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UNIT 1EQUILIBRIUM

(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) Gr. 12 Chemistry Equilibrium

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).

Gr. 12 Chemistry Equilibrium

Interpreting Equations

1. In your own words, write down your interpretation of the following equations:a. H2O(liquid) H2O(gas) Rate of H2O gas formation is greater than the rate

of H2O liquid formation.b. N2O4 2 NO2 Rates are equal

c. H2(g) + I2(g) 2 HI(g) Rate of HI consumption is greater than rate or H2 and I2 consumption.

d. PCl5(g) PCl(g) + Cl2(g) Rates are equal

e. COCl2(g) CO(g) + Cl2(g) Rate of COCl2 consumption is greater than rate of CO and Cl2 consumption.

2. Which of the above is a physical change rather than a chemical change?A is a physical change from liquid to gas.

3. Write an equation for the following word descriptions:a. The rate of consumption of dinitrogen tetroxide to nitrogen dioxide is

higher than the rate of formation of dinitrogen tetroxide.N2O4 2 NO2

b. 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)

c. 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)



Vapor-Liquid EquilibriumSituation 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?


Add heat If system remains at this new temperature, it will go back to 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?Gr. 12 Chemistry Equilibrium

Add heat

KCl KCl

KClK+

Cl-

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.

AnalogyImagine Polo Park 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 Gr. 12 Chemistry Equilibrium

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)


DifferenceChemical Physical

Similarities

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)Gr. 12 Chemistry Equilibrium

Equilibrium Constant Expressions for Reverse ReactionsThe 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 MC2H6O = 0.428MH2O = 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 SO 2 + O 2 2 SO 3

Initial (mol/L) 1.00 1.00 0

C hange (mol/L) -2x - x + 2x

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

I – Initial concentration (mol/L)C – Change to get to equilibriumE – Equilibrium concentration (mol/L)


Equilibrium Calculations

Variation 1: Given initial 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)



Variation 2 : Given initial 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)



Variation 3: Given initial 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)



Variation 4: Given initial 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 reactionN2(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. Enthalpy2. 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 enthalpy and 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 CompletionEnthalpy 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 OccurEnthalpy and Entropy favour REACTANTS.

3C(s) + 3H2(g) + heat C3H6(g)

Entropy is highest for: reactants

Enthalpy is lowest for: reactants


Reactions that Establish EquilibriumOne 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.

a. 2 Al(s) + 6 HCl(aq) 2 AlCl3(aq) + 3 H2(g)

b. CaCO3(s) CaO(s) + CO2(g)

c. N2(g) + 3 H2(g) 2 NH3(g)

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

e. I2(s) I2(alcohol solution)

f. H2O(l) H2O(s)

2. 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).

a. Cl2(g) Cl2(aq) + 25 kJ

b. Na(s) + H2O(l) Na+(aq) + OH-

(aq) + ½ H2(g) ∆H = - 184 kJ

c. ½ N2(g) + O2(g) NO2(g) ∆H = + 33 kJ

d. P4(s) + 6 H2(g) 4 PH3(g) ∆H = + 37 kJ

e. Na2CO3(s) + 2 HCl(aq) 2 NaCl(aq) + CO2(g) + H2O(l) + 27.7 kJ


Le Chatelier’s Demonstration

Reaction: CoCl42- + 6H2O Co(H2O)6

2+ + 4Cl- + heat

1. Adding a reactant, H2OPredict 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)6

2+ + 4Cl- + heat

4. Adding heatPredict Observe Explain

5. Removing heat i.e. cooling the solutionPredict 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 volumeIncrease in pressure

Pressure decreases Shifts toward fewer gas molecules No change

Increase in volumeDecrease 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?

2. 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?

3. 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 increasesremains 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 CHANGE10. Decrease

Pressure

Le Chatelier’s Principle Worksheet cont.Gr. 12 Chemistry Equilibrium

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 CHANGE8. Decrease

Pressure


Reaction QuotientWhen reactants and products are added into a con-tainer 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 re-verse, 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 expres-sion 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 com-pared to Keq:

If Q = K, the system is at equilibrium. The for-ward 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 reac-tion is favoured to bring the reactant-product ra-tio to equal K by increasing reactant concentra-tion.

If Q < K, the system is NOT at equilibrium. The concentration of reactants is too large, so the for-


ward reaction is favoured. This results in de-creasing reactant concentrations and increasing product concentrations, bringing their ratio to a value equal to K.

Reaction Quotient ExamplesGr. 12 Chemistry Equilibrium

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 con-stant 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 re-action rate is fastest? Which concentrations are in-creasing?


Graphing the Effects of Stresses on an Equilibrium System

2 NO2 N2O4 ∆H = -58.0kJ

Addition of NO2 Removal of NO2

Addition of N2O4 Removal of N2O4

Increase in temperature Decrease in temperature


time

Graphing the Effects of Stresses on an Equilibrium System

Example: Show the following situations graphically in the spaces provided.

Equilibrium is: H2(g) + I2(g) ↔ 2 HI(g) + 52 kJ

a. Increase in temperatureb. Decrease volumec. Inject some H2(g)

d. Add a catalyst

Equilibrium is: 2 SO2(g) + O2(g) ↔ 2 SO3(g) ∆H = -197 kJ

a. Inject some SO2(g)

b. Decrease temperaturec. Increase volumed. Increase [SO3]

Interpreting Graphs Worksheet

1. The graph below shows concentration versus time for a system containing carbon monoxide (CO), chlorine gas (Cl2), and phosgene (COCl2).


a. Write a balanced equation to represent the reaction studied.

b. How much time was required for the system to reach equilibrium?

c. Calculate the approximate value for the equilibrium constant, Keq, using the concentrations at 60 s.

d. Explain the changes 70 s after the initiation of the reaction.

e. What changes in conditions might have been imposed on the system 120 s after the initiation of the reaction?

Interpreting Graphs Worksheet Cont.

f. Are any events taking place at 320 s?


g. What differences would you have noticed if a catalyst had been present during the entire course of this reaction?

h. List the changes you might impose on this system if you wanted to produce a maximum amount of phosgene?

i. How could you account for the differences in the value calculated for the equilibrium constant, Keq, from the concentrations at different time points on the graph?

2. 2 SO2(g) + O2(g) 2 SO3(g) + 180 kJ

a. List 4 ways in which you could increase the amount of SO3 produced.

b. What affect does temperature have on this reaction.

Interpreting Graphs Worksheet Cont.

3. Consider the reaction: Heat + 2 NOCl(g) 2 NO(g) + Cl2(g)

At 462˚C, the reaction has a Keq of 8.0 x 10-2. What is the Keq at 462˚C for the following reaction: 2 NO(g) + Cl2(g) 2 NOCl(g) + Heat

4. With reference to the first equation given in question 3 answer the following (True or False). Explain your answer.

a. The reaction is exothermic.


b. After equilibrium is established, increasing the concentration of NO causes an increase in [Cl2].

c. After equilibrium is established, decreasing the volume favours the formation of NOCl.

d. Increasing the temperature favours the formation of NO and Cl2.

e. Decreasing the [Cl2] causes the equilibrium to shift to the left.

f. Adding argon gas to the equilibrium system will cause an increase in the yield of products.

g. Adding a catalyst decreases the time required for the reaction to reach equilibrium.

h. Adding more NOCl to the system, causes an increase in Keq.

i. Adding a catalyst causes a change in Keq.

j. Increasing temperature, increases the rate of the forward reaction.

k. Increasing temperature, increases the rate of the reverse reaction.

l. At a given temperature, only one set of product and reactant concentrations satisfies Keq.


Practical Applications of Le Chatelier’s Principle

Blood pHBlood contains dissolved carbonic acid in equilibrium with carbon dioxide and water.

H2CO3(aq) CO2(aq) + H2O(l)

To keep carbonic acid at safe concentrations in the blood, the CO2 product is exhaled. The removal of a product causes the forward reaction to be favoured, reducing the amount of carbonic acid to keep blood pH within a safe range.

Haemoglobin Production and AltitudeIn the body haemoglobin in readily used to transport oxygen to tissues.

Hb(aq) + O2(g) HbO2(aq)

In a place such as Mexico City, where the elevation is 2.3 km, atmospheric pressure and oxygen concentration are low. To offset the stress equilibrium favours the reverse direction. This results in hypoxia, which can cause headache, nausea, and extreme fatigue. In serious cases, if a victim is not treated quickly, they may slip into a coma and die. Individuals living at high altitudes for extended periods of time adapt to reduced oxygen concentrations by producing more haemoglobin. This shifts equilibrium to the right once more so that the symptoms of hypoxia disappear.Studies have shown that the Sherpas, long-time residents of the mountains, have adapted to high altitude conditions by maintaining high levels of haemoglobin in their blood, sometimes as much as 50% more than individuals living at sea level.

Carbonated BeveragesSoft drinks are carbonated under high pressure to create the following equilibrium system:

CO2(g) CO2(aq) + heatWhen a bottle of soda is opened, the pressure above the carbon dioxide decreases. The system shifts to the left, the solubility of the carbon dioxide drops, and carbon dioxide bubbles out of solution. If the bottle is left open for long periods of time, the pop will go “flat” due to the reduced pressure. Shaking a pop bottle will increase the pressure on the system, which will shift to relieve the stress byfavouring the forward reaction. Increasing the temperature of a pop bottle (i.e. leaving it in a warm car on a summer’s day) will cause equilibrium to shift in the reverse direction, creating more carbon dioxide gas. This generates a pressure that could potentially cause the pop bottle to burst.


Solubility Equilibrium

Soluble: all of the substance dissolves in water.

Partially Soluble: dissolves to a great extent in water (some precipitate left).

Insoluble: will dissolve in water, but the amount that dissolves is extremely small,

almost ALL the substance precipitates.

*Solubility Equilibrium looks at the partially or almost insoluble solids in equilibrium with their ions.

Solubility Equilibrium: When the rate of dissolving is equal to the rate of

precipitation.

Example: Draw the solubility equilibrium of AgI(s).

Dissociation Equation: AgI(s) Ag+(aq) + I-

(aq)

Explain what is occurring at equilibrium.Rate of dissolving = rate of precipitation

Using your equation, write the equilibrium constant expression for it.Keq=[Ag+][I-]

Whenever you write a chemical equation for the dissolving of a solid ionic substance,

you always obtain an equilibrium expression in which there is no denominator. Why?

Because reactants are always solid.


When reactions involve solubilities, the equilibrium constants for these reactions is called a solubility product constant referred to as Ksp.


Ag+ I-

AgI AgIAgI

Solubility product constant: The constant value obtained at standard temperature

when the concentration of the ions in a saturated solution are multiplied together and

raised to their coefficients.

Why Ksp is useful: Ksp tells you the relative solubilities of slightly soluble electrolytes. Smaller Ksp’s are less soluble than larger Ksp’s.

Compare NaCl (very soluble) to AgI (insoluble) using a picture.


Cl-Na+

Cl-

Na+

Cl- Na+

Ag+

Ag+

I-I-

AgI AgI AgIAgI AgI AgI AgI


Example:The solubility of calcium sulphate, CaSO4, is 4.9x10-3 mol/L. Calculate the Ksp for CaSO4.

Step 1: Write the dissociation equation for CaSO4.CaSO4(s) Ca2+

(aq) + SO42-

(aq)

Step 2: Write the Ksp expression for CaSO4.Ksp=[Ca2+][SO4

2-]

Step 3: Substitute the molar concentrations of the ions, Ca2+ and SO42- into the Ksp

expression and solve.Ksp=(4.9x10-3mol/L) (4.9x10-3mol/L)Ksp=2.4x10-5

Example:Calculate the molar solubility of PbCl2 in pure water at 25C. Ksp for PbCl2 is 2 x 10-5.

Step 1: Write the dissociation equation for PbCl2.PbCl2(s) Pb2+

(aq) + 2 Cl-(aq)

Step 2: Set up an ICE box and fill in the values for the unknown ions.Equation: PbCl 2(s) Pb 2+ (aq) + 2 Cl - (aq)

Initial (mol/L) --- 0 0

C hange (mol/L) --- + x + 2x

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

Step 3: Write the Ksp expression and substitute the known values into the expression. Then solve for x.

Ksp=[Pb2+][Cl-]2

2x10-5 = 4x3

x3 = 5x10-6

x = 1.7x10-2 mol/L

Ksp Problems

1. If the Ksp of CaCO3 is 8.70 x 10-9, what is the calcium ion concentration?


2. Calculate the Ksp of Na2SO4 if the sulphate ion concentration is 0.0032 mol/L in a saturated solution.

3. Determine the [IO3-] in a saturated Cu(IO3)2 solution if the Ksp of 1.47 x 10-7.

4. Calculate the [H+] of a saturated solution of H3PO4 with a Ksp of 4.78 x 10-4 assuming 100% ionization of the acid.

5. The Ksp of Al(OH)3 is 1.26 x 10-3. Determine the aluminium ion concentration and hydroxide ion concentration.

1. 9.3 x 10-5 mol/L 2. 1.31 x 10-7 3. 6.6 x 10-3 mol/L 4. 0 .195 mol/L 5. [Al3+] = .083 mol/L, [OH-] = 0.249


The Common Ion Effect

So far, we have discussed the solubility and Ksp of substances in pure water. Now we will examine what happens when we add a common ion.

The Ksp of AgCl is 1.8 x 10-10.

AgCl(s) Ag+(aq) + Cl-(aq)

In pure water, the solubility of the Ag+ and Cl- ions are:

AgCl is considered insoluble and most will precipitate, only a very small concentration of ions are present in solution.

What happens if we add a solution of NaCl to the beaker?

NaCl(s) Na+(aq) + Cl-(aq)

This equilibrium will react according to Le Chatelier’s Principle. If we add ions to the right hand side, the system will shift to produce more precipitate on the left hand side.

The result will be more AgCl(s) in the beaker, and less Ag+ ions in solution.


The Common Ion Effect

Example: Calculate the molar solubility of silver chloride in a 1.5 x 10 -3 mol/L silver nitrate solution. Ksp for AgCl is 1.6 x 10 -10.

Note: The common ion is Ag+, which is present in AgCl and AgNO3. The presence of the common ion affects the solubility of AgCl but not the Ksp value because it is an equilibrium constant.

AgNO3 dissociates completely as given by the equation, AgNO3(s) Ag+(aq) + NO3

-(aq).

Since the concentration of AgNO3 is given as 1.5 x 10-3 mol/L, then [Ag+] is 1.5 x 10-3 mol/L.

Step 1: Write the dissociation equation for AgCl.

AgCl(s) Ag+(aq) + Cl-(aq)

Step 2: Set up an ICE box. (Remember that there are two sources for Ag+)

Equation: AgCl (s) Ag + (aq) + Cl - (aq)

Initial (mol/L) --- 1.5x10-3 0

C hange (mol/L) --- + x + x

Equilibrium (mol/L) --- 1.5x10-3+ x x

Step 3: Write the Ksp equation, and substitute in the known values. Then solve for x.

Ksp=[Ag+][Cl-]1.6x10-10 = (1.5x10-3 + x)(x)1.6x10-10 = (1.5x10-3 )(x)x = 1.1x10-7

[AgCl] = 1.1x10-7 mol/L

The molar solubility of AgCl in a 1.5x10-3 mol/L solution AgNO3(aq) is 1.1x10-7 mol/L

Common Ion Effect Problems

1. What is the [Ba2+] in a saturated BaCO3 solution if 0.100 mol/L Na2CO3 is added? The Ksp of BaCO3 = 8.1 x 10-8.

2. Determine the [Cl-] in a saturated AgCl solution if 0.200 mol/L AgNO3 is added. The Ksp of AgCl = 1.56 x 10-10.


This ‘x’ can be ignored because the amount of Ag+ ion that can dissolve from AgCl is very small compared to the amount of Ag+ generated from AgNO3.

3. To a saturated LiBr solution, 0.03 mol/L NaBr is added. If the Ksp of LiBr = 1.3 x 10-11, what is the [Li+]?

4. Calculate the solubility (in mol/L) of Ag+ ions when Ag2CrO4 is dissolveda) in pure waterb) in a 0.005 M solution of K2CrO4

(The Ksp of Ag2CrO4 is 9.0 x 10-12)

1) 8.1 x 10-7 M 2) 7.8 x 10-10 M 3) 4.3 x 10-10 4) a) 2.6 x 10-4 M b) 4.2 x 10-5 M


Ksp Worksheet

1. Write the equilibrium dissociation equations for each of the following ionic solids. a. Lead iodide, PbI2 d. silver carbonate, Ag2CO3

b. Calcium hydroxide, Ca(OH)2 e. gold(III) iodate, Au(IO3)3

c. Barium chromate, BaCrO4 f. iron(II) phosphate, Fe3(PO4)2

2. Calculate the Ksp for:a. SrCO3 which has a molar solubility of 1.0 x 10-5 M.b. Ag2CO3 which has a molar solubility of 1.3 x 10-4 M.c. CrF3 given that a saturated solution contains 0.00751 g per 100.0 mL of

solution.

3. Calculate the molar solubility of:a. CuCrO4 given the Ksp of 3.6 x 10-6 M2.b. La(IO3)3 given the Ksp of 6.1 x 10-12M4.

4. The Ksp of Ag2CrO4 is 1.1 x 10-12 M3. If you were to heat 500.0 mL of a saturated solution to dryness, how many grams of Ag2CrO4 would remain behind?

5. The Ksp of PbI2 is 7.9 x 10-9 M3. Calculate its molar solubility in:a. Waterb. 0.10 M NaI solutionc. 0.10 M Pb(NO3)2

d. 0.10 M MgI2 solution

6. Use appropriate calculations to determine whether a precipitate will form when:a. 100.0 mL of a 0.0010 M BaCl2 solution is mixed with 25.0 mL of a

0.000010 M AgNO3 solution. Ksp(AgCl) = 1.8 x 10-10 M2

b. 10.0 mL of a 2.5 x 10-3 M Al2(SO4)3 solution is added to 85.0 mL of a 3.5 x 10-4 M AgNO3 solution. Ksp(Ag2SO4) = 1.5 x 10-5M3

c. 0.500 g of MgCl2 is added to 500.0 mL of a 0.0100 M NaF solution. Ksp(MgF2) = 6.6 x 10-9 M3


Using Ksp to Determine if a Precipitate Will Form

Example: The Ksp for Ca3(PO4)2 in water is 2.0 x 10-29. The approximate [Ca2+] in the blood is 1.2 x 10-3 M, and the approximate [PO4

3-] is 6.7 x 10-9 M. Do you expect a precipitate of Ca3(PO4)2 to form in the blood?

Determine the Ksp of Ca3(PO4)2 using data from the question.

Ca3(PO4)2(s) 3 Ca2+(aq) + 2 PO4

3-(aq)

Ksp = [Ca2+]3[PO43-]2

Ksp = (1.2 x 10-3)3(6.7 x 10-9)2

Ksp = 7.8 x 10-26

This Ksp value is greater than the Ksp of Ca3(PO4)2 given in the question, therefore a precipitate will form.

A larger Ksp means there are more ions in solution than are necessary for saturation, “supersatured” – tendency to precipitate.


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