13 Dec. 2010 Take Out Homework: Lab Notebook and Kinetics AP Questions. Objective: SWBAT Do now: 1.

13 Dec. 2010

Take Out Homework: Lab Notebook and Kinetics AP Questions.

Objective: SWBATDo now:

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Agenda

I. Do nowII. Kinetics Problem Set SolutionsIII. Introduction to EquilibriumIV. DemoHomework: p. #

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Chemical EquilibriumChapter 14

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Equilibrium is a state in which there are no observable changes as time goes by.

Chemical equilibrium is achieved when:

• the rates of the forward and reverse reactions are equal and

• the concentrations of the reactants and products remain constant

• However, there is a lot of activity at the molecular level!

• Reactants continue to form products, while products continue to yield reactants!

Physical vs. Chemical Equilibrium

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Physical equilibrium

H2O (l)

Chemical equilibrium

N2O4 (g)

H2O (g)

2NO2 (g)

NO2

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N2O4 (g) 2NO2 (g)

Start with NO2 Start with N2O4 Start with NO2 & N2O4

equilibrium

equilibrium equilibrium

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constant

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N2O4 (g) 2NO2 (g)

= 4.63 x 10-3K = [NO2]2

[N2O4]

aA + bB cC + dD

K = [C]c[D]d

[A]a[B]bLaw of Mass Action

Equilibrium Constant

Law of Mass Action

• For a reversible reaction at equilibrium and a constant temperature, a certain ratio of reactant and product concentration has a constant value, K (the equilibrium constant)– concentrations may vary– but as long as temperature stays the same and the

reaction is at equilibrium, K will not change.

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K >> 1

K << 1

Lie to the right Favor products

Lie to the left Favor reactants

Equilibrium Will

K = [C]c[D]d

[A]a[B]baA + bB cC + dD

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Homogenous equilibrium applies to reactions in which all reacting species are in the same phase.

N2O4 (g) 2NO2 (g)

Kc = [NO2]2

[N2O4]Kp =

NO2P 2

N2O4P

aA (g) + bB (g) cC (g) + dD (g)

Kp = Kc(RT)n

n = moles of gaseous products – moles of gaseous reactants

= (c + d) – (a + b)

In most cases

Kc Kp

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Homogeneous Equilibrium

CH3COOH (aq) + H2O (l) CH3COO- (aq) + H3O+ (aq)

Kc =′[CH3COO-][H3O+][CH3COOH][H2O]

[H2O] = constant

Kc = [CH3COO-][H3O+]

[CH3COOH]= Kc [H2O]′

General practice not to include units for the equilibrium constant.

Example 1

Write expressions for Kc, and KP if applicable, for the following reversible reactions at equilibrium:

a)HF(aq) + H2O(l) H3O+(aq) + F-(aq)b)2NO(g) + O2(g) 2NO2(g)c) CH3COOH(aq) + C2H5OH(aq) CH3COOC2H5(aq) + H2O(l)

Hints: KP applies only to gaseous reactions; the concentration of the solvent (usually water) does not appear in the equilibrium constant expression.

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Problem 1

Write Kc and KP for the decomposition of dinitrogen pentoxide:2N2O5(g) 4NO2(g) + O2(g)

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Example 2

The following equilibrium process has been studied at 230oC:2NO(g) + O2(g) 2NO2(g)

In one experiment, the concentrations of the reacting species at equilibrium are found to be [NO]=0.0542 M, [O2]=0.127 M and [NO2]=15.5 M. Calculate the equilibrium constant (Kc) of the reaction at this temperature.

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Problem 2

Carbonyl chloride (COCl2), also called phosgene, was used in WWI as a poisonous gas. The equilibrium concentrations for the reaction between carbon monoxide and molecular chlorine to form carbonyl chloride

CO(g) + Cl2(g) COCl2(g)

at 74oC are [CO]=1.2x10-2 M, [Cl2]=0.054 M and [COCl2]=0.14 M. Calculate the equilibrium constant (Kc). 16

Example 3

• At the equilibrium constant KP for the decomposition of phosphorus pentachloride to phosphorus trichloride and molecular chlorine

PCl5(g) PCl3(g) + Cl2(g)is found to be 1.05 at 250oC. If the equilibrium

partial pressures of PCl5 and PCl3 are 0.875 atm and 0.463 atm, respectively, what is the equilibrium partial pressure of Cl2 at 250oC?

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Problem 3

The equilibrium constant KP for the reaction2NO2(g) 2NO(g) + O2(g)

is 158 at 1000 K. Calculate PO2 if PNO2 = 0.400 atm and PNO = 0.270 atm.

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Example 4

Methanol (CH3OH) is manufactured industrially by the reaction

CO(g) + 2H2(g) CH3OH(g)

The equilibrium constant Kc for the reaction is 10.5 at 220oC. What is the value of KP at this temperature?

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Problem 4

For the reaction N2(g) + 3H2(g) 2NH3(g)

KP is 4.3x10-4 at 375oC. Calculate Kc for the reaction.

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14 Dec. 2010

• Objective: SWBAT write equilibrium constant expressions for reactions with heterogeneous equilibria, and calculate reaction quotients.

• Do now: The equilibrium constant KP for the reaction

2SO3(g) 2SO2(g) + O2(g)

is 1.8x10-5 at 350oC. What is KC for this reaction?

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Agenda

I. Do nowII. Homework solutionsIII. Heterogeneous equilibrium constantsIV. Reaction quotientsp. 649 #23, 25, 27, 29, 31, 35, 37

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Heterogenous equilibrium applies to reactions in which reactants and products are in different phases.

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

[CaCO3] = constant[CaO] = constant

Kc = [CO2] = Kp = PCO2

The concentration of solids and pure liquids are not included in the expression for the equilibrium constant.

[CaO][CO2][CaCO3]

Kc =′

[CaCO3][CaO]

Kc x′

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PCO 2= Kp

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

PCO 2 does not depend on the amount of CaCO3 or CaO

Example 5

Write the equilibrium constant expression Kc, and KP if applicable, for each of the following heterogeneous sytems:

a)(NH4)2Se(s) 2NH3(g) + H2Se(g)

b)AgCl(s) Ag+(aq) + Cl-(aq)c) P4(s) + 6Cl2(g) 4PCl3(l)

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Problem 5

Write equilibrium constant expressions for Kc and KP for the formation of nickel tetracarbonyl, which is used to separate nickel from other impurities:

Ni(s) + 4CO(g) Ni(CO)4(g)

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Example 6

Consider the following heterogeneous equilibrium:

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

At 800oC, the pressure of CO2 is 0.236 atm. Calculatea) Kp

b) Kc

for the reaction at this temperature.27

Problem 6

Consider the following equilibrium at 395 K:NH4HS(s) NH3(g) + H2S(g)

The partial pressure of each gas is 0.265 atm. Calculate KP and Kc for the reaction.

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For which of the following reactions is Kc equal to KP?

a)4NH3(g) + O2(g) 4NO(g) + 6H2O(g)

b)2H2O2(aq) 2H2O(l) + O2(g)

c) PCl3(g) + 3NH3(g) 3HCl(g) + P(NH2)3(g)

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• What if the product molecules of one reversible reaction are involved in a second reaction as the products?

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A + B C + D

C + D E + F

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A + B C + D

C + D E + F

A + B E + F

Kc =′[C][D][A][B]

Kc =′′[E][F][C][D]

[E][F][A][B]

Kc =

Kc ′

Kc ′′

Kc

Kc = Kc ′′Kc ′ x

If a reaction can be expressed as the sum of two or more reactions, the equilibrium constant for the overall reaction is given by the product of the equilibrium constants of the individual reactions.

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N2O4 (g) 2NO2 (g)

= 4.63 x 10-3K = [NO2]2

[N2O4]

2NO2 (g) N2O4 (g)

K = [N2O4]

[NO2]2′ =

1K

= 216

When the equation for a reversible reaction is written in the opposite direction, the equilibrium constant becomes the reciprocal of the original equilibrium constant.

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Writing Equilibrium Constant Expressions

1. The concentrations of the reacting species in the condensed phase are expressed in M. In the gaseous phase, the concentrations can be expressed in M or in atm.

2. The concentrations of pure solids, pure liquids and solvents do not appear in the equilibrium constant expressions.

3. The equilibrium constant is a dimensionless quantity.

4. In quoting a value for the equilibrium constant, you must specify the balanced equation and the temperature.

5. If a reaction can be expressed as a sum of two or more reactions, the equilibrium constant for the overall reaction is given by the product of the equilibrium constants of the individual reactions.

15 Dec. 2010

• Objective: SWBAT calculate reaction quotient and calculate an equilibrium concentration.

• Do now: If the equilibrium constant for the reaction

N2O4 (g) 2NO2 (g) is 4.63 x 10-3

calculate the equilibrium constant for the reaction

2NO2(g) N2O4(g)

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Agenda

I. Do nowII. Homework solutionsIII. Reaction quotientIV. Equilibrium concentrationsHomework: p. 650 #23, 30, 36, 39

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The Relationship betweenChemical Kinetics and Chemical Equilibrium

A + 2B AB2

kf

kr

ratef = kf [A][B]2

rater = kr [AB2]

Equilibriumratef = rater

kf [A][B]2 = kr [AB2]

kf

kr

[AB2]

[A][B]2= Kc =

What does the Equilibrium Constant tell us?

• We can use the equilibrium constant to calculate unknown equilibrium concentrations of products or reactants (at a constant temperature)

• We can predict the direction in which a reaction will proceed to achieve equilibrium

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What if the reaction is not yet at equilibrium?

• Calculate reaction quotient (Qc) by substituting the initial concentrations into the equilibrium constant expression.

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The reaction quotient (Qc) is calculated by substituting the initial concentrations of the reactants and products into the equilibrium constant (Kc) expression.

IF

• Qc > Kc system proceeds from right to left to reach equilibrium

• Qc = Kc the system is at equilibrium

• Qc < Kc system proceeds from left to right to reach equilibrium

Example 7

At the start of the reaction, there are 0.249 mol N2, 3.21x10-2 mol H2 and 6.42x10-4 mol NH3 in a 3.50 L reaction vessel at 375oC. If the equilibrium constant Kc for the reaction

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

is 1.2 at this temperature, decide whether the system is at equilibrium. If it is not, predict which way the net reaction will proceed.

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Problem 7

The equilibrium constant (Kc) for the formation of nitrosyl chloride, an orange-yellow compound, from nitric oxide and molecular chlorine

2NO(g) + Cl2(g) 2NOCl(g)

is 6.5x104 at 35oC. In a certain experiment, 2.0x10-

2 mole of NO, 8.3x10-3 mole of Cl2, and 6.8 moles of NOCl are mixed in a 2.0 L flask. In which direction will the system proceed to reach equilibrium?

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What do you do with the Kc and KP?

• If we know the equilibrium constant for a particular reaction, we can calculate the concentrations in the equilibrium mixture from the initial concentrations!

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Calculating Equilibrium Concentrations

1. Express the equilibrium concentrations of all species in terms of the initial concentrations and a single unknown x, which represents the change in concentration.

2. Write the equilibrium constant expression in terms of the equilibrium concentrations. Knowing the value of the equilibrium constant, solve for x.

3. Having solved for x, calculate the equilibrium concentrations of all species.

At 1280oC the equilibrium constant (Kc) for the reaction

Is 1.1 x 10-3. If the initial concentrations are [Br2] = 0.063 M and [Br] = 0.012 M, calculate the concentrations of these species at equilibrium.

Br2 (g) 2Br (g)

Example 8

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At 1280oC the equilibrium constant (Kc) for the reaction

Is 1.1 x 10-3. If the initial concentrations are [Br2] = 0.063 M and [Br] = 0.012 M, calculate the concentrations of these species at equilibrium.

Br2 (g) 2Br (g)

Br2 (g) 2Br (g)

Let x be the change in concentration of Br2

Initial (M)

Change (M)

Equilibrium (M)

0.063 0.012

-x +2x

0.063 - x 0.012 + 2x

[Br]2

[Br2]Kc = Kc =

(0.012 + 2x)2

0.063 - x= 1.1 x 10-3 Solve for x

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Kc = (0.012 + 2x)2

0.063 - x= 1.1 x 10-3

4x2 + 0.048x + 0.000144 = 0.0000693 – 0.0011x

4x2 + 0.0491x + 0.0000747 = 0

ax2 + bx + c =0-b ± b2 – 4ac

2ax =

Br2 (g) 2Br (g)

Initial (M)

Change (M)

Equilibrium (M)

0.063 0.012

-x +2x

0.063 - x 0.012 + 2x

x = -0.00178x = -0.0105

At equilibrium, [Br] = 0.012 + 2x = -0.009 M or 0.00844 M

At equilibrium, [Br2] = 0.062 – x = 0.0648 M

Example 9

A mixture of 0.500 mol H2 and 0.500 mol I2 was placed in a 1.00 L stainless steel flask at 430oC.

The equilibrium constant KC for the reaction

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

is 54.3 at this temperature. Calculate the concentrations of H2, I2 and HI at equilibrium.

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Example 10

For the reaction below, the initial concentrations of each species are H2 = 0.00623 M, I2 = 0.00414 M and HI = 0.224 M at 430oC.

The equilibrium constant KC for the reaction is still 54.3. Calculate the concentrations of these species at equilibrium.

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

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Homework

• p. 650 #23, 30, 36, 39

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Problem 8

Consider the reaction below: H2(g) + I2(g) 2HI(g)

with an equilibrium constant of 54.3. Starting with a concentration of 0.040 M for HI,

calculate the concentrations of HI, H2 and I2 at equilibrium.

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Problem 9

At 1280oC the equilibrium constant (Kc) for the reaction

Br2(g) 2Br(g)

is 1.1x10-3. If the initial concentrations are [Br2] = 6.3x10-2 M and [Br] = 1.2x10-2 M, calculate the concentrations of these species in equilibrium.

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3 January 2011

• Take Out AP Problem Set: Equilibrium• Objective: SWBAT review equilibrium

concentration calculations and describe factors that affect chemical equilibrium.

• Do now: What does a very large Kc indicate?

What does it mean if Qc is much smaller than Kc?

Agenda

I. Do nowII. AP Problem Set Questions?III. Review ICE boxes and equilibrium

concentrationsIV. Factors that affect chemical equilibrium:

DemoV. NotesHomework: p. 651 #45, 49, 51, 52, 53-61 odds

Demo

• Solid copper metal reacts with nitric acid to produce dinitrogen tetraoxide gas, water vapor and a solution of copper (II) nitrate.

• Dinitrogen tetraoxide gas forms nitrogen dioxide gas.

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If an external stress is applied to a system at equilibrium, the system adjusts in such a way that the stress is partially offset as the system reaches a new equilibrium position.

Le Châtelier’s Principle

• Changes in Concentration

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

AddNH3

Equilibrium shifts left to offset stress

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Le Châtelier’s Principle

• Changes in Concentration continued

Change Shifts the Equilibrium

Increase concentration of product(s) left

Decrease concentration of product(s) right

Decrease concentration of reactant(s)

Increase concentration of reactant(s) right

left

aA + bB cC + dD

AddAddRemove Remove

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Le Châtelier’s Principle

• Changes in Volume and Pressure

A (g) + B (g) C (g)

Change Shifts the Equilibrium

Increase pressure Side with fewest moles of gas

Decrease pressure Side with most moles of gas

Decrease volume

Increase volume Side with most moles of gas

Side with fewest moles of gas

Example 1At 720oC, the equilibrium constant Kc for the reaction

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

is 2.37x10-3. In a certain experiment, the equilibrium concentrations are [N2]=0.683 M, [H2]=8.80M and [NH3]=1.05 M. Suppose some NH3 is added to the mixture so that its concentration is increased to 3.65 M.

a) Use Le Châtelier’s principle to predict the shift in direction of the net reaction to reach a new equilibrium.

b) Confirm your prediction by calculation the reaction quotient Qc and comparing its value to Kc.

Problem 1

At 430oC, the equilibrium constant (KP) for the reaction

2NO(g) + O2(g) 2NO2(g)

is 1.5x105. In one experiment, the initial pressures of NO, O2 and NO2 are 2.1x10-3 atm, 1.1x10-2 atm and 0.14 atm respectively. Calculate QP and predict the direction that the net reaction will shift to reach equilibrium.

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Le Châtelier’s Principle

• Changes in Temperature (gases only!)

Change Exothermic Rx

Increase temperature K decreases

Decrease temperature K increases

Endothermic Rx

K increases

K decreases

colder hotter

N2O4 (g) 2NO2 (g)

Example 2

Consider the following equilibrium systems:a)2PbS(s) + 3O2(g) 2PbO(s) + 2SO2(g)

b)PCl5(g) PCl3(g) + Cl2(g)

c) H2(g) + CO2(g) H2O(g) + CO(g)

Predict the direction of the net reaction in each case as a result of increasing the pressure (decreasing the volume) on the system at constant temperature.

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Problem 2

Consider the equilibrium reaction involving nitrosyl chloride, nitric oxide and molecular chlorine:

2NOCl(g) 2NO(g) + Cl2(g)

Predict the direction of the net reaction as a result of decreasing the pressure (increasing the volume) on the system at a constant temperature.

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5 January 2011

• Objective: SWBAT describe how stress on a system shifts equilibrium.

• Do now: Complete this table:

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Stress to system (exothermic) EffectIncrease in temperatureIncrease in pressureincrease in concentration of reactantincrease in volume

Agenda

I. Do nowII. Homework answersIII. Finish Le Chatlier’s examplesIV. Equilibrium problem set (work time

tomorrow, too)Test Monday, 2nd half of class

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Catalyst lowers Ea for both forward and reverse reactions.

Catalyst does not change equilibrium constant or shift equilibrium.

• Adding a Catalyst• does not change K• does not shift the position of an equilibrium system• system will reach equilibrium sooner

Le Châtelier’s Principle

Example 3Consider the following equilibrium process between

dinitrogen tetrafluoride and nitrogen difluoride:N2F4(g) 2NF2(g) ∆Ho=38.5 kJ/mol

Predict the changes in the equilibrium if a)the reacting mixture is heated at a constant

volume.b)some N2F4 gas is removed from the reacting mixture

at constant temperature and volumec) the pressure on the reacting mixture is decreased at

constant temperatured)a catalyst is added to the reacting mixture.

Problem 3

Consider the equilibrium between molecular oxygen and ozone:

3O2(g) 2O3(g) ΔHo=284 kJ/mol

What would be the effect ofa)increasing the pressure on the system by decreasing

the volume?b)adding O2 to the system at constant volume?

c) decreasing the temperature?d)adding a catalyst?

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Chemistry In Action

Life at High Altitudes and Hemoglobin Production

Kc = [HbO2]

[Hb][O2]

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

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Chemistry In Action: The Haber Process

N2 (g) + 3H2 (g) 2NH3 (g) H0 = -92.6 kJ/mol

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Le Châtelier’s Principle - Summary

Change Shift EquilibriumChange Equilibrium

Constant

Concentration yes no

Pressure yes* no

Volume yes* no

Temperature yes yes

Catalyst no no

*Dependent on relative moles of gaseous reactants and products

Homework

p. 651 #45, 49, 51, 52, 53-61 odds

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20 Dec. 2010

Objective: SWBAT determine the stoichiometric relationship between reactants and products in a chemical reaction.

Do now: Grab your lab notebook and complete the pre-lab.

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Changes to the lab

• Use HALF the volume of each solution called for in the lab.

• Complete reactions in a 50 mL graduated cylinder.

• Measure reagents in a 25 mL graduated cylinder.

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Notes

• Pour or transfer pipette OUT of reagent bottles.

• Never pour or transfer pipette INTO reagent bottles.

• Pour excess down the sink while running the water.

• Pour products into the waste beaker.

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Homework

• Post lab calculations and conclusions due tomorrow.

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Equilibrium Review Problem Set

• p. 648 #8d, 9a, 16, 18, 21, 28, 31, 40, 42, 48, 54, 58, 65, 71, 84, 88, 98

• In-class work time: today and Thursday.• Due Monday• Test on Equilibrium: Monday

• Next unit: Acid-Base Reactions