would have to proceed to the left (increasing reactants and decreasing product concentrations)

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")

• If Q = Kc, then the system is already at equilibrium
• If Q > Kc, then essentially we have too much product and the reaction will proceed to the left (to reduce the concentration of product and increase the concentration of product)
• If Q < Kc, then essentially we have too little product and the reaction will proceed to the right (to produce more product and decrease the concentration of reactant)

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.

• For this reaction, the definition of Kp is given as

• The equilibrium partial pressures of H2(g) and N2(g) are given, as is the value for Kp

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

• We can solve for the partial pressure of NH3(g)

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:

• For equations of the form ax2 + bx + c = 0, the solution(s) for x are given by the quadratic equation:

• We may have more than one possible solution for x, but only one will make sense

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 472°C 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:

• N2(g): 1.8mol/2.0L = 0.9M
• H2(g): 0.2mol/2.0L = 0.1M
• NH3(g): 0.5mol/2.0L = 0.25M

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

• Q = 69.4 is a lot larger than the value of Kc = 0.105. Therefore, we are starting out with too much product and the reaction will proceed to the left, to produce more reactant.

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, for every mole of NH3(g) that decomposes, 0.5 mole of N2(g) will be produced and 1.5 mole of H2(g) will be produced
• We know that a certain amount of NH3(g) will have to break down in order for the reaction to achieve equilibrium. We can set this unknown amount to a variable, x

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

• Equilibrium concentration of NH3(g) = (starting concentration - amount that decomposes)
= 0.25M - x
• Equilibrium concentration of H2(g) = (starting concentration + (1.5 * amount of NH3(g) that decomposes))
= 0.1M + 1.5x
• Equilibrium concentration of N2(g) = (starting concentration + (0.5 * amount of NH3(g) that decomposes))
= 0.9M + 0.5x

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:

• The first solution (a negative value for x) does not make any sense because product must be degraded not increased. So we can exclude this solution
• The starting concentration of NH3(g) is 0.25M. The equilibrium concentration is (0.25M - x). The equilibrium concentration cannot be less than 0. Therefore, any solution greater than 0.25 is meaningless (i.e. incorrect)
• This leaves only one possible solution, x = 0.178M. Thus the equilibrium concentrations of the various components in the reaction are:
• Equilibrium concentration of NH3(g) = (0.25M - 0.178) = 0.072M
• Equilibrium concentration of H2(g) = (0.1M + 1.5*0.178) = 0.367M
• Equilibrium concentration of N2(g) = (0.9M + 0.5*0.178) = 0.989M
• Finally, we can check our answer by plugging these values into the equation for the equilibrium constant and making sure we get the right answer (i.e. 0.105 in this case)

© 2000 Dr. Michael Blaber