UNIT 1

EQUILIBRIUM

(Text p.558 – 593)

  • Physical Equilibrium
  • Chemical Equilibrium
  • Equilibrium Law Expression
  • Le Chatelier’s Principle
  • Interpreting Concentration Versus Time Graphs
  • Practical Applications of Le Chatelier’s Principle
  • Solubility Product (Ksp) Expressions
  • Practical Applications of Salts with Low Solubilities

Reversible Reactions


When a reaction results in almost a complete conversion of reactants to products, chemists say that the reaction goes to completion. But most reactions do not go to completion. They appear to stop. This is because these reactions are reversible.

Reversible Reaction: A reaction that can take place in both the forward and reverse directions. May lead to equilibrium if forward and reverse rates are equal.

Consider the physical changes of evaporation and condensation. These are opposing processes and reverse reactions of each other.

H2O(liquid) H2O(gas)EVAPORATION

H2O(gas)  H2O(liquid) CONDENSATION

which is the same as

H2O(liquid) H2O(gas)

Thus, chemists combine these two equations into a single equation that uses a double arrow to show that both reactions occur at the same time.

H2O(liquid) H2O(gas)

When a reaction is reversible, sometimes the formation of reactants are favoured over the products, and vice versa. This also relates to the rate of the reaction. Both the forward and reverse reactions have a rate. How the double arrow is drawn can inform you as to which has a greater rate, the reactants or products.

Example:

The arrow for the reverse reaction is larger than the arrow for the forward reaction. This indicates that the rate of consumption of NaCl(s) is greater than the rate of consumption of NaCl(aq).

Interpreting Equations

  1. In your own words, write down your interpretation of the following equations:
  2. H2O(liquid) H2O(gas)Rate of H2O gas formation is greater than the rate of H2O liquid formation.
  3. N2O4 2 NO2Rates are equal
  1. H2(g) + I2(g) 2 HI(g) Rate of HI consumption is greater than rate or H2 and I2 consumption.
  1. PCl5(g) PCl(g) + Cl2(g)Rates are equal
  1. COCl2(g) CO(g) + Cl2(g)Rate of COCl2 consumption is greater than rate of CO and Cl2 consumption.
  1. Which of the above is a physical change rather than a chemical change?

A is a physical change from liquid to gas.

  1. Write an equation for the following word descriptions:
  2. The rate of consumption of dinitrogen tetroxide to nitrogen dioxide is higher than the rate of formation of dinitrogen tetroxide.

N2O4 2 NO2

  1. The rate of formation of PCl5(g) is equal to the rate of formation of PCl(g) + Cl2(g).

PCl(g) + Cl2(g) PCl5(g)

  1. The rate of consumption of HI(g) is greater than the rate of consumption of H2(g) + I2(g).

2 HI(g) H2(g) + I2(g)

Physical Equilibrium

Vapor-Liquid Equilibrium

Situation 1: Uncovered beaker half filled with water.

Situation 2: Covered beaker half filled with water

Eventually rate of evaporation = rate of condensation

What happens when the water in situation 2 is heated?

Physical Equilibrium

Solution Equilibrium

Situation 1: A solution of KCl in 100 g of water at 20˚C.

Rate of dissolving

=

Rate of crystallization

Situation 2: The solution is heated.

If system remains at this new temperature, it will go back to equilibrium.

Question: Which solution; unsaturated, saturated, or supersaturated is at solution equilibrium?

Equilibrium: At equilibrium the concentration of reactants and products are CONSTANT NOT EQUAL. Forward rate = reverse rate.

Chemical Equilibrium

Macroscopic (what you see) Definition: A reaction occurring in a closed system, all reactants and products are present and the observable properties remain the same

Microscopic (what you can’t see) Definition:The reactants are forming products at the same rate as the products are forming reactants.

Analogy

Imagine PoloPark first thing in the morning. At first there is no one in the building, but when the doors open, people stream in through the main floor doors. Since the second floor doors are closed, everyone is at first on the main floor. Gradually people make their way up to the second floor. Now let’s pretend that someone has locked all the entrances. We now have a closed system. However, the shoppers haven’t noticed and are happily moving between the first and second floors. Some people are moving up and others are moving down; but at any one time the number of people on each floor is constant. This situation is an example of a dynamic equilibrium.

Equilibrium is a ‘dynamic’ process because it is in a state of action i.e. reactants are changing to products as products are changing to reactants.

*At equilibrium the concentration of reactants does not change AND the concentration of products does not change. They do NOT have to be equal to each other*

Equilibrium

Conditions Required for Equilibrium:

  1. Constant observable macroscopic properties
  2. A closed system
  3. Constant temperature and pressure
  4. Reversibility

Compare and Contrast Physical and Chemical Equilibrium

chemical reaction/ physical change

change

  • forward rate = reverse rate
  • dynamic
  • constant observable macroscopic properties (temperature, pressure, concentration)
  • closed system

Graphical Representation

How systems achieve equilibrium can be demonstrated through rate vs. time and concentration vs. time graphs.

Equilibrium is attained when the rates of the two opposing reactions are equal. Thus, at equilibrium, the two rates are EQUAL.

Note: At equilibrium, concentration of the reactants DO NOT have to equal the concentration of products. The concentrations must be constant over time.

Equilibrium Constant (Keq)

Equilibrium Constant: A mathematical expression giving the ratio of the products to the reactants.

Homogeneous v.s. Heterogeneous Equilibria

Homogeneous system: products and reactants are in the same phase.

Heterogeneous system: products and reactants are in different phases.

Note: Since the concentrations of solids and liquids are fixed by their densities and do NOT change during a chemical reaction, it is NOT necessary to include them in the mass action expression.

General Equation for Equilibrium Constant (Mass Action Expression):

mA + nB  sP + rQ

ex. 2 HI  H2 + I2

Example:2 C(s) + O2(g) ↔ 2 CO(g)

Example: H2(g) + S(l) ↔ H2S(g)

Equilibrium Constant (Keq)

Equilibrium Constant Expressions for Reverse Reactions

The constant (Keq) for the reverse reaction is the reciprocal of the constant for the forward reaction.

Example:

If Keq for the forward reaction is 10, then Keq for the reverse reaction is 1/10 = 0.10

Example:

H2 + I2 2 HI Keq = 15

2 HI  H2 + I2 Keq = 1/15 = 0.067

Meaning of the Magnitude of Equilibrium Constants

If Keq < 1 – more reactants than products at equilibrium

If Keq > 1 - more products than reactants at equilibrium

How Equilibrium Constants are Determined

To determine the equilibrium constants, we need the concentration of products and reactants at equilibrium.

To determine these concentrations, we could:

1. Measure the pressure of the gases formed with a manometer.

2.If there is a color change involved, measure the intensity of the substance to determine the concentration at equilibrium.

3. If there is ionization occurring, measure the electrical conductivity of the solution.

Keq Calculations

Example 1:

What is the value of the equilibrium constant for the following reaction if the final concentrations are:

C2H4O2 = 0.302 M

C2H6O = 0.428M

H2O = 0.654 M and

C4H8O2 = 0.655 M?

C2H4O2 + C2H6O ↔ C4H8O2 + H2O

Example 2:

What is the equilibrium concentration of SO3 in the following reaction if the concentration of SO2 and O2 are each 0.0500 M and Keq is 85.0?

2 SO2 + O2 ↔ 2 SO3

Equilibrium Constant (Keq) Worksheet

Write the expression for the equilibrium constant Keq for the reactions below in the left hand column and then solve for the equilibrium constant (Keq).

Keq Expression / Solve for Keq
N2(g) + 3H2(g) ↔ 2NH3(g)
/ [NH3] = 0.01 M, [N2] = 0.02 M, [H2] = 0.02 M
2KClO3(s) ↔ 2KCl(s) + 3O2(g)
/ [O2] = 0.05 M
H2O(l) ↔ H+(aq) + OH-(aq)
/ [H+] = 1 x 10-8 M, [OH-] = 1. x 10-6 M
2CO(g) + O2(g) ↔ 2CO2(g)
/ [CO] = 2.0 M, [O2] = 1.5 M, [CO2] = 3.0 M
Li2CO3(s)  2Li+(aq) + CO32-(aq)
/ [Li+] = 0.2 M, [CO32-] = 0.1 M

Calculating an Equilibrium Constant

When the values of the concentrations of all the reactants and products are known AT EQUILIBRIUM, calculating an equilibrium constant simply involves substituting the data into the equilibrium constant expression.

Often an experiment will only provide information on the initial quantities of reactants and the concentration at equilibrium of only one of the reactants or one of the products.

Example:

Consider the oxidation of sulfur dioxide to sulfur trioxide:

2 SO2(g) + O2 ↔ 2 SO3(g)

Suppose that, in an experiment to determine Keq for this reaction, you place 1.00 mol of SO2 and 1.00 mol of O2 into a 1L flask. Concentration of reactants can be determined.

The balanced chemical equation helps to define the situation at equilibrium.

The changes can be expressed in terms of ‘x’, and the known coefficients in the balanced equation. These numbers can be displayed in an equilibrium table (ICE CHART):

Equation:2 SO2 + O2 2 SO3

Initial (mol/L)1.00 1.00 0

Change (mol/L) -2x - x + 2x

Equilibrium (mol/L)1-2x 1 - x 2x

I – Initial concentration (mol/L)

C – Change to get to equilibrium

E – Equilibrium concentration (mol/L)

Equilibrium Calculations

Variation 1: Giveninitial concentrations of reactants and concentration of products formed at equilibrium.

1.00 mol of SO2 and 1.00 mol O2 are placed in a 1.00 L flask at 1000K. When equilibrium has been achieved, 0.925 mol of SO3 has formed. Calculate Keq for the reaction.

2 SO2(g) + O2(g) ↔ 2 SO3(g)

Equilibrium Calculations

Variation 2:Giveninitial concentrations of reactants and the concentration of reactant lost to get to equilibrium.

1.00 mol of ethanol and 1.00 mol of acetic acid are dissolved in water and kept at 100ºC. The volume of the solution is 250 mL. At equilibrium 0.25 mol of acetic acid has been consumed in producing ethyl acetate. Calculate Keq at 100ºC for the reaction.

C2H5OH(aq) + CH3CO2H ↔ CH3CO2C2H5(aq) + H2O(l)

Equilibrium Calculations

Variation 3: Giveninitial concentration of reactants and Keq.

If 1.00 mol each of H2 and I2 are placed in a 0.500 L flask at 425ºC, what are the equilibrium concentrations of H2, I2, and HI? At 425ºC Keq is 55.64 for the reaction.

H2(g) + I2(g) ↔ 2 HI(g)

Equilibrium Calculations

Variation 4: Giveninitial concentration of reactants and Keq, but must use quadratic equation to solve.

If 1.00 mol of Mg+2 and 1.50 mol of IO4- are placed in a 0.500 L flask at 425ºC. At 425ºC Keq is 2.5x10-13 for the reaction. These ions react according to the following chemical equation:

Mg+2(g) + IO4-(g) ↔ Mg(IO4)2(g)

What will the concentration of Mg+2 be at equilibrium?

Equilibrium Calculations Assignment

Total: / 22 marks

Complete the following questions on a piece of loose leaf. Due: ______

1. For the reaction

N2(g) + 3 H2(g)2 NH3(g)

at 225°C, a 2.0 L container holds 0.040 moles of N2, 0.15 moles of H2 and 0.50 moles of NH3. If the system is at equilibrium, calculate Keq. (3 marks)

2. For the reaction:H2(g) + CO2(g) H2O(g) + CO(g)

Initially 1.00 mol of H2 and 1.00 mol CO2 were placed in a sealed 5.00 L vessel.
What would be the equilibrium concentrations if Keq is 0.772? (5 marks)

3. A 10.0 L flask is filled with 0.200 mol of HI at 698 K. What will be the concentration of H2, I2 and HI at equilibrium? The value of Keq is 0.0184. (4 marks)

H2(g) + I2(g)  2 HI(g)

4. Phosphorus pentachloride, PCl5(g), dissociates at high temperature into phosphorous trichloride, PCl3(g), and chlorine, Cl2(g). Initially, 0.200 mol of PCl5 were placed in a 5.00 L flask at 200ºC, and at equilibrium the concentration of PCl5 was found to be 0.015 M. Calculate the value of the equilibrium constant at 200ºC. (5 marks)

5. Initially, there is 6.30g of CO and 9.11g of O2 in a 2.00L container. These react together to form carbon dioxide. If at equilibrium, the carbon dioxide concentration is 1.20M, what is the value of the equilibrium constant? (5 marks)

A solution was prepared such that the initial concentrations of Zn+2(g) and NO3-(g) were 0.0300M and 0.0240M, respectively. These ions react according to the following chemical equation:

Zn+2 + NO3-Zn(NO3)2 Keq=2.5 x 1023

What will be the concentration of Zn+2 at equilibrium? (5 marks)

Recall:

5. The equilibrium constant at 490ºC for the reaction 2 HI(g)  H2(g) + I2(g) Keq is 0.022. What are the equilibrium concentrations of HI, H2, and I2, when an initial amount of 2.00 mol of HI gas is placed in a 4.3 L flask at 490ºC.

6. Calculate Keq for the decomposition of H2S from this reaction:

2 H2S(g)  2 H2(g) + S2(g)

A tank initially contained H2S with a concentration of 10.00 mol/L at 800 K.
When the reaction had come to equilibrium, the concentration of S2 gas was
2.0 x 10-2 mol/L.

7. The equilibrium constant, Keq, for the reaction:N2O4(g) 2 NO2(g) at 25ºC is 5.88 x 10-3. Suppose 15.6 g of N2O4 is placed in a 5.00 L flask at 25ºC.

Calculate the number of moles of NO2 present at equilibrium.

Enthalpy and Entropy

Two factors influence whether or not a system reaches equilibrium:

  1. Enthalpy
  2. Entropy

Enthalpy: heat content or PE of a substance. A system tends to go from a state of higher energy to one of lower energy.

Entropy: degree of randomness or disorder.

*Natural processes tend to go from an orderly state to a disorderly one (entropy).*

A reaction proceeds in the direction that favours lowest enthalpyand highest entropy. Equilibrium is a compromise between these two factors.

When both factors favour products, the reaction goes to completion.

When both factors favour reactants, the reaction does not happen.

When one favours reactants and the other favours products, EQUILIBRIUM is reached.

Predict which substances have higher entropy

  1. Gases or liquids.
  2. Solids or liquids.
  3. Dissolved solids and liquids or pure solids and liquids.

Gases have the highest entropy. Entropy of gases and dissolved substances increases as their amounts are increased.

Entropy increases as you raise the temperature.

Enthalpy and Entropy

Reactions Going to Completion

Enthalpy and Entropy favour PRODUCTS.

Zn(s) + 2HCl(aq)  ZnCl2(aq) + H2(g) + heat

Entropy is highest for:products

Enthalpy is lowest for:products

Reactions that Do Not Occur

Enthalpy and Entropy favour REACTANTS.

3C(s) + 3H2(g) + heat  C3H6(g)

Entropy is highest for:reactants

Enthalpy is lowest for:reactants

Reactions that Establish Equilibrium

One factor favours reactants, the other favours products.

e.g. N2O4(g) + heat  2NO2(g)

Entropy is highest for:products

Enthalpy is lowest for:reactants

Practice Problems

  1. For each of the following reactions, decide whether the reactants or the products have the greater entropy. Indicate the cases in which no change occurs.
  2. 2 Al(s) + 6 HCl(aq)  2 AlCl3(aq) + 3 H2(g)
  1. CaCO3(s)  CaO(s) + CO2(g)
  1. N2(g) + 3 H2(g)  2 NH3(g)
  1. H2(g) + I2(g)  2 HI(g)
  1. I2(s)  I2(alcohol solution)
  1. H2O(l)  H2O(s)
  1. For each of the following reactions, decide on the basis of entropy and enthalpy whether a reaction in the direction shown will go to completion, reach a state of equilibrium, or not occur at all. (Assume a closed system).
  2. Cl2(g)  Cl2(aq) + 25 kJ
  1. Na(s) + H2O(l)  Na+(aq) + OH-(aq) + ½ H2(g)∆H = - 184 kJ
  1. ½ N2(g) + O2(g)  NO2(g)∆H = + 33 kJ
  1. P4(s) + 6 H2(g)  4 PH3(g)∆H = + 37 kJ
  1. Na2CO3(s) + 2 HCl(aq)  2 NaCl(aq) + CO2(g) + H2O(l) + 27.7 kJ

Le Chatelier’s Demonstration

Reaction:CoCl42- + 6H2O  Co(H2O)62+ + 4Cl- + heat

1. Adding a reactant, H2O

Predict / Observe / Explain

2. Adding a product, Cl-(remember: when HCl is put in solution it dissociates to form H+ and Cl- ions)

Predict / Observe / Explain

3. Removing a product, Cl-(AgNO3 dissociates into Ag+ + NO3- ; Ag+ combines with Cl- to form an AgCl precipitate)

Predict / Observe / Explain

Le Chatelier’s Demonstration

Reaction:CoCl42- + 6H2O  Co(H2O)62+ + 4Cl- + heat

4. Adding heat

Predict / Observe / Explain

5. Removing heat i.e. cooling the solution

Predict / Observe / Explain

Le Chatelier’s Principle

What happens to an equilibrium system when you impose a change on it?

It constitutes a “stress” and the system wants to oppose that stress ie. undo it.

There are four common ways a chemical reaction at equilibrium may be disturbed:

  1. Change in temperature
  2. Change in concentration of a reactant or product
  3. Change in pressure (gaseous systems only)
  4. Adding a catalyst

According to Le Chatelier: When a stress is applied to an equilibrium system, changes occur in the system that remove the stress (at least partially).

This is called “Le Chatelier’s Principle.”

It means that the system tends to undo (at least partially) what is being done to it.

Examples:

  • If you add heat, the system will remove the heat (partially)
  • If you increase the concentration of a substance, the system will tend to lower it again.
  • If you increase the pressure, the system will lower it (partially again).

Effects of Disturbances

on Equilibrium and Keq

Disturbance / Change That Occurs as System Returns to Equilibrium / Effect on Equilibrium / Effect on Keq
Addition of reactant / Some added reactant is consumed / Right / No change
Addition of product / Some added product is consumed / Left / No change
Decrease in volume
Increase in pressure / Pressure decreases / Shifts toward fewer gas molecules / No change
Increase in volume
Decrease in pressure / Pressure increases / Shifts toward more gas molecules / No change
Rise in temperature / Energy will be consumed / Shifts toward endothermic reaction / K will increase if products are formed
Drop in temperature / Energy will be produced / Shifts toward exothermic reaction / K will decrease if reactants are formed

Practice Problems

N2(g) + 3 H2(g)  2 NH3(g)ΔH = -22 kcal

  1. Does equilibrium shift to the left or the right when extra H2 is added? When extra NH3 is added?
  1. What is the effect on the position of equilibrium when the volume of the system is increased? Does the equilibrium shift to the left or to the right, or is the system unchanged?
  1. List all the stresses that can be imposed on this system to maximize the production of NH3.

Le Chatelier’s Principle Worksheet

Complete the following chart by writing left, right, or none for equilibrium shift, and decreases, increases, or remains the same for the concentrations of reactants and products, and for the value of Keq.

N2(g) + 3 H2(g)  2 NH3(g) + 22 kcal

Stress / Equilibrium Shift / [N2] / [H2] / [NH3] / Keq
1. Add N2 / right / ------/ decreases / increases / remains the same
2. Add H2 /  /  / ------/ 
3. Add NH3 /  /  /  / ------
4. Remove N2 /  / ------/  / 
5. Remove H2 /  /  / ------/ 
6. Remove NH3 /  /  /  / ------
7. Increase Temperature /  /  /  /  / Smaller
8. Decrease Temperature /  /  /  /  / Larger
9. Increase Pressure /  /  /  /  / no change
10. Decrease Pressure /  /  /  /  / no change

Le Chatelier’s Principle Worksheet cont.

12.6 kcal + H2(g) + I2(g) ↔ 2 HI(g)

Stress / Equilibrium Shift / [H2] / [I2] / [HI] / Keq
1. Add H2 / right / ------/ decreases / increases / remains the same
2. Add I2 /  /  / ------/ 
3. Add HI /  /  /  / ------
4. Remove H2 /  / ------/  / 
5. Remove I2 /  /  / ------/ 
6. Remove HI /  /  /  / ------
7. Increase Temperature /  /  /  /  / 
8. Decrease Temperature /  /  /  /  / 
9. Increase Pressure / NO CHANGE
10. Decrease Pressure

Le Chatelier’s Principle Worksheet cont.

NaOH(s) ↔ Na+(aq) + OH-(aq) + 10.6 kcal

*remember that pure solids and liquids do not affect equilibrium values

Stress / Equilibrium Shift / Amount NaOH(s) / [Na+] / [OH-] / Keq
1. Add NaOH(s) / / ------/ / /
2. Add NaCl (adds Na+) /  /  / ------/  / remain the same
3. Add KOH (adds OH-) /  /  /  / ------
4. Add H+ (removes OH-) /  /  /  / ------
5. Increase Temperature /  /  /  /  / 
6. Decrease Temperature /  /  /  /  / 
7. Increase Pressure / NO CHANGE
8. Decrease Pressure

Reaction Quotient

When reactants and products are added into a container it is good to know whether equilibrium has been reached. If equilibrium has not been reached it is helpful to know which reaction, forward or reverse, is favoured in order for equilibrium to be achieved.

The reaction quotient, Q, or trial Keq is determined by using the equilibrium law or mass action expression and substituting either initial concentrations or those determined during experimental trials.

To determine which reaction is favoured and in which direction the system is moving, Q is compared to Keq:

  • If Q = K, the system is at equilibrium. The forward and reverse rates are equal and the reactant and product concentrations remain constant.
  • If Q > K, the system is NOT at equilibrium. There is too much product, so the reverse reaction is favoured to bring the reactant-product ratio to equal K by increasing reactant concentration.
  • If Q < K, the system is NOT at equilibrium. The concentration of reactants is too large, so the forward reaction is favoured. This results in decreasing reactant concentrations and increasing product concentrations, bringing their ratio to a value equal to K.

Reaction Quotient Examples

Example 1:

For the reaction

N2(g) + O2(g) ↔2 NO(g)

It was found that 8.50 moles of nitrogen, 11.0 moles of oxygen and 2.20 moles of nitrogen monoxide were in a 5.00 L container. If the equilibrium constant is 0.035, are the following concentrations at equilibrium? If not, which reaction is favoured and which concentrations are increasing and which are decreasing?

Example 2:

For the following imaginary equilibrium system the value of K is 0.22

2 A(g) + B (g) ↔ 3 C(g)

1.50 moles of A and 3.20 moles of B are placed into a 1.0 L container and allowed to react. After several minutes, a sample is taken and found to contain 1.00 moles of A. Is the system at equilibrium? Which reaction rate is fastest? Which concentrations are increasing?