ofb chapter 7 lecture notes - georgia institute of...

9/22/2004 OFB Chapter 7 1

OFB Chapter 7Chemical Equilibrium

7-1 Chemical Reactions in Equilibrium7-2 Calculating Equilibrium Constants7-3 The Reaction Quotient7-4 Calculation of Gas-Phase Equilibrium7-5 The effect of External Stresses on Equilibria:

Le Châtelier’s Principle7-6 Heterogeneous Equilibrium7-7 Extraction and Separation Processes


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All exercisesHomework:Chapter 7:-8, 14, 24, 42, 54

Note: this is all that’s onWebCT… Be familiar withequilibrium problems – theseare very important

9/22/2004 OFB Chapter 7 3~220 K (-53oC)

dD cC bBaA + ← →

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7-1 Chemical Reactions and Equilibrium

The equilibrium condition for every reaction can be summed up 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 0.0277225.0 0.0312630.0 0.0418750.0 0.1217

H2O(l) ↔ H2O(g) @ 30oC

K = 0.03126

K = 0.04187


Chemical Reactions and Equilibrium

PH2O

Pref= K Pref is numerically equal to 1

The convention in the book 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)



2 NO2(g) → N2O4(g)

N2O4(g) → 2 NO2 (g)

An equilibrium reaction



2 NO2(g) → N2O4(g)

N2O4(g) → 2 NO2 (g)↓T↑T

↑P↓P

2 NO2(g) ↔ N2O4(g); K = 8.8 @ 25oC



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. They display no macroscopic evidence of change.2. They are reached through spontaneous processes.3. They show a dynamic balance of forward and backward processes.4. They are the same regardless of the direction from which they are approached.

Four fundamental characteristics of equilibrium states in isolated systems:


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The Form of 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

provided that all species are present as low-pressure gases.


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9/22/2004 OFB Chapter 7 10

Exercise 7-1

Write equilibrium expressions for the reactions defined by the following equations:

3 H2(g) + SO2(g) ⇔ H2S(g) + 2 H2O(g)

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

9/22/2004 OFB Chapter 7 11

7-2 Calculating Equilibrium ConstantsExample 7-2

Consider 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)

9/22/2004 OFB Chapter 7 12

4 NO2(g) ↔ 2 N2O(g) + 3 O2(g)

initial partial pressure (atm)

change in partial pressure (atm)

equilibrium partial pressure (atm)

K =(PN2O)2(PO2)

3

(PNO2)4

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

=(5.7)2(8.9)3

(2.4)4 =

9/22/2004 OFB Chapter 7 13

Exercise 7-2

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)



9/22/2004 OFB Chapter 7 14

K =(PGeWO2)

2

(PGeO)2(PW2O6)=

2 GeO(g) + W2O6(g) ⇔ 2 GeWO4(g) initial 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

=(0.980)2

(0.020)2(0.510)

9/22/2004 OFB Chapter 7 15

Relationships Among the K’s of Related ReactionsRule 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

#1

#2

9/22/2004 OFB Chapter 7 16

Rule 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.

H2(g) + ½ O2(g) ↔ H2O(g) Rxn 3

(PH2O)(PH2)(PO2)

½ = K3 K3 = K1½ = √K1

Relationships Among the K’s of Related Reactions

2 H2(g) + O2(g) ↔2 H2O(g) Rxn 1#1

#3

9/22/2004 OFB Chapter 7 17

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

2 BrCl(g) ↔ Br2(g) + Cl2(g) (PBr2)(PCl2)(PBrCl)2 = K1 = 0.45 @ 25oC

Br2(g) + I2(g) ↔ 2 IBr(g) (PIBr)2

(PBr2) (PI2)= K2 = 0.051 @ 25oC

= K1K2

Relationships Among the K’s of Related Reactions

= (0.45)(0.051)

=0.023 @ 25oC

9/22/2004 OFB Chapter 7 18

7-3 The Reaction Quotient

Note that K (the Equilibrium Constant) uses equilibrium partial pressuresNote that Q (the reaction quotient) uses prevailing partial pressures, not necessarily at equilibrium


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( ) ( )( ) ( )

QPPPP

bB

aA

dD

cC =

( ) ( )( ) ( )

KPPPP

bB

aA

dD

cC =

9/22/2004 OFB Chapter 7 19

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

The Reaction Quotient

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

If Q = K, the reactions stops because the reaction is in equilibrium.


+ reverse

forward

( ) ( )( ) ( )

QPPPP

bB

aA

dD

cC =

9/22/2004 OFB Chapter 7 20

Exercise 7-4

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

( )( )

39.114

22 == K

PP

P

P nRTPV =KQ=?

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

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

Q = (2.39)2/(6.08) = 0.939 0.939 < 1.39; Q < K

9/22/2004 OFB Chapter 7 21

7-4 Calculations of Gas-Phase EquilibriaExercise 7-5Carbon monoxide reacts with water to give hydrogen:

CO (g) +H2O (g) ↔ CO2 (g) + H2 (g)

At 900 K, the equilibrium constant for this reaction, the so-called shift reaction, equals 0.64. suppose the partial pressures of three gases at equilibrium at 900 K are

PCO = 2.00 atm, PCO2 = 0.80 atm, and PH2 =0.48 atm

Calculate the partial pressure of water under these conditions.

9/22/2004 OFB Chapter 7 22

7-4 Calculations of Gas-Phase EquilibriaExercise 7-5Carbon monoxide reacts with water to give hydrogen:

CO (g) +H2O (g) ↔ CO2 (g) + H2 (g)

At 900 K, the equilibrium constant for this reaction, the so-called shift reaction, equals 0.64. suppose the partial pressures of three gases at equilibrium at 900 K are

PCO = 2.00 atm, PCO2 = 0.80 atm, and PH2 =0.48 atm

Calculate the partial pressure of water under these conditions.

9/22/2004 OFB Chapter 7 23

• Solving quadratic equations

9/22/2004 OFB Chapter 7 24

7-5 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.

Effects of Changing the Volume of the System

Effects of Changing the Temperature

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…

9/22/2004 OFB Chapter 7 25

Effects of Adding or Removing Reactants or Products

PCl5(g) ↔ PCl3(g) + Cl2(g)

( ) ( )( ) QP

PP

PCl

ClPCl =1

11

5

23

add extra PCl5(g)

add extra PCl3(g)

remove some PCl5(g)

remove some PCl3(g)

9/22/2004 OFB Chapter 7 26

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

9/22/2004 OFB Chapter 7 27

Equilibrium not affectedEquilibrium not affectedV reactants = V products

Equilibrium shift right (toward products)

Equilibrium Shifts left (toward reactants)V reactants < V products

Equilibrium Shifts left (toward reactants)

Equilibrium shift right (toward products)V reactants > V products

(Pressure Decreased)(Pressure Increased)

Volume IncreasedVolume Decreased

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

CO (g) +H2O (g) ↔ CO2 (g) + H2 (g)

9/22/2004 OFB Chapter 7 28

PCl5(g) ⇔ PCl3(g) + Cl2(g)


Let’s decrease the temperature of the reaction

Let’s increase the temperature of the reaction

9/22/2004 OFB Chapter 7 29



Exothermic Reaction (liberate heat)



Endothermic Reaction (absorb heat)

Temperature LoweredTemperature Raised

If a forward reaction is exothermic,

Then the reverse reaction must be endothermic

9/22/2004 OFB Chapter 7 30

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


Let’s decrease the temperature of the reaction

Let’s increase the temperature of the reaction

9/22/2004 OFB Chapter 7 31

Driving Reactions to Completion


Equilibrium shift right (toward products)V reactants > V products

(Pressure Decreased)(Pressure Increased)

Volume IncreasedVolume Decreased



Exothermic Reaction (liberate heat)

Temperature LoweredTemperature Raised

Industrial Synthesis of Ammonia

N2 (g) + 3H2 (g) ↔ 2NH3 (g)

Forward reaction exothermic

9/22/2004 OFB Chapter 7 32


+ reverse

forward

( ) ( )( ) ( )

QPPPP

bB

aA

dD

cC =

( ) ( )( ) ( )

KPPPP

bB

aA

dD

cC = KQ=

?

Review: Chemical Equilibrium

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

9/22/2004 OFB Chapter 7 33

Exercise 7-10

State the effect of an increase in temperature and also of a decrease in volume on the equilibrium yield of the products in each of the following reactions.

a) CH3OCH3 (g) + H2O (g) ↔ 2 CH4 (g) + O2 (g)

b) H2O (g) + CO (g) ↔ HCOOH (g)

Increase Temp?

Decrease Volume?Increase Temp?

Decrease Volume?

9/22/2004 OFB Chapter 7 34

Heterogeneous Equilibrium

Solids

Liquids

Dissolved species

H2O(l) ↔ H2O(g)

CaCO3(s) ↔ CaO(s) + CO2(g)

I2(s) ↔ I2(aq)

9/22/2004 OFB Chapter 7 35

Law of Mass Action1. Gases enter equilibrium expressions as partial pressures, inatmospheres. E.g., PCO2

2. Dissolved species enter as concentrations, in 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.,

][1

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

)()(

22

2

2

22

IaqIsI

aqIK

aqIsI

===

↔

9/22/2004 OFB Chapter 7 36

Law of Mass Action

4. 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

9/22/2004 OFB Chapter 7 37

Exercise 7-11

Write equilibrium-constant equations for the following Equilibria:

(a) Si3N4(s) + 4 O2(g) ⇔ 3 SiO2(s) + 2 N2O(g)

(b) O2(g) + 2 H2O(l) ⇔ 2 H2O2(aq)

(c) CaH2(s) + 2 C2H5OH(l) ⇔ Ca(OC2H5)2(s) + 2 H2(g)

9/22/2004 OFB Chapter 7 38

Exercise 7-12

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.

9/22/2004 OFB Chapter 7 39

Exercise 7-12

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))(PP

[Ni(s)])(PP

K 4CO

Ni(CO)4

CO

Ni(CO) 44 ==

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

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

9/22/2004 OFB Chapter 7 40

7-7 Extraction and Separation Processes• Extraction

– Partitioning of a solute between two immiscible solvents

• Chromatography– Solute is partitioned between a mobile phase

and a stationary phase• Column Chromatography• Gas-liquid Chromatography

ofb chapter 7 lecture notes - georgia institute of...

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