Part 01 - Notes: Equilibrium & Equilibrium ExpressionsHonors Chemistry Unit 13 - Equilibrium

Objectives: Use a Mass Action Expression to derive an equilibrium expression. Differentiate between a mass action expression and an equilibrium expression. Calculate equilibrium constants based on lab data. Use equilibrium constants and mass action expression to determine if a system is in equilibrium.

Text Reference: Chang and Goldsby (Chemistry: The Essential Concepts, McGraw-Hill, 2014) pp. 514−524.

Text Vocabulary: • Chemical equilibrium (p514) - a state in which the forward and reverse reactions are equal and no net changes can be observed • Physical equilibrium (p514) - an equilibrium in which only physical properties change • Equilibrium constant (p516) - a number equal to the ratio of the equilibrium concentration of the products to the equilibrium con-

centrations of reactants, each raised to the power of its stoichiometric coefficient • Homogeneous equilibrium (p517) - an equilibrium condition in which all reacting species are in the same phase • Heterogeneous equilibrium (p521) - an equilibrium condition in which the reacting species are not all in the same phase • Reaction quotient (Qc) (p525) - a number equal to the ratio of product concentrations to reactant concentrations, each raised to the

power of its stoichiometric coefficient. It comes from the Law of Mass Action and does NOT require a state of equilibrium.

The Law of Mass Action • first published in 1864 • Cato Guldberg (Norwegian chemist) and Peter Waage (Norwegian mathematician) - primarily qualitative work • “re-discovered” by Jacobus Henricus van’t Hoff (1877) - quantitative work • lead to a formula being deduced for the effect of temperature on the reaction rate constant by Svante Arrhenius

Law of Mass Action - describes how the rate of a chemical reaction is related to the concentrations of the reactants. Two parts of the Law of Mass Action: (i) the velocity of a reaction depends on the molecular concentrations of the reactants, and (ii) when a chemical reaction reaches equilibrium, the concentrations of the chemicals involved have a constant relation to each other, which is described by an equilibrium constant.

Mass Action Expression: Consider the general balanced equation: n A + m B . . . ⇄ p C + q D . . .

The Mass Action Expression is where square brackets mean the MOLAR concentration of the substance. (Note: it’s the product of the product side concentrations over the product of the reactant side concentrations raised to the coefficients.) Also referred to as reaction quotient.

Example 1: Write the mass action expression for N2 + 3 H2 ⇄ 2 NH3. 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 MAE?

Example 2: Reconsider POGIL: Equilibrium. The equation is A(g) ⇄ B(g). Write the expression for the reaction quotient and calcu-late the Q value after 1 minute and also after 3 minutes. Assume a 1 liter container.

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Q =C⎡⎣ ⎤⎦

pD⎡⎣ ⎤⎦

q

A⎡⎣ ⎤⎦nB⎡⎣ ⎤⎦

m

Part 01 - Notes: Equilibrium & Equilibrium ExpressionsHonors Chemistry Unit 13 - Equilibrium

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. When a system is at equilibrium, there is no net change in the concentrations or pressures of the reactants or products.

Equilibrium may be chemical or physical:

Consider the general balanced equation: n A + m B . . . ⇄ p C + q D . . .

The equilibrium expression for the general equation above is where Keq is called the equilibrium constant.

Significance of the Size of the Equilibrium Constant Large value of Keq (i.e. 10n)

Small value for Keq (i.e. 10−n)

15.1 Review of Concept: Consider the equilibrium X ⇌ Y, where the forward reaction rate constant is greater than the reverse reaction rate constant. Which of the following is true about the equilibrium constant? Why? (A) K > 1 (B) K < 1 (C) K = 1

Homogeneous Equilibrium homogeneous equilibrium system - only one state of matter is present.

• KC = concentration equilibrium constant; the [ ] indicate molar concentrations, subscript “c” indicates concentration

• KP = pressure equilibrium constant; the pressure is usually in atmospheres (or kilopascals), subscript “p” indicates pressure The expressions for KC and KP are determined the same way, they are just written to indicate the specific type of constant.

Consider the homogeneous reversible chemical system that is in equilibrium: N2O4(g) ⇄ 2 NO2(g)

KC = KP =

Relationship between KC and KP KP = KC(RT)∆n

where ∆n = moles of gaseous products − moles of gaseous reactants. If the pressures are in atmosphere, R = 0.0821.

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Keq=

C⎡⎣ ⎤⎦pD⎡⎣ ⎤⎦

q

A⎡⎣ ⎤⎦nB⎡⎣ ⎤⎦

m

Part 01 - Notes: Equilibrium & Equilibrium ExpressionsHonors Chemistry Unit 13 - Equilibrium

Question: When does KC equal KP? How do you know?

What about Units??? “Note that it is general practice not to include units for the equilibrium constant.” Text - page 519

Example 3: Consider the reversible reaction equilibrium system: 4 NH3(g) + 5 O2(g) ⇄ 4 NO(g) + 6 H2O(g) At equilibrium, in a 2.50L container, there are 5.00mol NH3, 1.875mol O2, 27.5mol NO, and 6.25mol H2O at 27oC. (a) Write the concentration equilibrium expression for the system. (b) Write the pressure equilibrium expression for the system. (c) Calculate the KC value. (d) Calculate the KP value. (e) If, in a 1.00L container, there are 3.00mol NH3, 2.00mol O2, 10.0mol NO, and 1.00mol H2O, is the system in equilibrium? Justify your answer.

15.2 Review of Concept: In which of the following reactions is the value of KC equal to the value of KP? (a) 2 H2O2(aq) ⇄ 2 H2O(l) + O2(g) (b) PCl3(g) + 3 NH3(g) ⇄ 3 HCl(g) + P(NH2)3(g) (c) C(s) + O2(g) ⇄ 2 CO(g)

Heterogeneous Equilibrium In a heterogeneous equilibrium system, there are multiple states of matter present.

However, since the concentration of any pure solid or pure liquid is fairly constant in a heterogeneous equilibrium system, the concentra-tion of pure solids and/or liquids is left out of the equilibrium expression. It is part of the constant, Keq.

For example, for the following reaction: C(s) + O2(g) ⇄ CO2(g), the equilibrium expression is

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

Example 4: Consider the following heterogeneous equilibrium system: CaCO3(s) ⇄ CaO(s) + CO2(g). At 822oC, the pressure of

CO2(g) is 0.236atm. Calculate KP and KC for this system at 822oC.

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Keq=

CO2

⎡⎣ ⎤⎦O

2⎡⎣ ⎤⎦

Part 01 - Notes: Equilibrium & Equilibrium ExpressionsHonors Chemistry Unit 13 - Equilibrium

Changing the System - Changing the Keq Value Consider the Born-Haber process for the production of ammonia through the reaction of nitrogen with hydrogen.

N2 + 3 H2 ⇄ 2 NH3 all substances are gases

The Keq value for the above system at 573K is 9.60.

Q1:Write the equilibrium expression for the above system.

Q2:Consider a formation equation for ammonia. ½ N2 + 3/2 H2 ⇄ NH3 at 573K

Write the equilibrium expression for the above system. Calculate the value of Keq for the system.

Q3:Consider the following system. 2 NH3 ⇄ N2 + 3 H2 at 573K

Write the equilibrium expression for the above system. Calculate the value of Keq for the system.

Q4:What is the relationship between these values of Keq? Why?

Halve the equation . . .

Reverse the equation . . .

Take 1/3 of the equation . . .

Summary of the Rules for Writing Equilibrium Constant Expressions (Text - page 524) 1. The concentration of the reacting species in the condensed phase are expressed in molar concentrations; in the gaseous phase,

can be expressed in molar concentrations or atmospheres. KC is related to KP by KP = KC (RT)∆n. 2. The concentrations of pure solids and pure liquids (heterogeneous equilibriums) and pure solvents (homogeneous equilibrium)

do NOT appear in the Keq expressions. 3. The equilibrium constant (KC or KP) is dimensionless. (The unit for pressure may be designated.) 4. In quoting a value for the equilibrium constant, we must specify the balanced equation and temperature.

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Part 02 - Notes: More with Equilibrium & ExpressionsHonors Chemistry Unit 13 - Equilibrium

Objectives: Determine equilibrium concentrations of a dynamic system from initial concentration or equilibrium constants. Determine equilibrium constants from initial system data using RICE tables. Determine if a system is in equilibrium and, if not, what the system is doing to get to equilibrium.

Text Reference: Chang and Goldsby (Chemistry: The Essential Concepts, McGraw-Hill, 2014) pp. 524-529.

15.2-2 Review of Concept: From the equilibrium constant expression at the right, write a balanced chemical equation for the gas-phase reaction. Does KP = KC for this reaction system? Justify.

Determining Equilibrium Concentrations and Equilibrium Constants from Stoichiometry Q1: Consider the gaseous reaction system represented by the following equation: 4 NH3 + 5 O2 ⇄ 4 NO + 6 H2O. Initially, in

a 2.500 L container, there are 7.50 mol NH3, 5.00 mol O2, 25.00 mol NO, and 2.50 mol H2O. Once the system has reached equilib-rium, the concentration of NI has increased to 11.00M. Calculate the the value of Keq for this system. (Use a RICE table.)

Q2: For the system 4 H2 (g) + Fe3O4 (s) ⇄ 3 Fe(s) + 4 H2O(g), find Keq if, initially, there is only 1.60 M H2 and 100. g Fe3O4 and if,

at equilibrium, 37.5% of the hydrogen has reacted.

Determining Equilibrium Concentrations from Equilibrium Constants Q3: How many grams of CO2 would exist in equilibrium with 1.0 mole H2, 2.0 moles CO, and 3.0 moles H2O at 986oC if Keq = 1.60 at 986oC for the system H2(g) + CO2(g) ⇄ H2O(g) + CO(g) in a 500. mL container?

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KC=

NH3

⎡⎣ ⎤⎦2H

2O⎡⎣ ⎤⎦

4

NO2

⎡⎣ ⎤⎦2H

2⎡⎣ ⎤⎦

7

Part 02 - Notes: More with Equilibrium & ExpressionsHonors Chemistry Unit 13 - Equilibrium

Determining Equilibrium Concentrations from Initial Concentrations Steps: • (Most likely), select the substance in the equation with the lowest coefficient in front of it and . . .

• if it is on the left side of the equation, make it into a statement, “Let ‘x’ M of it be used up”. or • 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. Remember - stoichiometry is key.

• 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 Keq 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’.

Q 4: A mixture of 1.500mol H2 and 1.500mol I2 was placed in a 2.00-L stainless steel flask at 430.oC. The KC for the reaction system H2(g) + I2(g) → 2 HI(g) is 54. at this temperature. (a) Determine the equilibrium concentrations of H2 and HI. (b) Calculate the number of moles of I2 present at equilibrium.

Q 5: Keq = 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?

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Part 02 - Notes: More with Equilibrium & ExpressionsHonors Chemistry Unit 13 - Equilibrium

Predicting the Direction of a Reaction System Q6: At the start of a reaction, there are 0.351mol N2, 3.67x10−2 mol H2, and 7.51x10−4 NH3 in a 3.75-L reaction vessel at 375oC. If the KC for N2(g) + 3 H2(g) ⇄ 2 NH3(g) is 1.2 at this temperature, decide whether the system is at equilibrium. If it is not

at equilibrium, predict in which direction the net reaction will proceed to reach equilibrium.

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Part 03 - Notes: Solubility Product ConstantHonors Chemistry Unit 13 - Equilibrium

Objectives: Write Ksp expressions and determine Ksp values from experimental data. Determine the solubility of a given salt based on the Ksp value. Determine is a precipitate forms when two solutions are mixed.

Text Reference: Chang and Goldsby (Chemistry: The Essential Concepts, McGraw-Hill, 2014) pp. 609−615.

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 an equilibrium constant, Keq. 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:

Since this is a heterogeneous equilibrium system, the concentration of a pure solid, such as AgCl, is relatively constant. Since both terms, AgCl and Keq, are constants, they may be multiplied together to form a new constant, which is known as the solubility product constant, or the Ksp.

The solubility product constant, Ksp, 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.

Q1: Write the equation representing the equilibrium system of relatively insoluble PbCl2. Also write the solubility product con-stant expression.

Using the equation for Ksp it is possible to calculate the solubility of a salt if its Ksp is known, or to calculate the Ksp from the solubili-ty. The Ksp is an experimental value and depends upon the temperature, as does Keq.

Q 2: A 1.00 L saturated solution of AgCl at 25oC is evaporated to dryness and the residue is equivalent to 1.34 x 10−5 mole of AgCl. What is the experimental Ksp of the silver chloride at 25oC?

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Keq=

products⎡⎣ ⎤⎦reactants⎡⎣ ⎤⎦

=Ag+⎡⎣ ⎤⎦ Cl−⎡⎣ ⎤⎦AgCl⎡⎣ ⎤⎦

Ag+⎡⎣ ⎤⎦ Cl−⎡⎣ ⎤⎦ = Keq

AgCl⎡⎣ ⎤⎦ = Ksp

Part 03 - Notes: Solubility Product ConstantHonors Chemistry Unit 13 - Equilibrium

Q3: 500.mL of a saturated solution of aluminum hydroxide at 28oC is evaporated to dryness and the residue is equal to 9.22x10−10 mol of aluminum hydroxide. What is the experimental Ksp of the salt at 28oC?

Q4: At 25oC, the Ksp of Cu(OH)2 is 2.24 x 10−20. Determine the solubility of the Cu(OH)2 in g/L.

Try it out . . . Q5: 425-mL of a saturated cobalt(II) hydroxide solution is evaporated to dryness and 4.85x10−6 mole cobalt(II) hydroxide salt is recovered. Determine the Ksp of cobalt(II) hydroxide at the temperature of the experiment.

Q6: At 25oC the Ksp of thallium(III) hydroxide is 1.68x10−44. Determine the solubility of thallium(III) hydroxide in g/L.

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Part 04 - Notes: More Solubility Product ConstantHonors Chemistry Unit 13 - Equilibrium

Objectives: Write Ksp expressions and determine Ksp values from experimental data. Determine the solubility of a given salt based on the Ksp value. Determine is a precipitate forms when two solutions are mixed.

Text Reference: Chang and Goldsby (Chemistry: The Essential Concepts, McGraw-Hill, 2014) pp. 609−615.

Using Ksp Values A useful application of Ksp 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 Ksp value.

• Ion product < Ksp no precipitation will occur, unsaturated solution • Ion product = Ksp no precipitation will occur, saturated solution • Ion product > Ksp 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 mo-larity of the salt then needs to be calculated using the 2.00 L volume.

Q1: Will precipitation occur when 50.0 mL of a 3.00 x 10-2 M Pb(NO3)2 solution is mixed with 85.0 mL of a 2.00 x 10-3 M CaCl2? The Ksp of PbCl2 is 1.62 x 10-5.

Write some steps to solving this type of problem.

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Part 04 - Notes: More Solubility Product ConstantHonors Chemistry Unit 13 - Equilibrium

Q2: 200.0mL 0.00400M BaCl2 are added to 600.0mL 0.00800M K2SO4. Will a precipitate form? The Ksp of BaSO4 is 1.1x10−10.

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Part 05 - Notes: Justifying Shifts in a SystemHonors Chemistry Unit 13 - Equilibrium

Objectives: Predict and justify the shift of a system based on its current conditions and its Keq. Justify and explain how a system will achieve equilibrium.

Text Reference: Chang and Goldsby (Chemistry: The Essential Concepts, McGraw-Hill, 2014) pp. 529−536.

Predicting If a System Is at Equilibrium with Certain Conditions Basic Idea: If you substitute the value for Keq and the equilibrium concentrations into the equilibrium expression, the left and right sides MUST be EQUAL if there is equilibrium. If they are not equal, i.e. if Keq does not equal the product of the concentrations of the products divided by the product of the reactant concentrations, the system is NOT in equilibrium.

There are two possibilities: (1) The mass action expression may be less than Keq. 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 Keq and there is equilibrium.

(2) The mass action expression may be greater than Keq. 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 Keq and there is equilibrium.

Q1: 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. The equation is 2 CO(g) + O2(g) ⇄ 2 CO2(g). (a) Find Keq. (b) Would there be equilibrium at the same

temperature if the concentrations were [CO2] = 15.5 M, [CO] = 10.0 M, and [O2] = 0.250 M? (c) If the system is not in equilibrium, which reaction is faster, forward or reverse? Which concentrations are increasing and which concentra-tions are decreasing?

Reaction Kinetics You should also consider the following information related to kinetics:

• Temperature - All reactions, exothermic and endothermic, are faster at higher temperatures. Although all reactions are sped up by an increase in temperature, endothermic reactions are sped up MORE than exothermic reactions, and they are also slowed down more by a decrease in temperature. Temperature is the only thing that changes the equilibrium constant.

• Catalysis - Catalysts increase the rate of a reaction usually by decreasing the activation energy barrier and thereby allowing a greater fraction of molecules to have the necessary minimal energy to have a chemical change when they collide. However, catalysts lower the activation energy of both the forward and reverse reactions equally so they cause no change to the equilibrium system or the equilibrium constant.

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Part 05 - Notes: Justifying Shifts in a SystemHonors Chemistry Unit 13 - Equilibrium

Q2: What is the effect of decreasing the temperature of the following: N2 + 3 H2 ⇄ 2 NH3 + 22 kcal (all gases)

Recall: With regard to decreasing the volume of a reaction of a mixture of gases in equilibrium: Although the con-centration of all substances in the system increase with a decrease in volume, and although this tends to increase both the forward and the 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 sped up more than the reverse as fewer moles of a gas are produced which would attempt to lower pressure and provide as much “space” as possible to each molecule.

Q3: What is the effect of compressing the following system to a smaller volume: N2O4(g) ⇄ 2 NO2(g)

Q4: What is the effect of compressing the following system to a smaller volume: C(s) + O2(g) ⇄ CO2(g)

The 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 Keq and the Mass Action Expression (Q expression) and (b) if the equality is upset by (1) changing the value of Keq by raising of lowering the temperature or (2) changing the value of Q by changing the concentrations of reactants or products, the system will proceed to restore equality between Keq and the Q expression.

Q5: What is the effect of removing some I2 from the system: H2(g) + I2(g) ⇄ 2 HI(g)

Q6: Explain the effect of compressing the following system: N2 + 3 H2 ⇄ 2 NH3 + 22 kcal (all gases)

Q7: What is the effect of increasing the temperature of the following system: 43kcal + N2(g) + O2(g) ⇄ 2 NO(g)

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Part 06 - Notes: A Little Class PracticeHonors Chemistry Unit 13 - Equilibrium

Task: Show all work, units, set-ups, etc. Did you answer the question that was asked or leave it incomplete? 1. For the system 4 H2(g) + Fe3O4(s) ⇄ 3 Fe(s) + 4 H2O(g), find Keq if, initially, there is only 2.25M H2 and 85.0 g

Fe3O4 and if, at equilibrium, the concentration of H2 is 0.750M.

2. Keq = 16.0 at a certain temperature for the reaction 4 A(g) + B(s) ⇄ 2 C(g) + 2 D(g). If, initially, there were 2.25

moles of A and 2.25 moles of B in a 750.0 mL container, what would be the equilibrium concentration of C?

3. How many grams of XCl2 will dissolve to form 765 mL of a saturated solution if the Ksp is 5.87x10-12? The molar mass of XCl2 is 156.43 g/mol.

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Part 06 - Notes: A Little Class PracticeHonors Chemistry Unit 13 - Equilibrium

4. 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. (A) Is there equilibrium? (B) Which reaction is faster, the forward or the reverse? (C) Which concentrations are increasing? Which concentrations are decreasing?

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