Chapter 11 Chemical Equilibrium 11.1 The Equilibrium Condition 11.2 The Equilibrium Constant 11.3...

Chapter 11Chemical Equilibrium

11.1 The Equilibrium Condition 11.2 The Equilibrium Constant 11.3 Equilibrium Expressions Involving Pressures 11.4 The Concept of Activity 11.5 Heterogeneous Equilibria 11.6 Applications of the Equilibrium Constant 11.7 Solving Equilibrium Problems 11.8 Le Chatelier's Principle 11.9 Equilibria Involving Real Gases

The Equilibrium Condition (General)

Thermal equilibrium indicates two systems in thermal contact with each do not exchange energy by heat. If two bricks are in thermal equilibrium their temperatures are the same.

Chemical equilibrium indicates no unbalanced potentials (or driving force). A system in equilibrium experiences no change over time, even infinite time.

The opposite of equilibrium systems are non-equilibrium systems that are off balance and change with time.

Example 1 atm O2 + 2 atm H2 at 298K

aA + bB cC + dD

The same equilibrium state is achieved whether starting with pure reactants or pure products.

The equilibrium state can change with temperature.

The Equilibrium Condition (Chem Rxn)

The Equilibrium State (Chem Rxn)

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

Change [CO] to PCO

[H2O] to PH2O

etc

As the equilibrium state is approached, the forward and backward rates of reaction approach equality. At equilibrium the rates are equal, and no further net change occurs in the partial pressures of reactants or products.

1. No macroscopic evidence of change.2. Reached through spontaneous processes.3. Show a dynamic balance of forward and backward processes.

4. Same regardless of the direction from which they are approached.

Fundamental characteristics of equilibrium states:

Chemical Reactions and Equilibrium

5. No change over time.

↔

Use this in an equilibrium expression.

Use this to indicate resonance.

Arrows: Chemical Symbolism

Chemical Reactions and Equilibrium

The equilibrium condition for every reaction can be described in a single equation in which a number, the equilibrium constant (K) of the reaction, equals an equilibrium expression, a function of properties of the reactants and products.

H2O(l) H2O(g) @ 25oC Temperature (oC) Vapor Pressure (atm) 15.0 0.01683 17.0 0.0191219.0 0.0216821.0 0.0245423.0 0.02772

25.0 0.03126

30.0 0.04187

50.0 0.1217

H2O(l) H2O(g) @ 30oC

K = 0.03126

K = 0.04187

Partial pressures and concentrations of products appear in the numerator and those of the reactants in the denominator. Each is raised to a power equal to its coefficient in the balanced chemical equation.

aA + bB cC + dD

€

if gases

PC( )c

PD( )d

PA( )a

PB( )b = K

€

if concentrations

C[ ]c

D[ ]d

A[ ]a

B[ ]b = K

Law of Mass Action (1)

1. Gases enter equilibrium expressions as partial pressures, in atmospheres. E.g., PCO2

2. Dissolved species enter as concentrations, in molarity (M) moles per liter. E.g., [Na+]

3. Pure solids and pure liquids are represented in equilibrium expressions by the number 1 (unity); a solvent taking part in a chemical reaction is represented by unity, provided that the solution is dilute. E.g., I2(s) ↔ I2(aq) [I2 (aq) ] = K

)]([1

)]([)]([)]([

)()(

22

2

2

22

aqIaqI

sIaqI

K

aqIsI

===

↔

Law of Mass Action (2)

PH2O

Pref

= K Pref is numerically equal to 1

The convention is to express all pressures in atmospheres and to omit factors of Pref because their value is unity. An equilibrium constant K is a pure number.

H2O (l) H2O (g) Kp = P H2O

The concept of Activity (i-th component)

= ai = Pi / P reference

@ 25oC Kp = 0.03126 atm

K = 0.03126

Activities

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

COOH

COHp PP

PPK

2

22=

The Equilibrium State

The Equilibrium Expressions

In a chemical reaction in which a moles of species A and b moles of species B react to form c moles of species C and d moles of species D,

The partial pressures at equilibrium are related through

K = PcCPd

D/PaAPb

B

aA + bB cC + dD

Write equilibrium expressions for the following reactions

3 H2(g) + SO2(g) H2S(g) + 2 H2O(g)

2 C2F5Cl(g) + 4 O2(g) Cl2(g) + 4 CO2(g) + 5 F2(g)

Gases and Solids

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

K=PCO2

K is independent of the amounts

of CaCO3(s) or CaO(s)

Heterogeneous Equilibrium

Liquids

Solutions

H2O(l) H2O(g)

K=PH2O

I2(s) I2(aq)

K=[I2]

Heterogeneous Equilibrium

Relationships Among the K’s of Related Reactions

#1: The equilibrium constant for a reverse reaction is always the reciprocal of the equilibrium constant for the corresponding forward reaction.

2 H2 (g) + O2 (g) 2 H2O (g) (PH2O)2

(PH2)2(PO2

) = K1

2 H2O (g) 2 H2 (g) + O2 (g) (PH2)2(PO2

)(PH2O)2 = K2

K1 = 1/K2

1KKor K

1K revfor

revfor ==

#1

#2

aA + bB cC + dD cC + dD aA + bB versus

# 2: When the coefficients in a balanced chemical equation are all multiplied by a constant factor, the corresponding equilibrium constant is raised to a power equal to that factor.

(PH2O)

(PH2)(PO2

)½= K3 K3 = K1

½


2 H2 (g) + O2 (g) 2 H2O (g) Rxn 1#1

H2 (g) + ½ O2 (g) H2O (g) Rxn 3 = Rxn 1 times 1/2#3

(PH2O)2

(PH2)2(PO2

) = K1

# 3: when chemical equations are added to give a new equation, their equilibrium constants are multiplied to give the equilibrium constant associated with the new equation.

2 BrCl (g) ↔ Br2 (g) + Cl2 (g)

(PBr2)(PCl2

)(PBrCl)2

Br2 (g) + I2 (g) ↔ 2 IBr (g)

(PIBr)2

(PBr2) (PI2

)

2 BrCl (g) + I2 (g) ↔ 2 IBr (g) + Cl2(g) = K1K2

= (0.45)(0.051)

=0.023 @ 25oC


(PBr2)(PCl2

)

(PBrCl)2 = K1 = 0.45 @ 25oC

X

(PIBr)2

(PBr2) (PI2

) = K2 = 0.051 @ 25oC

= K1K2 = K3

wrong arrow

wrong arrow

wrong arrow

Calculating Equilibrium ConstantsConsider the equilibrium

4 NO2(g)↔ 2 N2O(g) + 3 O2(g) The three gases are introduced into a container at partial pressures of 3.6 atm (for NO2), 5.1 atm (for N2O), and 8.0 atm (for O2) and react to reach equilibrium at a fixed temperature. The equilibrium partial pressure of the NO2 is measured to be 2.4 atm. Calculate the equilibrium constant of the reaction at this temperature, assuming that no competing reactions occur.

4 NO2(g) ↔ 2 N2O(g) + 3 O2(g)

initial partial pressure (atm) change in partial pressure (atm) equilibrium partial pressure (atm)

4 NO2(g) ↔ 2 N2O(g) + 3 O2(g)

initial partial pressure (atm) 3.6 5.1 8.0

change in partial pressure (atm) – 4x +2x +3x

equilibrium partial pressure (atm) 2.4 5.1 + 2x 8.0 + 3x

3.6 – 4x = 2.4 atm NO2; x = 0.3 atm

5.1 + 2(0.3 atm) = 5.7 atm N2O

8.0 + 3(0.3 atm) = 8.9 atm O2

K =(PN2O)2(PO2

)3

(PNO2)4 =

Calculate the equilibrium constant of the reaction at this temperature,

assuming that no competing reactions occur.

The compound GeWO4(g) forms at high temperature in the reaction 2 GeO (g) + W2O6(g) ↔ 2 GeWO4(g)

Some GeO (g) and W2O6 (g) are mixed. Before they start to react, their partial pressures both equal 1.000 atm. After their reaction at constant temperature and volume, the equilibrium partial pressure of GeWO4(g) is 0.980 atm. Assuming that this is the only reaction that takes place, (a) determine the equilibrium partial pressures of GeO and W2O6, and (b) determine the equilibrium constant for the reaction.

2 GeO (g) + W2O6 (g) ↔ 2 GeWO4(g) initial partial pressure (atm) 1.000 1.000 0

change in partial pressure (atm) – 2x – x +2x

equilibrium partial pressure (atm)

0 + 2x = 0.980 atm GeWO4; x = 0.490 atm

1.000 – 2(0.490) = 0.020 atm GeO

1.000 – 0.490 = 0.510 atm W2O6

K =(PGeWO4

)2

(PGeO)2(PW2O6) =

2 GeO(g) + W2O6(g) ↔ 2 GeWO4(g) initial partial pressure (atm) 1.000 1.000 0

change in partial pressure (atm) –2x –x +2x

equilibrium partial pressure (atm) 1.000 – 2x 1.000 – x 0.980

(a) determine the equilibrium partial pressures of GeO and W2O6, and

(b) determine the equilibrium constant for the reaction.

• Skip Solving quadratic equations

• Will utilize approximation method– Systems that have small equilibrium

constants.– Assume “x” (the change in concentration)

is small (less than 5%) of the initial concentration.

A vessel holds pure CO (g) at a pressure of 1.282 atm and a temperature of 354K. A quantity of nickel is added, and the partial pressure of CO (g) drops to an equilibrium value of 0.709 atm because of the reaction

Ni (s) + 4CO (g) ↔ Ni(CO)4 (g)

Compute the equilibrium constant for this reaction at 354K.

(1))(P

P

[Ni(s)])(P

PK

4CO

Ni(CO)

4CO

Ni(CO) 44 ==

initial partial pressure (atm) 1.282 0 change in partial pressure (atm) -4x +1x equilibrium partial pressure (atm) 0.709 x

P CO (atm) P Ni(CO)4 (atm)

At equil. Pco= x = PNi(CO)4 =

Construct an “ICE” table

Ni (s) + 4CO (g) ↔ Ni(CO)4 (g)

Equilibrium Calculations

At a particular temperature, K = 2.0 x 10-6 mol/L for the reaction

2CO2 (g) 2CO (g) + O2 (g)

If 2.0 mol CO2 is initially placed into a 5.0-L vessel, calculate the equilibrium concentrations of all species.

initial partial pressure (mol/L) 0.4 0 0change in partial pressure (mol/L) – 2x +2x +1xequilibrium partial pressure (mol/L) 0.4 -2x 2x 1x

2CO2 (g) 2CO (g) + O2 (g)


2CO2 (g) 2CO (g) + O2 (g)



2CO2 (g) 2CO (g) + O2 (g)

2 2-62

p 2 22

2-6 -3

2

[CO] [O ] (2 ) ( )K = = 2.0x10

[CO ] (0.40 2 )

assume 2x << 0.40

(2x) (x)2.0x10 = x = 4.3x10 M

(0.40)

x x

x=

−

⇒


2CO2 (g) 2CO (g) + O2 (g)



2CO2 (g) 2CO (g) + O2 (g)

2-62

p 22

2

2

-3

[CO] [O ]K = = 2.0x10

[CO ]

(2 ) ( )

(0.40 2 )

x = 4.3x10 M

x x

x=

−

2

-3

[CO ] = 0.40 - 2x =

= 0.40 - 2(4.3x10 ) = 0.39M

-3

-3

[CO] = 2x = 2(4.3 x 10 )

= 8.6 x 10 M-3

2[O ] = x = 4.3 x 10 M

K (the Equilibrium Constant) uses equilibrium partial pressures

Q (the reaction quotient) uses prevailing partial pressures, not necessarily at equilibrium

dD cC bBaA +⏐⏐ ⏐←⏐⏐⏐ →⏐

+reverse

forward

( ) ( )( ) ( )

QPP

PPb

Ba

A

dD

cC =( ) ( )

( ) ( )K

PP

PPb

Ba

A

dD

cC = KQ=

?

wrong arrow

Non-Equilibrium Conditions: The Reaction Quotient (1)

If Q < K, reaction proceeds in a forward direction (toward products);

If Q > K, reaction proceeds in a backward direction (toward reactants);

If Q = K, the reaction is in equilibrium.

KQ=?

dD cC bBaA +⏐⏐ ⏐←⏐⏐⏐ →⏐

+reverse

forward

( ) ( )( ) ( )

QPP

PPb

Ba

A

dD

cC =

( ) ( )( ) ( )

KPP

PPb

Ba

A

dD

cC =

wrong arrow

The Reaction Quotient (2)

The equilibrium constant for the reaction P4(g) ↔ 2 P2(g) is 1.39 at 400oC. Suppose that 2.75 mol of P4(g) and 1.08 mol of P2(g) are mixed in a closed 25.0 L container at 400oC. Compute

Q(init) (the Q at the moment of mixing) and state the direction in which the reaction proceeds.

K = 1.39 @ 400oC; nP4(init) = 2.75 mol; nP2(init) = 1.08 mol

PP4(init) = nP4(init)RT/V

PP2(init) = nP2(init)RT/V

= [(2.75mol)(0.08206 atm L mol-1 K-1)(273.15+400oC)]/(25.0L)

= [(1.08mol)(0.08206 atm L mol-1 K-1)(273.15+400oC)]/(25.0L)

= 6.08 atm

= 2.39 atm

Q =

( )( )

39.114

22 == K

P

P

P

P nRTPV =KQ=?

Henri Louis Le Châtelier (1850-1936)

Highlights– 1884 Le Chatelier's Principle: A system in

equilibrium that is subjected to a stress reacts in a way that counteracts the stress

– If a chemical system at equilibrium experiences a change in concentration, temperature or total pressure the equilibrium will shift in order to minimize that change.

– Industrial chemist involved with industrial efficiency and labor-management relations

Moments in a Life– Le Chatelier was named "chevalier"

(knight) of the Légion d'honneur in 1887, decoration established by Napoléon Bonaparte in 1802.

Effects of External Stresses on Equilibria: Le Châtelier’s Principle

A system in equilibrium that is subjected to a stress reacts in a way that counteracts the stress.

2. Effects of Changing the Volume (or Pressure) of the System

3. Effects of Changing the Temperature

1. Effects of Adding or Removing Reactants or Products

Le Châtelier’s Principle provides a way to predict the response of an equilibrium system to an external perturbation, such as…

Effects of Adding or Removing Reactants or Products

PCl5(g) PCl3(g) + Cl2(g) K = 11.5 @ 300oC = Q

add extra PCl5(g)

add extra PCl3(g)

remove some PCl5(g)

remove some PCl3(g)

A system in equilibrium that is subjected to a stress reacts in a way that counteracts the stress. In this case adding or removing reactants or products

Effects of Changing the Volume of the System

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

Let’s decrease the volume of the reaction container

Let’s increase the volume of the reaction container

Less room :: less amount (fewer moles)

Shifts reaction to restore equilibrium

More room :: more amount (greater moles)

Shifts reaction to restore equilibrium

1 mole 1+1 = 2 moles

A system in equilibrium that is subjected to a stress reacts in a way that counteracts the stress. In this case a change in volume

Volume Decreased Volume Increased

(Pressure Increased) (Pressure Decreased)

V reactants > V productsEquilibrium shift right

(toward products)Equilibrium Shifts left

(toward reactants)

V reactants < V productsEquilibrium Shifts left

(toward reactants)Equilibrium shift right

(toward products)

V reactants = V products Equilibrium not affected Equilibrium not affected

2 P2(g) P4 (g)PCl5(g) PCl3(g) + Cl2(g)

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

Boyles Law:

PV = Constant

A system in equilibrium that is subjected to a stress reacts in a way that counteracts the stress. In this case a change in volume (or pressure)

Effects of Changing the Temperature

Endothermic: heat is aborbed by a reaction

Reactants + heat gives Products

Exothermic: heat is liberated by a reaction

Reactants gives Products + heat

A system in equilibrium that is subjected to a stress reacts in a way that counteracts the stress. In this case a change in temperature


Let’s decrease the temperature of the reaction

Let’s increase the temperature of the reaction, what direction does the equilibrium reaction shift

Endothermic: absorption of heat by a reaction

Reactants + heat gives Products

Let’s decrease the temperature of the reaction

Let’s increase the temperature of the reaction


Exothermic: heat liberated by a reaction

Reactants Products + heat

A system in equilibrium that is subjected to a stress reacts in a way that counteracts the stress. In this case a change in temperature

Temperature Raised Temperature Lowered

Endothermic Reaction

(absorb heat)

Equilibrium shift right (toward products)

Equilibrium Shifts left (toward reactants)

Exothermic Reaction

(liberate heat)

Equilibrium Shifts left (toward reactants)

Equilibrium shift right (toward products)

If a forward reaction is exothermic,

Then the reverse reaction must be endothermic

Heat dD cCbB aA forward ++⏐ →⏐+

dD cC bBaA +⏐⏐ ⏐←⏐⏐⏐ →⏐

+reverse

forward

bB aAHeat dD cC reverse +⏐ →⏐++

Driving Reactions to Completion/ Increasing Yield

Industrial Synthesis of Ammonia (Haber)

N2 (g) + 3H2 (g) ↔ 2NH3 (g)

Forward reaction is exothermic

What conditions do we need to increase the yield, i.e., produce more ammonia?

Volume Decreased Volume Increased

(Pressure Increased) (Pressure Decreased)

V reactants >

V products

Equilibrium shift right

(toward products)

Equilibrium Shifts left (toward

reactants) Temperature Raised Temperature Lowered

Exothermic Reaction

(liberate heat)

Equilibrium Shifts left (toward

reactants)

Equilibrium shift right (toward

products)

Chapter 11 Chemical Equilibrium 11.1 The Equilibrium Condition 11.2 The Equilibrium Constant 11.3...

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Transcript of Chapter 11 Chemical Equilibrium 11.1 The Equilibrium Condition 11.2 The Equilibrium Constant 11.3...