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:

- How to figure out in which direction a reaction will go (i.e. towards making product, or more reactant)
- Calculating equilibrium concentrations. This may involve knowing equilibrium values for some of the reactants and products and determining the concentration of an unknown.
- 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 theis large. In this case, the reaction as written will*products*(resulting in an increase in the concentration of products)*proceed to the right* - 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(resulting in an increase in the concentration of reactants)*proceed to the left*

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 K_{c} for the Haber reaction at 472°C is 0.105. If we place the following amounts of H_{2}(*g*), N_{2}(*g*) and NH_{3}(*g*) in a 3.0L container at 472°C will the N_{2}(*g*) and H_{2}(*g*) react to form more NH_{3}(*g*)?

N_{2}(*g*) + 2H_{2}(*g*) ó 2NH_{3}(*g*)

- H
_{2}(*g*) = 0.5 mol - N
_{2}(*g*) = 8.3 mol - NH
_{3}(*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 H
_{2}(*g*) = 0.5mol/3.0L = 0.167M - Concentration of N
_{2}(*g*) = 8.3mol/3.0L = 2.77M - Concentration of NH
_{3}(*g*) = 1.8mol/3.0L = 0.600M

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

N_{2}(*g*) + 3H_{2}(*g*) <=> 2NH_{3}(*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 "
" is 27.9, at equilibrium we know that this must decrease to 0.105*initial value of the equilibrium constant* - For the value of K
_{c}to decrease to the experimentally expected value at equilibrium, the concentration of NH_{3}(*g*) must*decrease*, and/or the concentrations of N_{2}(*g*) and H_{2}(*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 = K
_{c}, then the system is*already*at equilibrium - If Q > K
_{c}, 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 < K
_{c}, 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 K_{p} and partial pressures for our friend, the Haber reaction:

At 500° K_{p} = 1.45 x 10^{-5} for the Haber reaction:

N_{2}(*g*) + 3H_{2}(*g*) <=> 2NH_{3}(*g*)

In a sample of N_{2}(*g*), H_{2}(*g*) and NH_{3}(*g*) at equilibrium in a Haber reaction the partial pressure of the H_{2}(*g*) is 1.32 atm, and the partial pressure of the N_{2}(*g*) is 0.648 atm. What is the partial pressure of the NH_{3}(*g*) in the equilibrium mixture?

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

- For this reaction, the definition of K
_{p}is given as

- The equilibrium partial pressures of H
_{2}(*g*) and N_{2}(*g*) are given, as is the value for K_{p}

1.45 x 10^{-5} = (P_{NH3})^{2} / ((0.648)*(1.32)^{3})

- We can solve for the partial pressure of NH
_{3}(*g*)

P_{NH3} = (2.16 x 10^{-5})^{1/2}

P_{NH3} = 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 H_{2}(*g*) and N_{2}(*g*)) into the equilibrium expression and make sure that we get the correct value for K_{p}

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:

N_{2}(*g*) + O_{2}(*g*) ó 2NO(*g*)

K_{c} = 2.1 x 10^{-3} (2500K)

A reaction is setup with 5.75M N_{2}(*g*), 4.89M O_{2}(*g*) and 0M NO(*g*). What will be the equilibrium concentrations of all components?

XM = conc of N_{2}(*g*) that reacts as reaction goes to equilibrium. Therefore:

Equilibrium conc of N_{2}(*g*) = (5.75M - XM)

From the stoichiometry of the balanced equation, if XM of N_{2}(*g*) reacts, then an equivalent concentration of O_{2}(*g*) must react. Therefore:

Equilibrium conc of O_{2}(*g*) = (4.89M - XM)

Furthermore, from the stoichiometry of the balanced equation, XM of N_{2}(*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} / ([N_{2}][O_{2}]) = (2X)^{2} / {(5.75 - X)(4.89 - X)}

2.1 x 10^{-3} = 4X^{2} / (28.1 - 10.6X + X^{2})

(2.1 x 10^{-3})*(28.1 - 10.6X + X^{2}) = 4X^{2}

0.0590 - 0.0223X + 0.0021X^{2} = 4X^{2}

4.00X^{2} + 0.0223X - 0.059 = 0

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

- For equations of the form ax
^{2}+ 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:

N_{2}(*g*) = (5.75 - 0.119) = 5.63M

O_{2}(*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:

K_{c} = [NO]^{2} / ([N_{2}][O_{2}]) = (0.238)^{2} / (5.63)(4.77)

K_{c} = 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 K_{c} for the Haber reaction at 472°C is 0.105. A 2.0L flask is filled with 0.2 mol of H_{2}(*g*), 1.8 mol of N_{2}(*g*) and 0.5 mol of NH_{3}(*g*). What are the concentrations of H_{2}(*g*), N_{2}(*g*) and NH_{3}(*g*) at equilibrium?

The equilibrium constant K_{c} 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:

- N
_{2}(*g*): 1.8mol/2.0L = 0.9M - H
_{2}(*g*): 0.2mol/2.0L = 0.1M - NH
_{3}(*g*): 0.5mol/2.0L = 0.25M

K_{c} 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 K
_{c}= 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 NH_{3}(*g*) product to produce N_{2}(*g*) and H_{2}(*g*) reactants will follow the stoichiometry of the balanced equation:

N_{2}(*g*) + 3H_{2}(*g*) <=> 2NH_{3}(*g*)

- Thus, for every mole of NH
_{3}(*g*) that decomposes, 0.5 mole of N_{2}(*g*) will be produced and 1.5 mole of H_{2}(*g*) will be produced - We know that a certain amount of NH
_{3}(*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 H_{2}(*g*), N_{2}(*g*) and NH_{3}(*g*) at equilibrium will equal the following:

- Equilibrium concentration of NH
_{3}(*g*) = (starting concentration - amount that decomposes)

= 0.25M - x - Equilibrium concentration of H
_{2}(*g*) = (starting concentration + (1.5 * amount of NH_{3}(*g*) that decomposes))

= 0.1M + 1.5x - Equilibrium concentration of N
_{2}(*g*) = (starting concentration + (0.5 * amount of NH_{3}(*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 K_{c}:

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 NH
_{3}(*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 NH
_{3}(*g*) = (0.25M - 0.178) = 0.072M - Equilibrium concentration of H
_{2}(*g*) = (0.1M + 1.5*0.178) = 0.367M - Equilibrium concentration of N
_{2}(*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