Chapter Fifteen

Chemical Equilibrium

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The Concept of Equilibrium

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Dynamic EquilibriumOpposing processes occur at equal rates

Forward and reverses reaction proceed at equal ratesNo outward change is observedRatio of reactants to products is constant.A double sided arrow ( ) is used Often temperature dependent

Physical equilibriumEquilibrium between phases

H2O(l) H2O(g)

Chemical equilibriumEquilibrium between reactants and products

N2O4(g) 2NO2(g) colorless brown

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Equilibrium Constant Expression, Kc

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At equilibrium, [N2O4] and [NO2] are constant N2O4 2NO2

Rate (forward) = Rate (reverse) k1 [N2O4] = k-1 [NO2]2

cKON

NOkk

][

][

42

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1

1

Initial vs. Equilibrium ConcentrationsProof of Kc Calculation

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The Equilibrium Constant Expression, Kc

For a reaction of the type:

The equilibrium expression is:Equilibrium Constant Expression:

Concentrations of the products appear in the numerator Concentrations of the reactants appear in the denominator.Exponents match coefficients in chemical equation.Units are not included!Also known as a Mass Action Expression

Chemical Equation must be balanced to calculate Kc

K is based on HOW the equation is balanced6

....... dDcCbBaA

ba

dc

c BADCK][][][][

Chemical Reactions and Kc

a. SO2(g) + ½O2(g) SO3(g) b. 2O2(g) + O2(g) 2SO3(g)

Equilibrium constants change if reaction balanced differently.Numerical values for Kc are differentKc (reaction b) = [Kc (reaction a)] 2

You must know how the reaction was balanced

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2/122

3

]][[][

OSOSOK c ][][

][

22

2

23

OSOSOKc

Rules for Modifying A Chemical EquationaA + bB cC + dD

Reversing a chemical equation: Invert Kc

Multiplying coefficients by n: Raise Kc to n power

Adding equations: Multiply Kcs

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bBaAdDcC c

dc

ba

c KDCBAK 1][][][][

)( bBaAdDcCn ncndnc

nbna

c KDCBAK ][][][][

dDcCfFbBaAeEdDcCbBaA

fFeE

21][][][

][][][ KKEBAFDCK eba

fdc

c

Meaning of Equilibrium Constants

At Equilibrium

More products than reactants More reactants than products

As K goes to infinity reaction goes to completionAs K goes to zero, no reaction occurs 9

dDcCbBaA

ba

dc

c BADCK][][][][

At a given temperature, 0.100 moles of NO were added to a 2.00 L vessel. At equilibrium, 0.044 moles of NO were

remaining. What is the value of Kc? Balanced Equation and Equilibrium Constant Expression:2NO(g) N2(g) + O2(g) Kc = [N2] [O2]

[NO]2Find equilibrium molarities of reactants and productsmolesN2 = molesO2 = ½ moles NO used up in reactionmolesN2 = molesO2 = ½ (0.100-0.044)= 0.028moles[N2] = [O2] = 0.028moles/2.00L = 0.014M

[NO]= 0.044moles/2.00L = 0.022M

Calculate KcKc = [N2] [O2] = [0.014][0.014] = 0.40

[NO]2 [0.022]2 10

Ways of Expressing Equilibrium Constants

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Homogeneous Equilibria, Kc and KpAll products and reactants are in an aqueous or gas

phase for ANY equilibrium constantKc All chemicals in units of molarity, moles/L

Can be either gas or aqueous solutionsKp 2CO(g) + O2(g)2CO2(g)

All chemicals in gas phase Partial pressure in atmospheres

Relationship between Kc and Kp Kp = Kc(RT) nDerived from ideal gas lawR is the gas constant, T is temperature in kelvinn = moles gaseous products - moles gaseous reactantsKp=Kc if moles gas of product = moles gas of reactantOtherwise, a decrease in moles of gas decreases pressure

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Calculate Kp for the following reactions2CO(g) + O2(g)2CO2(g) 2NO(g) N2(g) + O2(g)

Kc= 12800 Kc= 12800ngas= ngas = 0

Kp = Kc(RT) ngas

Kc = 12800R = 0.0821L atm/mol KT = 1500 oC= 1773Kngas = Moles productsgas phase - Moles reactantsgas phase

If ngas= 0 : Kc = Kp13

9.87)00687(.12800)17730821.0(12800 1 xKxmolK

LatmxK p

Heterogeneous Equilibrium and KeqAre the equilibrium constants for the 2 reactions the same?SO3(g) + H2O(g) H2SO4(g) SO3(g) + H2O(l) H2SO4(aq)

K1 = [H2SO4(g)] K2 = [H2SO4(aq)][SO3(g)][H2O(g)] [SO3(g)][H2O(l)]

K1 equals K2 if [H2SO4(g)]=[H2SO4(aq)] & [H2O(g)]=[H2O(l)]

[H2SO4(g)] variable dependent on partial pressure [H2SO4(aq)] variable # moles dissolved in water[H2O(g)] variable dependent on partial pressure[H2O(l)] 56M Density =1 g/mL = 1000 g/L

[H2O(l)] =1000g/L x mol/18g= 56mol/LThe two equilibrium constants are not the same

The phase of the material matters! 14

Equilibria Involving Mixed PhasesUse the subscript “eq” to designate general K

Keq = Kc= K=, etc. Value depends on equation

Do not include solids and liquids in Keq expressionSO3(g) + H2O(l) H2SO4(aq) Keq = [H2SO4(aq)]

[SO3(g)]

Kp for liquid-vapor equilibrium is vapor pressure of liquidH2O(l) H2O(g) Kp = P(H2O)Liquid will not be included

Reactions must show phase for each reactant and product. 15

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Kp= PCO2PCO2 does not depend on the amount of CaCO3 or CaO

CaCO3(s)CaO(s) + CO2(g)Equilibria Involving Pure Solids And Liquids

Exam 1 Feb. 23rd 7:30AM-9:15AMCovers Chapters 14 and 15

8 questions, 3 parts per question (see practice exam)Will be assigning seats tomorrow morning by 10am

Seat assignment in Sakai is randomYou cannot ask for a certain seat, just general preferenceRequest preferences immediatelyCheck seat before exam so you can get started

Bring your ID to the examTAs will check IDs once exam is started

Exam will be on your seat with your name on itSit in assigned seat and start exam

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Alternate Exam: Feb. 24th, 6:30AM-7:50AMIf you do not take the exam on Tuesday…….

Ideally email me by 5pm on the 22nd so I know you are comingIf you just miss the exam, email me as soon as possibleNo guarantee I’ll have an extra exam if I don’t know you are coming!!!

If you do not take exam 1 on either dayIt is your dropped exam, no further makeups given

DSS ExamsI need DSS paperwork by next Tuesday (1 week before exam)I will try to get a time and date between Tues-Fri that works for allOnce I know your schedules, I’ll get back to you with options 18

What Does the Equilibrium Constant Tell Us?

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Relative Meaning Of Equilibrium Constants

Large value of Kc or Kp[C] & [D] very large or [A] & [B] very small. Reaction goes to completionEquilibrium favors products.

Small value of Kc or Kp[C] & [D] very small or [A] & [B] very largeReaction proceeds slowly or not at all (NR: no reaction)Equilibrium favors reactants

Does not guarantee reaction will occurThermodynamically favored, but may be kinetically controlled.

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....... dDcCbBaA ba

dc

c BADCK][][][][

The Reaction Quotient, Qc or QpRepresents Kc or Kp at non-equilibrium conditions

Predicts direction of reaction to get to equilibrium.Qc < K

Concentration of products is less than at equilibriumReaction occurs in the forward direction

Qc > KcConcentration of products is greater than at equilibriumReaction proceeds in the reverse direction

Qc = KcConcentration of products equals that at equilibriumNo macroscopic changes are observed

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ba

dc

c BADCQ][][][][

....... dDcCbBaA

For the reaction of hydrogen with nitrogen to form ammonia, the equilibrium concentrations were found to be 1.0×10-3 M in both hydrogen and nitrogen and 0.020 M in ammonia. Now add 0.010 M nitrogen. Which direction does

the reaction shift?

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

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For the reaction of hydrogen with nitrogen to form ammonia, the equilibrium constant Kc = 4.0×108 at 25 oC. The equilibrium concentrations were found to be 1.0×10-3 M in both hydrogen and nitrogen and 0.020 M in ammonia.

Now add 0.010 M nitrogen. Which direction does the reaction shift?

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

Fill in Concentrations:[N2] = 1.0×10-3 M + 0.010M= 0.011 M[NH3] = 0.020 M[H2] = 1.0×10-3 M

Compare Qc to Kc: 3.6×107< 4.1×108 Qc<Kc

The reaction goes toward productsAdding Nitrogen forces reaction to the right

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HNNHKc

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106.3]100.1][011.0[

]020.0[ xx

Qc

Solving Equilibrium Problems

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Use of Tabulated Equilibrium Constants

1. Write the properly balanced chemical reaction. This gives the stoichiometry and the species involved in the reaction

2. Set up a table of concentrations for all componentsUse reaction stoichiometry to express all unknown concentrations in terms of a single reactant or product, x.

3. Write out the equilibrium constant expression for the reaction

This is the equation for K4. Substitute these concentrations into the equation for K 5. Solve the equation for the unknown concentration, x.6. Substitute the value you calculated for x into the expressions for the other equilibrium concentrations.

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The decomposition of BrCl to bromine and chlorine has a Kc, of 0.14 at 350 K. If the initial concentration of BrCl is 0.062 M, what are the equilibrium

concentrations of all components?

3. Equilibrium Constant Expression 5. Solve for x

4. Substitute equilibrium values

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x = 0.0232-0.748x x = 0.013

1. Equation 2BrCl(g) Br2(g) + Cl2(g)2. Table BrCl Br2 Cl2

Initial .062 0 0Change – 2x +x +xEquilibrium 0.062-2x +x +x

222

][]][[

BrClClBrK c

2]2062.0[]][[14.0

xxx

]2062.0[

][374.0

]2062.0[][14.0

2/1

2

22/1

xx

xx

6. Substitute back into table[BrCl] = 0.062–2x = 0.062–2(0.013) = 0.036 M[Br2] = [Cl2] = x = 0.013M

Carbon monoxide reacts with water at 1000 oC to give carbon dioxide and hydrogen with Kc = 0.58. A reaction was started with the following composition: CO2, 0.2M M; H2 1.20 M;

H2O, 0.50 M, and CO, 1.0 M. What are the equilibrium concentrations of all components?

3. Equilibrium Constant Expression

4. Substitute equilibrium values

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1. Equation CO(g) + H2O(g) CO2(g) + H2(g)2. Table CO H20 CO2 H2

Initial 1.0 0.50 0.2 1.2Change – x -x +x +xEquilibrium 1.0-x 0.50-x 0.2+x 1..2+x

]][[]][[

2

22

OHCOHCOK c

]50.0][00.1[]2.1][20.0[58.0xxxx

5. Solve for x:

Need quadratic equation x = 0.0226. Substitute back into table

[CO] = 1.00 – 0.022 = 0.98M [H2O] = 0.50 – 0.022 = 0.48M[CO2] = 0.20+ 0.022 = 0.22M [H2] = 1.20 + 0.022 = 1.22M

Solving Quadratic Equations

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Solving for X

x = 0.022 or –5.4

Only x = 0.022 makes sense

x = – 5.4 gives negative concentrations

From the previous problem:

For equations: aX2 + bX + c = 0

Factors that AffectChemical Equilibrium

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Le Châtelier’s PrincipleWhen stress is applied to a system at equilibrium, the system will shift to reduce the applied stress and reestablish equilibrium

2NO2(g) 1 N2O4(g)

Addition of pressure causes reaction to shift towards productsNumber of moles decreases (decreases the stress).

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General Rules for Le Châtelier’s Principle

5 types of stresses will affect the equilibriumChange in Components or Concentrations

Change in Partial Pressure

Change in External Pressure

Change in Total Volume

Change in Temperature

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Change in Components or Partial Pressures

If the concentration of a reactant increases:Fe3+(aq) + SCN-(aq)FeSCN2+(aq)yellow clear red

Denominator of the Qc increasesQc < Kc

To equalize Qc and Kc, Concentration of other reactants decreaseConcentration of the products increases Equilibrium is pushed right.

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]][[][

3

2

SCNFe

FeSCNKc

Change in TemperatureExothermic Reaction Endothermic ReactionHeat is released as a product Heat is needed as a reactant

Increase T: K decreases Increase T: K increasesDecrease T: K increases Decrease T: K Decreases

2NO2(g) N2O4(g) Exothermic Reactionred clear ΔH=-58kJ/mol

Cold: more N2O4K increases

Hot: More NO2K decreases

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dDcCbBaAHeat HeatdDcCbBaA

Change in External Pressure or Volume

When pressure is applied (increased)Equilibrium shifts to produce smaller # of moles of gas

When pressure is decreasedEquilibrium shifts to produce larger # of moles of gas

Reaction with same # of moles of gas on both sides Changes in external pressure do not affect equilibrium

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DCBA 1123

DCBA 2323

Adding A Catalyst or An Inhibitor

CatalystsThe role of a catalyst is to lower activation energy.Catalysts and inhibitors do not affect the equilibriumCatalysts cause a reaction to reach equilibrium faster

InhibitorsInhibitors slow down the reaction to prevent equilibrium from being reached as quickly

Inhibitors may slow the reaction rate to the point that there appears to be no reaction, but it is really just infinitely slow.

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Which way does the reaction shift if…N2O3(g) NO(g) + NO2(g) Ho = +39.7 kJ/mol

NO is added? Shifts left to use up additional chemical

Volume of reaction vessel reduced?Increased pressure: shifts left to reduce # moles

The total internal pressure is increased by adding He gas?No shift: Total pressure on system increases, but no change in partial pressures of reactants or products.

The temperature is increased?This an endothermic reaction, heat is absorbed. Increasing the temperature adds heat to the system. Reaction shifts right (endothermic) to remove heat.

A catalyst is added?No shift: A catalyst does not affect the equilibrium composition

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SummaryEquilibrium is a balance between products and reactants

Use stoichiometry to determine reactant or product ratios, but NOT reactant to product ratios.

Capital K is used to represent the equilibrium constantProducts over reactants raised to their stoichiometric coefficientsCalculated from balanced equation, subscript designates units used, Kc, KpNo units used in final written K

Equlibrium Calculations and Reaction QuotientsICE tables used to manipulate initial and equilibrium concentrations.

Factors influencing K Concentration of chemicals and temperature affect all equilibria

T, P, V changes affect Kp in gases Catalysis

Effect of catalysts and inhibitors, reasons they are used.

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