Chemical Equilibrium CHAPTER 15 Chemistry: The Molecular Nature of Matter, 6 th edition

Chemical Equilibrium

CHAPTER 15

Chemistry: The Molecular Nature of Matter, 6th editionBy Jesperson, Brady, & Hyslop

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CHAPTER 15 Chemical Equilibrium

Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Learning Objectives:

Reversible Reactions and Equilibrium Writing Equilibrium Expressions and the Equilibrium

Constant (K) Reaction Quotient (Q) Kc vs Kp

ICE Tables Quadratic Formula vs Simplifying Assumptions LeChatelier’s Principle van’t Hoff Equation

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Lecture Road Map:

① Dynamic Equilibrium

② Equilibrium Laws

③ Equilibrium Constant

④ Le Chatelier’s Principle

⑤ Calculating Equilibrium

4Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Dynamic Equilibrium



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Dynamic Eq Equilibrium

• Chemical equilibrium exists when– Rates of forward and reverse reactions are equal– Reaction appears to stop – Concentration of reactants and products do not

change over time• Remain constant• Both forward and reverse reaction never cease

• Equilibrium signified by double arrows ( )


N2O4 2 NO2

• Initially have only N2O4

– Only forward reaction

– As N2O4 reacts NO2 forms

• As NO2 forms

– Reverse reaction begins to occur

– NO2 collide more frequently as concentration of NO2 increases

• Eventually, equilibrium is reached

– Concentration of N2O4 does not change

– Concentration of NO2 does not change

Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 6



N2O4 2NO2

Closed system• Equilibrium can be

reached from either direction

• Independent of whether it starts with “reactants” or “products”

• Always have the same composition at equilibrium under same conditions



N2O4 2NO2

Reactants ProductsEquilibrium

Dynamic Eq Mass Action Expression


• Simple relationship among [reactants] and [products] for any chemical system at equilibrium

• Called the mass action expression– Derived from thermodynamics

• Forward reaction: A B • Reverse reaction: A B • At equilibrium: A B

Dynamic Eq Reaction Quotient


• Uses stoichiometric coefficients as exponent for each reactant

• For reaction: aA + bB cC + dD

Reaction quotient– Numerical value of mass action expression– Equals “Q ” at any time, and– Equals “K ” only when reaction is known to be at

equilibrium

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Ex. 1 H2(g) + I2(g) 2HI(g) 440˚C

Exp’t Initial Amts

Equil’m Amts

Equil’m [M]

I 1.00 mol H2

0.222 mol H2

0.0222 M H2

10 L

1.00 mol I2

0.222 mol I2

0.0222 M I2

0.00 mol HI

1.56 mol HI 0.156 M HIII 0.00 mol H2

0.350 mol H2

0.0350 M H2

10 L 0.100 mol I2

0.450 mol I2

0.0450 M I2

3.50 mol HI

2.80 mol HI 0.280 M HI

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Ex. 1 H2(g) + I2(g) 2HI(g) 440 ˚C

Exp’t Initial Amts

Equil’m Amts

Equil’m [M]

III 0.0150 mol H2

0.150 mol H2

0.0150 M H2

10 L 0.00 mol I2 0.135 mol I2 0.0135 M I21.27 mol HI 1.00 mol HI 0.100 M HI

IV 0.00 mol H2 0.442 mol H2

0.0442 M H2

10 L 0.00 mol I2 0.442 mol I2

0.0442 M I2

4.00 mol HI 3.11 mol HI 0.311 M HI

Equilibrium Concentrations (M )

Exp’t [H2] [I2] [HI]

I 0.0222 0.0222 0.156

II 0.0350 0.0450 0.280

III 0.0150 0.0135 0.100

IV 0.0442 0.0442 0.311

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Mass Action Expression

= same for all data sets at equilibrium

4.49)0222.0)(0222.0(

)156.0( 2

8.49)0450.0)(0350.0(

)280.0( 2

4.49)0135.0)(0150.0(

)100.0( 2

5.49)0442.0)(0442.0(

)311.0( 2

Average = 49.5


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GroupProblem

Write mass action expressions for the following:

• 2NO2(g) N2O4(g)

• 2CO(g) + O2(g) 2CO2(g)

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GroupProblem


Which of the following is the correct mass action expression for the reaction:

Cu2+(aq) + 4NH3(aq) [Cu(NH3)42+](aq)?


Equilibrium Laws



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Equilibrium Equilibrium Laws

• For reaction

H2(g) + I2(g) 2HI(g) at 440 ˚C

at equilibrium write the following equilibrium law

• Equilibrium constant = Kc = constant at given T

• Use Kc since usually working with concentrations in mol/L

• For chemical equilibrium to exist in reaction mixture, reaction quotient Q must be equal to equilibrium constant, Kc


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Equilibrium Predicting Equilibrium Laws

For general chemical reaction:• dD + eE fF + gG

– Where D, E, F, and G represent chemical formulas– d, e, f, and g are coefficients

• Mass action expression is

• Note: Exponents in mass action expression are stoichiometric coefficients in balanced equation.

• Equilibrium law is:


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Equilibrium Predicting Equilibrium Laws

• Only concentrations that satisfy this equation are equilibrium concentrations

• Numerator– Multiply concentration of products raised to their

stoichiometric coefficients• Denominator

– Multiply concentration reactants raised to their stoichiometric coefficients

is scientists’ convention


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Equilibrium Example

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

Kc = 4.26 × 108 at 25 °C

What is equilibrium law?


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Equilibrium Operations

Various operations can be performed on equilibrium expressions

1. When direction of equation is reversed, new equilibrium constant is reciprocal of original

A + B C + D

C +D A + B


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1. When direction of equation is reversed, new equilibrium constant is reciprocal of original

3H2(g) + N2(g) 2 NH3(g) at 25˚C

2NH3(g) 3H2(g) + N2(g) at 25 ˚C

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23

2

23 1026.4

][N][H

][NHcK


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2. When coefficients in equation are multiplied by a factor, equilibrium constant is raised to a power equal to that factor.

A + B C + D

3A + 3B 3C + 3D


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2. When coefficients in equation are multiplied by factor, equilibrium constant is raised to power equal to that factor

3H2(g) + N2(g) 2NH3(g) at 25 ˚C

Multiply by 3

9H2(g) + 3N2(g) 6NH3(g)


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3. When chemical equilibria are added, their equilibrium constants are multiplied

A + B C + D

C + E F + G

A + B + E D + F + G

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3. When chemical equilibria are added, their equilibrium constants are multiplied

][CO][NO]][CO[NO

][NO

][NO][NO

3

222

2

3

2 NO2(g) NO3(g) + NO(g)

NO3(g) + CO(g) NO2(g) + CO2(g)

NO2(g) + CO(g) NO(g) + CO2(g)

Therefore

][CO][NO][NO][CO

2

2


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GroupProblem For: N2(g) + 3H2(g) 2NH3(g)

Kc = 500 at a particular temperature.

What would be Kc for following?

• 2NH3(g) N2(g) + 3H2(g)

• 1/2N2(g) + 3/2H2(g) NH3(g)

22.4

0.002


Equilibrium Constant



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Equilibrium Constant Kc

• Most often Kc is expressed in terms of a ratio of concentrations of products and reactants as shown on previous slides

• Sometimes partial pressures, in atmospheres, may be used in place of concentrations


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Equilibrium Kp

• Based on reactions in which all substances are gaseous

• Gas quantities are expressed in atmospheres in mass action expression

• Use partial pressures for each gas in place of concentrations

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


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Equilibrium Relationship between Kp and Kc

• Start with ideal gas law

PV = nRT• Rearranging gives

• Substituting P/RT for molar concentration into Kc results in pressure-based formula

• ∆n = moles of gas in product – moles of gas in reactant


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GroupProblem

Consider the reaction: 2NO2(g) N2O4(g)

If Kp = 0.480 for the reaction at 25 ˚C, what is value of Kc at same temperature?

n = nproducts – nreactants = 1 – 2 = –1

Kc = 11.7


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GroupProblem

Consider the reaction A(g) + 2B(g) 4C(g) If the Kc for the reaction is 0.99 at 25 ˚C, what would be the Kp?

A. 0.99

B. 2.0

C. 24

D. 2400

E. None of these

Δn = (4 – 3) = 1

Kp = Kc(RT)Δn

Kp= 0.99 × (0.082057 × 298.15)1

Kp = 24


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Equilibrium Homogeneous and Hetergeneous

Homogeneous reaction/equilibrium– All reactants and products in same phase– Can mix freely

Heterogeneous reaction/equilibrium– Reactants and products in different phases– Can’t mix freely– Solutions are expressed in M– Gases are expressed in M

– Governed by Kc


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Equilibrium Heterogeneous

2NaHCO3(s) Na2CO3(s) + H2O(g) + CO2(g)

• Equilibrium Law =

• Can write in simpler form• For any pure liquid or solid, ratio of moles to

volume of substance (M ) is constant– e.g. 1 mol NaHCO3 occupies 38.9 cm3

2 mol NaHCO3 occupies 77.8 cm3

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2NaHCO3(s) Na2CO3(s) + H2O(g) + CO2(g)

– Ratio (n/V ) or M of NaHCO3 is constant (25.7 mol/L) regardless of sample size

– Likewise can show that molar concentration of Na2CO3 solid is constant regardless of sample size

• So concentrations of pure solids and liquids can be incorporated into equilibrium constant, Kc

• Equilibrium law for heterogeneous system written without concentrations of pure solids or liquids


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Write equilibrium laws for the following:

Ag+(aq) + Cl–(aq) AgCl(s)

H3PO4(aq) + H2O H3O+(aq) + H2PO4

–(aq)

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GroupProblem


Given the reaction:

3Ca2+(aq) + 2PO43–(aq)

Ca3(PO4)2(s)

What is the mass action expression?


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GroupProblem

Given the reaction:

3Ca2+(aq) + 2PO43–(aq) Ca3(PO4)2(s)

What is mass action expression for the reverse reaction?

Chemical Equilibrium CHAPTER 15 Chemistry: The Molecular Nature of Matter, 6 th edition

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