ofb chapter 7 chemical equilibrium

04/19/23 OFB Chapter 7 1

OFB Chapter 7Chemical Equilibrium

7-1 Chemical Reactions in Equilibrium

7-2 Calculating Equilibrium Constants

7-3 The Reaction Quotient

7-4 Calculation of Gas-Phase Equilibrium

7-5 The effect of External Stresses on Equilibria:

Le Châtelier’s Principle

7-6 Heterogeneous Equilibrium

7-7 Extraction and Separation Processes


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04/19/23 OFB Chapter 7 3~220 K (-53oC)

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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 0.02772

25.0 0.03126

30.0 0.04187

50.0 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)

TT

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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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)


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)


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 =


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)




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)


Relationships Among the K’s of Related Reactions

Rule 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


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 = K1

½ = K1


2 H2(g) + O2(g) ↔2 H2O(g) Rxn 1#1

#3


Rule 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 = K1 = 0.45 @ 25oC

Br2(g) + I2(g) ↔ 2 IBr(g) (PIBr)2

(PBr2) (PI2

) = K2 = 0.051 @ 25oC

= K1K2


= (0.45)(0.051)

=0.023 @ 25oC


7-3 The Reaction Quotient

Note that K (the Equilibrium Constant) uses equilibrium partial pressures

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

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QPP

PPb

Ba

A

dD

cC

KPP

PPb

Ba

A

dD

cC


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.

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QPP

PPb

Ba

A

dD

cC


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

P

P

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


7-4 Calculations of Gas-Phase Equilibria

Exercise 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.


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.


• Solving quadratic equations


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…


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)


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


Volume Decreased Volume Increased

(Pressure Increased) (Pressure Decreased)

V reactants > V products

Equilibrium shift right (toward products)

Equilibrium Shifts left (toward reactants)

V reactants < V 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)


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


Let’s decrease the temperature of the reaction

Let’s increase the temperature of the reaction


Temperature Raised Temperature Lowered

Endothermic Reaction (absorb heat)



Exothermic Reaction (liberate heat)



If a forward reaction is exothermic,

Then the reverse reaction must be endothermic


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


Driving Reactions to Completion

Volume Decreased Volume Increased

(Pressure Increased) (Pressure Decreased)

V reactants > V products



Temperature Raised Temperature Lowered

Exothermic Reaction (liberate heat)



Industrial Synthesis of Ammonia

N2 (g) + 3H2 (g) ↔ 2NH3 (g)

Forward reaction exothermic


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QPP

PPb

Ba

A

dD

cC

KPP

PPb

Ba

A

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


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?


Heterogeneous Equilibrium

Solids

Liquids

Dissolved species

H2O(l) ↔ H2O(g)

CaCO3(s) ↔ CaO(s) + CO2(g)

I2(s) ↔ I2(aq)


Law of Mass Action

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

IaqI

sI

aqIK

aqIsI


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


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)


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.


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))(P

P

[Ni(s)])(P

PK

4CO

Ni(CO)

4CO

Ni(CO) 44

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

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


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

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