Chapter 19b Equilibrium

Chapter 19b: Equilibrium

Part I - Introduction to Equilibrium	Part II - Equilibrium Expressions	Part III - More Equilibrium Expressions
Part IV - Concentrations at Equilibrium	Part V - Solubility Product Constant	Part VI: Predicting Effects of Disturbing Equilibrium
Part VII - More Stresses on a System in Equilibrium

Chapter 19b- Equilibrium
Notes: Part 1 – Introduction to Equilibrium

Chemical Equilibrium

Chemical Equilibrium is a condition in which two opposing reactions are going on at equal rates producing no overall change in the concentrations of the substances involved. For example, if H₂ and N₂ are placed in a reaction chamber they react and join together according to the reaction N₂ + 3 H₂ --> 2 NH₃ and, as the N₂ and H₂ get used up in the reaction, the reaction slows down. However, the N₂ and the H₂ are never completely used up because as soon as some of NH₃ is produced by the above reaction, the reverse reaction: 2 NH₃ --> N₂ + 3 H₂ starts to take place and it speeds up as more and more NH₃ is produced. This “production of and using up of” NH₃ goes on until eventually the reverse reaction is using the NH₃ as quickly as the forward reaction is producing it. At this point, the concentrations of the NH₃, N₂, and H₂ remain constant. This is called the equilibrium point. During the time when the rates of the forward and reverse reactions are NOT equal, the system is said to be approaching equilibrium. If the forward reaction is faster than the reverse reaction, it is said to be proceeding to the right or that there is a net forward reaction.

How Equilibrium Reactions are Different from Other Reactions

If you were asked how many moles of water can be formed from 0.25 moles of H₂ when it burns according to the equation: 2 H₂ + O₂ --> 2 H₂O, you could use the equation to say that 2 moles of hydrogen produce 2 moles of H₂O and so, 0.25 moles of H₂ would produce 0.25 moles of H₂O. These statements are true because all of the 0.25 moles of H₂ are used up since there is no reverse reaction. However, if there is a reverse reaction and the equations involve an equilibrium system, the situation is different because not all of the reactants are used up.

For example, if you were asked the following question:
How many moles of NH₃ could be formed from 0.25 moles of H₂ where the reaction is as follows:

N₂ + 3 H₂ <--> 2 NH₃ (Note: the double arrows indicate an equilibrium system where there is a forward and a reverse reaction.)

You could not say “3 moles of H₂ produce 2 moles of NH₃ and so, 0.25 moles of H₂ would produce 0.25 mol H₂ x 2 mol NH₃/3 mol H₂ = 0.17 moles of NH₃ because not all of the 0.25 moles of H₂ would be used up. HOWEVER, THE EQUATION DOES TELL THE RELATION BETWEEN THE NUMBERS OF MOLES THAT ARE USED UP OR PRODUCED.

Using the above situation, for example, if you know that, at equilibrium, there was still 0.10 moles of H₂ remaining, this would tell you that 0.25 mol starting total – 0.10 mol remaining = 0.15 moles used up. Then you would be able to calculate that 0.15 moles H₂ used x 2 mol NH₃/3 mol H₂ = 0.10 moles of NH₃ produced.

Example: If 0.50 moles of H₂ and 1.50 moles of I₂ are placed in a container and react according to the equation: H₂ + I₂ <--> 2 HI, and if, at equilibrium, there is still 1.25 moles I₂ remaining, how many moles of H₂ and HI are in the equilibrium mixture?

Answer:

Chapter 19b: Equilibrium
Assignment: Part 1 – Introduction to Equilibrium

Answer the following questions on a separate sheet of paper, showing all work, set-ups, units, etc.

1. If 1.7 mole SO₂ and 1.9 mole O₂ are placed in a reaction chamber, and if, at equilibrium, there is still 1.0 mol SO₂ remaining, how many moles of O₂ and SO₃ are in the equilibrium mixture?

The equation is: 2 SO₂ + O₂ <--> 2 SO₃

2. 2.6 mol P₄ and 3.6 mol H₂ are reacted until the following equilibrium is established: P₄ + 6 H₂ <--> 4 PH₃. If, at equilibrium, 1.6 mol PH₃ are formed, how many moles of P₄ and H₂ are in the equilibrium mixture?

3. 1.0 mol of NH₃, 1.0 mol of O₂, 1.0 mol of NO, and 1.0 mol of H₂O are mixed and the following equilibrium is reached: 4 NH₃ + 5 O₂ <--> 4 NO + 6 H₂O. How many moles of (a) O₂ (b) NO and (c) H₂O remain, if, at equilibrium, there is still 0.80 moles of NH₃ remaining?

4. 1.5 mol N₂ and 1.5 mol O₂ are mixed. The equation is N₂ + O₂ <--> 2 NO. If the number of moles of NO at equilibrium is ‘x’, how many moles of N₂ are left (in terms of ‘x’)?

5. 1.5 moles of CO and 1.5 moles of O₂ are placed together and the following equilibrium is reached: 2 CO + O₂ <--> 2 CO₂. If the number of moles of O₂ is reduced by ‘x’ moles at equilibrium, how many moles of CO and CO₂ remain once the system has reached equilibrium?

6. 2.2 mol N₂, 3.3 mol H₂, and 1.0 mol NH₃ are approaching equilibrium as shown by the equation: N₂ + 3 H₂ <--> 2 NH₃. If, at equilibrium, the number of moles of NH₃ has increased by ‘x’ moles, what is the number of moles of N₂ and the number of moles of H₂ in the equilibrium mixture?

7. 1.2 mol HI is placed in a vessel and the following equilibrium is attained: H₂ + I₂ <--> 2 HI. If, at equilibrium, the number of moles of HI has been reduced by ‘x’ moles, what is the number of moles of H₂ and the number of I₂ in the equilibrium mixture?

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Chapter 19b - Equilibrium
Notes: Part 2 – Equilibrium Expressions

Mass Action Expression:

For the general balanced equation, n A + m B . . . <-->   p C + q D . . .
the Mass Action Expression is         _ _[C]^p [D]^q
                                                                [A]ⁿ [B]^m

where square brackets mean the concentration of the substance in MOLARITY.
(Note: it’s the product of the product side concentrations over the product of the reactant side concentrations.

Example 1: Write the mass action expression for N₂ + 3 H₂ <--> 2 NH₃.

Example 2: If 1.5 moles of N₂, 2.5 moles of H₂ and 0.50 moles of NH₃ are approaching equilibrium, according to the above equation, in a 500. mL container. What is the value of the mass action expression? (Remember that the concentration must be MOLARITY.)

The Law of Chemical Equilibrium

In a system at chemical equilibrium, the concentrations of the materials involved must be such that the mass action expression equals a certain constant called the equilibrium constant.

Equilibrium Condition refers to the expression              ___[C]^p [D]^q             = K_eq
                                                                                                        [A]ⁿ [B]^m
          where K_eq is called the equilibrium constant.

To Derive or Prove the Equilibrium Constant
- Consider the general equation as a 1 step reaction:           n A + m B <--> p C + q D
- The rate law for the forward reaction is:                               rate = k₁ [A]ⁿ [B]^m
- The rate law for the reverse reaction is:                                rate = k₂ [C]^p [D]^q
- Since at equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction:
                                                                                                      k₁ [A]ⁿ [B]^m = k₂ [C]^p [D]^q
- When we solve for a single constant:                                     k₁        =         [C]^p [D]^q
                                                                                                         k₂                    [A]ⁿ [B]^m
- So, letting k₁/k₂ = K_eq, we get the equilibrium condition of the above.

Significance of the Size of the Equilibrium Constant

A large value for K_eq (i.e. 10ⁿ) means that at equilibrium nearly all reactants have been used up and there are mainly products remaining. The reaction is said to go almost to completion.

A small value for K_eq (i.e. 10^-n) means that at equilibrium there is very little of the products and most of the reactants are left, i.e. it doesn’t proceed very far to the right.

Problems Involving the Equilibrium Constant

Evaluating K_eq

For the reaction, n A + m B <--> p C + q D, the usual equilibrium condition is [C]^p [D]^q
[A]ⁿ [B]^m

However, since the concentration of any solid (and in some cases, liquids) is fairly constant, it is left out of the mass action expression for the equilibrium constant and thereby it is included in the value of K_eq.
For example, the equilibrium condition for the following reaction:

C_(s) + O₂ _(g) <--> CO_{2 (g)} is [CO₂]
[O₂]

To determine the value of K_eq, you substitute the concentrations that are known to exist at equilibrium.

Example: It is known that at a certain high temperature, there exists in equilibrium, 2.5 mol CO₂, 1.5 mol O₂, and 2.5 mol C in a 2.5 L container. What is the value for K_eq?

We use the equilibrium expression from above and plug in the concentrations.
(Remember, the concentrations are in MOLARITY – which is moles/L.)
Answer:

(Note: the carbon is omitted because it is solid and its concentration does not change.)

Chapter 19b: Equilibrium
Assignment: Part 2 – Equilibrium Expressions

Answer the following questions on a separate sheet of paper, showing all work, set-ups, units, etc.

1.     Write the mass action expression for the following equilibrium systems:
                    a.     2 SO₂ + O₂ <--> 2 SO₃
                    b.     COCl₂   <--> CO + Cl₂
                    c.     P₄ + 6 H₂ <--> 4 PH₃
                    d.     4 NH₃ + 5 O₂ <--> 4 NO + 6 H₂O

2.     (A)     What is the value of the mass action expression for the equation in question 1(a) when 2.50 mol SO₂ and 1.00 mol O₂ are placed in a 500. mL container?
        (B)     What is the value of the expression after 0.500 mol of SO₂ have been used up?
        (C)     What is the value at equilibrium if, at equilibrium, there are still 1.50 mol of SO₂ remaining?

3. (A) What is the value of the mass action expression for the equation in question 1(d) when, in a 2.50 L container, there are 7.50 mol NH₃, 5.00 mol O₂, 25.0 mol NO, and 2.50 mol H₂O?
(B) What is the value of the mass action expression if, at equilibrium, the concentration of the NO has increased to 11 M?

4. Find the value of K_eq if, at equilibrium, there are 10.0 moles of P₄, 25.0 moles of H₂, and 5.00 moles PH₃ in a 5.00 L container.

The equation is: P_{4 (g)} + 6 H_{2 (g)} <--> 4 PH_{3 (g)}.

5. Find the value of K_eq for the equilibrium system, ZnO_(s) + CO_(g) <--> CO_{2 (g)} + Zn_(g) if, at equilibrium, there are 3.00 moles of CO, 4.00 moles Zn, 4.00 moles CO₂ in a 500. mL container.

6. Find K_eq for H_{2
(g)} + S_(l) <--> H₂S_(g) if, at equilibrium, the concentration of H₂ andH₂S are 1.5 M and 2.5 M, respectively.

7. For the system P_{4 (s)} + 5 O_{2 (g)} <--> P₄O_{10 (s)}, find the value of K_eq if initially there was 12.8 g P₄ and 1.00 mol O₂ in a 1.00 L container and if, at equilibrium, 0.400 g P₄ remained.

8. Initially the concentrations of N₂ and O₂ are 1.80 M and 1.80 M, respectively, and there is no NO. If, at equilibrium, the concentration of NO is 1.80 M, find K_eq.

The reaction N₂ _(g) + O_{2 (g)} <--> 2 NO_(g)

9. For the system 4 H_{2 (g)} + Fe₃O_{4 (s)} <--> 3 Fe_(s) + 4 H₂O_(g), find K_eq if, initially, there is only 1.60 M H₂ and 100. g Fe₃O₄and if, at equilibrium, the concentration of H₂ is 1.00 M.

10. Find K_eq for the equilibrium system 2CO_(g) + O_{2 (g)} <--> 2 CO_{2
(g)} if, initially, there are 5.00 moles of CO, 10.0 moles of O₂, and 1.00 mole of CO₂ in a 2.00 L container and if, at equilibrium, there is a CO₂ concentration of 2.50 M.

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Chapter 19b: Equilibrium
Notes: Part 3 – More Equilibrium Expressions

Problems Involving the Equilibrium Constant

Determining Equilibrium Concentrations from Equilibrium Constants

Example: How many moles of CO₂ would exist in equilibrium with 1.0 mole H₂, 2.0 moles CO, and 3.0 moles H₂O if K_eq = 1.60 for the system H_{2 (g)} + CO_{2 (g)} <--> H₂O_(g) + CO_(g) at 986^oC in a 500. mL container?

Answer:

Determining Equilibrium Concentrations from Initial Concentrations

Steps:

Select the substance in the equation with the lowest coefficient in front of it and

             (i) if it is on the left side of the equation, make it into a statement, “Let ‘x’ M of it be used up”.
             or
             (ii) if it is on the right side of the equation, make it into the statement, “Let ‘x’ M of it be produced”.

Write down the concentrations used up or produced of all the other substances, using the coefficients in the balanced equation.

For example: If the equation is 4 HCl + O₂ <--> 2 Cl₂ + 2 H₂O, you would “Let x M of O₂ be used up” and also “4x M of HCl would also be used up.” You would then say “Let 2x M of Cl₂ and 2x M of H₂O are then produced.”

In order to write the reaction equilibrium concentrations in terms of ‘x’, you either

             (i) subtract the concentration used up from the initial concentrations given
             or
             (ii) add the concentration produced to the initial concentration given (if any).

Substitute the above equilibrium concentrations and the given value for K_eq into the equilibrium condition expression and solve for ‘x’.
To determine the final answer for the equilibrium concentrations go back to the third step and put in the value for ‘x’.

Example: At a certain temperature, K_eq = 49.0 for the reaction A_(g) + B_(g) <--> 2C_(g). If, initially, 5.00 mol A and 5.00 mol B are placed in a 5.00 L container, what will be the equilibrium concentration of C? How many moles of A will be left?

Answer:

Chapter 19b: Equilibrium
Assignment – Part 3 – More Equilibrium Expressions

Answer the following on a separate sheet of paper, showing all work ,set-ups, units, etc.

1. If K_eq = 46.0 for H_{2 (g)} + I_{2 (g)} <--> 2 HI_(g), what concentration of I₂ would be in equilibrium with 0.500 M HI and 0.100 M H₂?

2. For the reaction A_(s) + 2 B_(g) <--> 2 C_(g), K_eq = 2.00 at 325 K. Find the equilibrium concentration of C if, at equilibrium, there are 3.00 moles of B and 1.00 moles of A in a 250. mL container.

3. At equilibrium in a 5.00 L container there are 5.00 mol HCl, 10.0 mol O₂, 15.0 mol Cl₂, and 20.0 mol H₂O. Find K_eq for 4 HCl_(g) + O_{2 (g)} <--> 2 Cl_{2 (g)} + 2 H₂O_(g). How many moles of O₂ could exists in this container with 0.500 moles of each HCl, Cl₂, and H₂O?

4. At a certain temperature there exists in equilibrium 10.0 M SO₂, 10.0 M O₂ and 3.50 M SO₃. The equation is 2 SO_{2 (g)} + O_{2 (g)} <--> 2 SO_{3 (g)}. At the same temperature, how many moles of SO₃ would be in equilibrium in a 2.50 L container with 25.0 mol O₂ and 5.00 mol SO₂?

5. K_eq = 81.0 at a certain temperature for the reaction 4 A_(g) + B_(s) <--> 2 C_(g) + 2 D_(g). If, initially, there were 7.00 moles of A and 7.00 moles of B in a 500.0 mL container, what would be the equilibrium concentration of C?

6. If K_eq = 1.00 x 10^-4 for the system N_{2 (g)} + O_{2 (g)} <--> 2NO_(g), find the number of moles of NO in a 10.0 L container if the initial concentrations of N₂ and O₂ are each 1.00 M.

7. K_eq = 25.0 for the reaction C_(s) + O_{2 (g)} <--> CO_{2 (g)}. Find the equilibrium concentration of CO₂ if initially, there are 100. moles of C, 50.0 moles of O₂, and 2.00 moles of CO₂ in a 2.00 L container.

8. If, for the system, NH₄Cl_(s) <--> NH_{3 (g)} + HCl_(g), K_eq = 1.00 x 10^-4. Find the moles of NH₃ in a 100.0 L container, if, initially, there were 10.0 mol NH₃ added to 10.0 mol HCl

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Chapter 19b: Equilibrium
Notes: Part 4 – Concentrations at Equilibrium

Predicting if Certain Concentrations will be at Equilibriums

Basic Idea: If you substitute the values for K_eq and the equilibrium concentrations into the equilibrium condition, the left and right sides MUST be EQUAL if there is equilibrium. If they are not equal, i.e. if K_eq does not equal the mass action expression, there are two possibilities:

(1) The mass action expression may be less than K_eq. In this case (a) the system is not at equilibrium, (b) there is a NET FORWARD REACTION, (c) the forward reaction is faster than the reverse and this causes an (d) increase in the concentration of products (on the right side) and this makes the mass action expression larger until it equals K_eq and there is equilibrium.

(2) The mass action expression may be greater than K_eq. In this case (a) the system is not at equilibrium, (b) there is a NET REVERSE REACTION, (c) the reverse reaction is faster than the forward one and this causes (d) an increase in the concentration of the reactants (on the left side) and this makes the mass action expression smaller until it equals K_eq and there is equilibrium.

Example: If K_eq = 0.90 for N₂ + 3 H₂ <--> 2 NH₃ (all gases), will the following be at equilibrium: 0.50 M of NH₃, 0.10 M of N₂, and 0.10 M of H₂?

Answer:

Since the mass action expression has a value of 2500 which is greater than the K_eq = 0.9, the system is not in equilibrium. There would be a net reverse reaction that would increase the concentration of N₂ and H₂ and decrease the concentration of NH₃ and so thereby make the value of the mass action expression smaller.

Chapter 19b: Equilibrium
Assignment: Part 4 – More Equilibrium Expressions

For each the following problems, answer each the following questions:
                    (A)      Is there equilibrium?
                    (B)      Which reaction is faster, the forward or the reverse?
                    (C)      Which concentrations are increasing? Which concentrations are decreasing?

1. K_eq = 25.0 for the system A_(s) + 2 B <--> C_(g). At a certain temperature, [B] = 2.00 M and [C] = 50.0 M.

2. At a certain temperature, in a 2.00 L container, there exists in equilibrium 5.00 mol CO₂. 5.00 mol CO, and 0.200 mol O₂. Find K_eq. Would there be equilibrium at the same temperature if the concentrations were [CO₂] = 15.8 M, [CO] = 10.0 M, and [O₂] = 0.250 M? The equation is 2 CO_(g) + O_{2 (g)} <--> 2 CO_{2 (g)}.

3. K_eq = 16.0 for the system 2 SO_{2 (g)} + O_2(g) <--> 2 SO_{3 (g)}. Initially, the concentrations are [SO₂] = 5.00 M, [O₂] = 10.0 M, and [SO₃] = 0.0 M. After 2 hours, the O₂ concentration has dropped to 7.90 M.

4. At a certain temperature, the following concentrations are at equilibrium: 0.100 M H₂O, 2.00 M O₂, 0.500 M NH₃, and 0.200 M NO. The equation is 4 NH_{3 (g)} + 5 O_{2 (g)} <--> 4 NO_(g) + 6 H₂O_(g). Answer the questions based on the following mixture: 0.750 mol H₂O, 12.0 mol NO, 30.0 mol O₂, and 0.300 mol NH₃ in a 3.00 L container.

5. K_eq = 46.0 for H_{2
(g)} + I_{2 (g)} <--> 2 HI_(g). Initially, there are 69.0 mol H₂ and 24.0 mol I₂ in a 10.0 L container. After 5 hours, there is still 1.00 mol I₂ left.

6. At a certain temperature, there exists an equilibrium when [A] = 1.00 M, [B] = 2.50 M, and [C] = 3.00 M. The equation is A_(g) + 2 B_(g) <--> 3 C_(g). Answer the question based on the following mixture: Initially, the concentrations of C = 18.0 M and B = 6.00 M and [A] = 0.0 M and after some time 2.00 M of A has been produced.

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Chapter 19b: Equilibrium
Notes: Part 5 – Solubility Product Constant

What is Ksp and How Do I Find It?

Soluble is a relative term. Most insoluble ionic solids are actually soluble in water to a limited extent. These solids dissociate slightly in water. The compound silver chloride dissociates slightly in water to silver ions (Ag⁺) and chloride ions (Cl^-). An equilibrium is established in the saturated solution between the solid and the ions in solution. The equation for this equilibrium is: AgCl_(s) <--> Ag⁺_(aq) + Cl^-_(aq).

          The above equation can be represented mathematically by a constant, K_eq, called the equilibrium constant. By definition, this constant is equal to the molar concentration of the products divided by the molar concentration of the reactants. The concentration of each ion is raised to a power equal to the coefficient of the ion in the balanced equation. This constant can be expressed as follows:
          Keq =      [products]     =        [Ag⁺] [Cl^-]
                           [reactants]                  [AgCl]
         The concentration of a pure solid, such as AgCl, is a constant. Since both terms, AgCl and K_eq, are constants, they may be multiplied together to form a new constant, which is known as the solubility product constant, or the K_sp.
               [Ag⁺] [Cl^-] = K_eq [AgCl] = K_sp

The solubility product constant, K_sp, is the product of the concentration of the ions in a saturated solution raised to the power of the coefficients in a balanced chemical equation. For example, the expression of the solubility product constant for PbCl₂ would be:

PbCl₂ <--> Pb⁺² + 2 Cl^-
K_sp = [Pb⁺²] [Cl^-]²

Using the equation for K_sp it is possible to calculate the solubility of a salt if its K_sp is known, or to calculate the K_sp from the solubility. The K_sp is an experimental value and depends upon the temperature, as does K_eq. The solubility of a salt must be expressed as the molar solubility of the salt, the Molarity, M.

Example 1: A 1.00 L saturated solution of AgCl is evaportaed to dryness and the residue is equivalent to 1.34 x 10^-5 mole of AgCl. What is the experimental K_sp of the silver chloride?

Example 2: At a specified temperature, the K_sp of AgCl is 1.56 x 10^-10. Determine the solubility of the AgCl in grams per Liter.

Using K_sp Values
A useful application of K_sp is to determine if precipitation will occur when a salt and a solution or when two solutions are mixed. Precipitation takes place when the ion product EXCEEDS the K_sp value.
                Ion product < K_sp      no precipitation will occur, unsaturated solution
                Ion product = K_sp      no precipitation will occur, saturated solution
                Ion product > K_sp           precipitation will occur, saturated with excess

Remember that if the final solution is formed by mixing two solutions it is necessary to consider the dilution. Each solute is diluted when the other solution is added. In other words, when a 1.00 mole sample that was dissolved to make 1.00 L of solution is mixed with another solution that also has a volume of 1.00 L, then the new volume is 2.00 L. So the molarity of the salt then needs to be calculated using the 2.00 L volume.

Example 3: Will precipitation occur when 50.0 mL of a 3.00 x 10^-2 M Pb(NO₃)₂ solution is mixed with 50.0 mL of a 2.00 x 10^-3 M KCl? The K_sp of PbCl₂ is 1.62 x 10^-5.

Chapter 19b: Equilibrium
Assignment: Part 5 – Solubility Product Constant

Answer the following problems on a separate sheet of paper, showing all work, set-ups, units, etc.

1.      From the given solubilities, determine the experimental values of the K_sp for the following compounds:
               a.     BaCO₃         solubility = 7.00 x 10^-5 M
               b.     Pb(OH)₂      solubility = 4.20 x 10^-6 M
               c.     CaF₂             solubility = 0.0170 grams / L

2.     Given the K_sp values, calculate the molar solubility, in Molarity, of the following compounds:
               a.     CuS              Ksp = 6.31 x 10^-36
               b.     Al(OH)₃      Ksp = 1.26 x 10^-33
               c.     SrC₂O₄        Ksp = 1.58 x 10^-7

3.     Determine if precipitation would occur in the following cases.
              a.     25.0 mL of a 6.00 x 10^-6 M Sr(NO₃)₂ solution is mixed with 25.0 mL of a 4.00 x 10^-7 M H₃PO₄ solution. The K_sp of Sr₃(PO₄)₂ is 4.07 x 10^-28. Will precipitation occur?
              b.     100.0 mL of 5.00 x 10^-3 M Ba(NO₃)₂ is mixed with 100.0 mL of 2.00 x 10^-2 M NaF. The K_sp of BaF² is 1.05 x 10^-6. Will precipitation occur?
              c.     50.0 mL of 6.00 x 10^-4 M AgNO₃ is mixed with 25.0 mL of 5.00 x 10^-4 M K₂CrO₄. The K_sp of Ag₂CrO₄ is 9.00 x 10^-12. Will precipitation occur?

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Chapter 19b: Equilibrium
Notes: Part 6 - Predicting the Effects of Disturbing a System in Equilibrium

Introduction:
In predicting the effects of disturbing a system in equilibrium, there are three principles that you should be able to use: Le Chatelier’s Principle, the principles of reaction kinetics, and the Principle of the Equilibrium Constant.

Le Chatelier’s Principle:
Le Chatelier’s Principle is not really an explanation of why an equilibrium system behaves as it does, but it is a handy, easy-to-use rule for predicting the effects of “stresses” on a system in equilibrium.

Le Chatelier’s Principle states that when a stress is applied to a system in equilibrium, the system proceeds either forward or backward so as to offset the stress. The stresses and the action that would offset them are listed below in the following table.

	*Stress*	*Action that Would Offset the Stress*
1	Increasing the concentration of a substance. (Another way of saying this is "the partial pressure of a gaseous substance in increased.")	Using up that substance (in either forward or reverse reaction) will offset the stress.
2	Decreasing the concentration of a substance. Remember, a solid reactant cannot increase or decrease in concentration.)	Producing more of that substance in a forward or reverse reaction (since this will reduce pressure).
3	Decreasing the enclosing volume of a gaseous mixture. This is the same as increasing the pressure of the gaseous mixture as a whole.	Producing fewer gas molecules in a forward or reverse reaction (since this will reduce pressure).
4	Increasing temperature	Using up heat in a forward or reverse reaction.
5	Decreasing temperature	Producing heat in a forward or reverse reaction.

Note: A catalyst is NOT considered a stress because it speeds up both reactions, the forward and the reverse, at equal rates.

Example 1: If the following system is in equilibrium: H_{2 (g)} + I_2(g) <--> 2 HI_(g), and then some more H₂ is added to a container of constant volume, what is the effect on equilibrium?

Answer: The stress:

Offset by:

Reaction:

[ ] Change:

Example 2:     The following system is in equilibrium: N_{2 (g)} + 3H_{2 (g)} <--> 2NH_{3 (g)} + 22 kcal
                           Then (a) the temperature is raised, and
                           (b) the total volume is compressed.
                           What is the effect on the equilibrium of each of these stresses?

Answer: (a) The stress:

Offset by:

Reaction:

[ ] Change:

(b) The stress:

Offset by:

Reaction:

[ ] Change:

Chapter 19b: Equilibrium
Assignment: Part 6 – Predicting the Effects of Disturbing a System in Equilibrium

Answer the following on a separate sheet of paper. For each of the following, state the stress, how the stress is offset, which reaction is favored, and the concentration changes.

1. What is the effect of reducing the concentration of Zn⁺² on the following:

Zn_(s) + Cu⁺² <--> Cu^(s) + Zn⁺² Explain in detail according to Le Chatelier’s Principle.

2. Explain in detail the effect of adding NaOH (forms Na⁺ and OH^-) to the following system in equilibrium:

NH₄⁺_(aq) + OH^-_(aq) <--> NH₃ _(g) + H₂O.

3. Explain in detail the effect of increasing the partial pressure of Cl₂ in the following equilibrium system:

PCl_{5 (g)} <--> PCl₃ _(g) + Cl₂ _(g).

4. Explain the effect of raising the temperature on the following:

N₂O_{4 (g)} + 14 kcal <--> 2 NO_{2 (g)}.

5. Explain the effect of reducing the temperature on the following:

2 C_(s) + O_{2 (g)} <--> 2 CO_(g) + 53 kcal

6. Explain the effect of raising the temperature on the following:

O_{2 (g)} <--> O_{2 (aq)} + 3 kcal

7. Explain the effect of decreasing the enclosing volume of:

N_{2 (g)} + O_{2 (g)} <--> 2 NO_(g)

8. Explain the effect of compressing the equilibrium mixture of the following:

2 NO_(g) + O_{2 (g)} <--> 2 NO_{2 (g)}

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Notes: Part 7 – More Stresses on a System in Equilibrium

Reaction Kinetics

You should also be able to use the principles of reaction kinetics in order to predict and explain the effects of disturbing a system in equilibrium. The important points of reaction kinetics are:

Concentration of Reactants – The rate law for a one step reaction, nA = mB ? products is: rate = k [A]n [B]m.
Temperature – All reactions, exothermic and endothermic, are faster at higher temperatures. Although all reactions are speeded up by a rise in temperature, endothermic reactions are speeded up MORE than exothermic ones, and they are also slowed down more by a decrease in temperature.
Catalysis – catalysts increase the rate by decreasing the activation energy barrier and thereby allowing a greater fraction of molecules to have the necessary energy to cause chemical change when they collide

Example 1: Explain by the principles of reaction kinetics the effect of removing some H₂ from the following system in equilibrium: H_{2 (g)} + I_{2 (g)} <--> 2 HI_(g).

Answer: Reaction:

Net reaction:

[ ] Change:

Example 2: What is the effect of decreasing the temperature of the following system:

N_{2 (g)} + 3 H_{2 (g)} <--> 2 NH_{3 (g)} + 22 kcal

Answer: Reaction:

Net reaction:

[ ] Change:

Recall: With regard to decreasing the volume of a reaction of a mixture of gases in equilibrium, although the concentration of all substances in the equilibrium increase with a decrease in volume and, although this tends to increase both forward and reverse reactions, if the number of molecules of reactants for the forward reaction is greater than that for the reverse reaction, the forward reaction is speeded up more than the reverse.

Example 3: What is the effect of compressing to a smaller volume, the system:

N₂O_{4 (g)} <--> 2 NO_{2 (g)}

Answer: Reaction:

Net reaction:

[ ]Changes:

Example 4: What is the effect of adding a catalyst to the following system in equilibrium:

H_{2 (g)} + I_{2 (g)} <--> 2 HI_(g)

Answer: Reaction:

Net reaction:

[ ] Change:

Principle of the Equilibrium Constant
The basic idea of the equilibrium constant is that (a) when a system is at equilibrium there is an equality between K_eq and the Mass Action Expression and (b) if the equality is upset by (1) changing the value of K_eq by raising or lowering the temperature or (2) changing the concentrations of the substances in the Mass Action Expression, the system will proceed to restore equality between K_eq and the Mass Action Expression.

Example 5: What is the effect of removing some I₂ from the system: H₂ + I₂ <--> 2 HI.

Answer: K_eq or MAE:

Restore equality:

Example 6: Explain the effect of compressing the following equilibrium system:

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

Answer:

Recall:
A RISE in temperature causes Keq to
               (i)      DECREASE if the forward reaction is exothermic
               (ii)     INCREASE if the forward reaction is endothermic
A DROP in temperature causes Keq to
               (i)      INCREASE if the forward reaction is exothermic
               (ii)     DECREASE if the forward reaction is endothermic

Example 7: What is the effect of raising the temperature of the following system:

N₂ + O₂ <--> 2 NO – 43 kcal

Answer:

Chapter 19b: Equilibrium
Assignment: Part 7 – More Stresses on a System in Equilibrium

Answer the following on a separate sheet of paper. Show all parts, including which reaction is favored to offset the given stress.

1. Explain in terms of kinetics the effect of each of the following stresses on the system in equilibrium:

PCl_{5 (g)} <--> PCl_{3 (g)} + Cl_{2 (g)}                (a)     more Cl₂ is added
               (b)     the enclosing volume is decreased
               (c)     a catalyst is added

2. Explain the effect of the following stresses on the system:

4 HCl + O₂ <--> Cl₂ + 2 H₂O + 28 kcal (all gases) (a) the concentration of H₂O is decreased
(b) the temperature is raised

3. Explain the effect of the stresses on the system:

N₂ + O₂ <--> 2 NO – 43,200 cal (all gases) (a) decreasing the enclosing volume
(b) lowering the temperature

4. Explain in terms of the equilibrium constant the effect of the following changes on the equilibrium of the system:

CO_(g) + 2 H_{2 (g)} <--> CH₃OH_(g) + 31 kcal                 (a)     adding more CO
                (b)    compression to a smaller volume
                (c)     raising the temperature

5. Explain the effect on the equilibrium of the system by the following changes:

CO_{2 (g)} + H_{2 (g)} <--> CO_(g) + H₂O – 10 kcal (a) compression to a smaller volume
(b) a rise in temperature

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