Chapter Thirteen - myweb.ntut.edu.twchpro/Chem/Chap13.pdf · Chemical Equilibrium The state where the concentrations of all reactants and products remain constant with time. On the

Chapter Thirteen:

CHEMICALEQUILIBRIUM

p578

Contents p578

13-1 The Equilibrium Condition p579

The violet solution in the center is at 25 ℃ and contains

significant quantities of both pink Co(H2O)62+ and blue

CoCl42-. When the solution is cooled, it turns pink because

the equilibrium is shifted to the left. Heating the solution

favors the blue CoCl42-.

The effect of temperature on the endothermic,

aqueous equilibrium:

Co(H2O)62+ + 4Cl- CoCl4

2- + 6H2O

The concept of chemical equilibrium is analogous to theflow of cars across a bridge connecting two island cities.The traffic flow is the same in both directions. So, theresult is no net change in the car population on the bridge

Figure 13.1

A molecular representation of the reaction 2NO2(g) → N2O4(g) over

time in a closed vessel. Note that the number of NO2 and N2O4 in the

container become constant (c and d) after sufficient time has passed.

p580

Chemical Equilibrium

The state where the concentrations of all reactants

and products remain constant with time.

On the molecular level, there is frantic activity.

Equilibrium is not static, but is a highly dynamic

situation.

p580

Equilibrium is: Macroscopically static.

Microscopically dynamic.

The changes in concentrations with time p580

Figure 13.3

(a) H2O and CO are mixed in equal numbers and begin to

react to form CO2 and H2. After time has passed,

equilibrium is reached (c) and the numbers of reactant

and product molecules then remain constant over time (d).

p581

The system has reached equilibrium.

p581The Changes with Time in the Rates ofForward and Reverse Reactions

The Characteristic of Chemical Equilibrium p582

Changes in Concentration

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

13-2 The Equilibrium Constant p582

Law of mass action

The square brackets indicate the concentrations of the chemical

species at equilibrium and K is a constant called the equilibrium

constant.

j A + k B lC + mD

)(6)(4)(7)(4 2223 gOHgNOgOgNH

P583Ex 13.1 Writing Equilibrium Expressions

Write the equilibrium expression for the following reaction:

solution

Applying the law of mass action gives

72

43

62

42

][O][NHO][H][NO

KCoefficient of O2

Coefficient of H2O

Ex 13.2 Calculating the Values of K

)(H3)(N)(2NH 223 ggg

P583The following equilibrium concentrations were observedfor the Haber process at 127℃:

[NH3] = 3.1 × 10 -2 mole/L; [N2] = 8.5 × 10 -1 mole/L

[H2] = 3.1 × 10 -3 mole/L

a. Calculate the value of K at 127℃ for this reaction.

b. Calculate the value of the equilibrium constant at

127℃ for the reaction.

c. Calculate the value of the equilibrium constant at

127℃ for the reaction given by the equation.

)(NH)(H23

)(N21

322 ggg

solutiona. The balanced equation for the Haber process is

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

b.

c.

p584

We can summarize theseconclusions about the equilibriumexpression as follows:

p584

Ex 13.3 Equilibrium Positions P585

The following results were collected for twoexperiments involving the reaction at 600℃ betweengaseous sulfur dioxide and oxygen to form gaseoussulfur trioxide:Show that the equilibrium constant is the samein both cases.

For Experiment 1,

For Experiment 2,

p586

13-3 Equilibrium Expressions InvolvingPressures

p586

For the ammonia synthesis, the equilibrium expressioncan be written in terms of concentrations, that is,

or in terms o the equilibrium partial pressures ofthe gases, that is,

3

2

))((22

3

HN

NHp PP

pK

)(2)()(2 2 gNOClgClgNO

P587Ex 13.4 Calculating Values of KP

The reaction for the formation of nitrosyl chloride

was studied at 25℃. The pressures at equilibrium

were found to be PNOCl = 1.2 atm; PNO = 5.0 × 10 -2

atm; PCl2= 3.0 × 10 -1 atm. Calculate the value

of KP for this reaction at 25℃.

Solution

3122

2

Cl2

NO

2NOCl 109.1

)100.3()100.5()2.1(

)()(22

PPP

k p

The relationship between K and Kp

p587

When the powers in the numerator is different fromthat in the denominator, and K does not equal Kp.

p587

p588

j A + k B lC + mD

For the general reaction

The relationship between K and Kp isn

p RTKK )(

P588Ex 13.5 Calculating K from KPUsing the value of KP obtained in SampleExercise 13.4, calculate the value of K at 25 ℃for the reaction

)(2)()(2 2 gNOClgClgNO Solution

Δn = 2 - (2 + 1)

Sum of product

Sum of reactant

13-4 Heterogeneous Equilibria

2

12'

C][CO C

K )(CO)CaO()(CaCO 23 gss

][CaCO][CaO][CO

3

2' K

p589

]CO[ 21

'2 KC

KC

P590Ex 13.6 Equilibrium Expressions forHeterogeneous Equilibria

Write the expressions for K and KP for thefollowing process:

a. Solid phosphorus pentachloridedecomposes to liquid phosphorustrichloride and chlorine gas.

b. Deep blue solid copper (II) sulfatepentahydrate is heated to drive off watervapor to from white solid copper (II) sulfate.

p590

13-5 Applications of the Equilibrium Constantp591

p592

The extent of a Reactionp592

The inherent tendency for a reaction to occur is indicated by

the magnitude of the equilibrium constant. A value of K

much than 1 means that equilibrium the reaction system will

consist of mostly products-the equilibrium lies to right.

Another way of saying this is that reactions with very large

equilibrium to completion. On the other hand, a very small

value of K means that the system at equilibrium will consist

of mostly reactants- the equilibrium position is far to the left.

The given reaction does not occur to any significant extent.

Reaction Quotientp593

Q is equal to K. The system is at equilibrium; no shiftwill occur.Q is greater than K. In this case, the ratio of initialconcentrations of products to initial concentrations ofreactants is too large. To reach equilibrium, a netchange of products to reactants must occur. The systemshifts to the left, consuming products and formingreactants, until equilibrium is achieved.Q is less than K. In this case, the ratio of initialconcentrations of products to initial concentrations ofreactants of reactants is too small. The system mustshift to the right, consuming reactants and formingproducts, to attain equilibrium.

P593Ex 13.7 Using the Reaction Quotient

For the synthesis of ammonia at 500℃, theequilibrium constant is 6.0 × 10 -2. Predictthe direction in which the system will shift toreach equilibrium in each of the followingcases:

a. [NH3]o = 1.0 × 10 -5 M;[N2]o = 1.0 × 10 -5 M;

[H2]o = 2.0 × 10 -3 M

b. [NH3]o = 2.00 × 10 -4 M;[N2]o = 1.5 × 10 -5 M;

[H2]o = 3.54 × 10 -1 M

c. [NH3]o = 1.0 × 10 -4 M;[N2]o = 5.0 M;

[H2]o = 1.0 × 10 -2 M

Solution:p594

(a)

Since Q > K the system will shift to left.

(b)

(c)

Since Q < K, so the system will shift to right.

In this case Q = K, so the system is at equilibrium.No shift will occur.

Ex 13.8 Calculating equilibrium Pressures Ip594

P595Ex 13.9 Calculating Equilibrium Pressures II

At a certain temperature a 1.00-L flask initially

contained 0.298 mol PCl3(g) and 8.70 × 10 -3 mol

PCl5(g). After the system had reached equilibrium, 2.00

× 10 -3 mol Cl2(g) was found in the flask. Gaseous PCl5

decomposes according to the reaction

)()()( 235 gClgPClgPCl

Calculate the equilibrium concentrations of all species and

the value of K.

p595

Solution

The equilibrium expression for this reaction is

][PCl]][PCl[Cl

3

32K

p596

)(H)(CO)O(H)CO( 222 gggg

P596Ex 13.10 Calculating EquilibriumConcentrations I

Carbon monoxide reacts with stream to produce

carbon. Dioxide and hydrogen. At 700K the

equilibrium concentrations of all species if 1.000

mol of each component is mixed in a 1000-L flask.Solution

The balanced equation for the reaction is

5.0O][CO][H

]][H[CO

2

22 Kand

p597

p598

Check:

Assume that the reaction for the formation of

gaseous hydrogen and fluorine has an equilibrium

constant of 1.15 × 102 at a certain temperature. In a

particular experiment, 3.000 mol of each component

was added to a 1.500-L flask. Calculate the

equilibrium concentrations of all species.

P598

Ex 13.11 Calculating EquilibriumConcentrations II

Solution: p598

p599

13-6 Solving Equilibrium Problems p600

P601Ex 13.12 Calculating Equilibrium Pressures

Assume that gaseous hydrogen iodide is synthesized

from hydrogen gas and iodine vapor at a temperature

where the equilibrium constant is 1.00 × 102. Suppose

HI at 5.000 × 10 -1 atm, H2 at 1.000 × 10 -2 atm, and I2 at

5.000 × 10 -3 atm are mixed in a 5.000-L flask.

Calculate the equilibrium pressures of all species.

p602Solution:

Le Châtelier’s Principle

If a change is imposed on a system at equilibrium,

the position of the equilibrium will shift in a

direction that tends to reduce that change.

Equilibrium Decomposition of N2O4

Treating Systems That Have Small EquilibriumConstants

)(Cl)(NO2 2 gg )(NOCl2 g

p603

and

The initial concentrations are

0][Cl0[NO]0.50L2.0

mol1.0NOCl][ 0200 M

522

2

106.1[NOCl]

][Cl[NO] K

P604

p604

Table 13.2 The percent by mass of NH3at

equilibrium in a mixture of N2, H2, NH3 as a

function of temperature and total pressure

13-7 LeChâtelier’s Principle

Effects of Changes on the System1. Concentration: The system will shift away from

the added component.

2. Temperature: K will change depending upon thetemperature (treat the energy change as areactant).

3. Pressure:(a) Addition of inert gas does not affect the

equilibrium position.(b) Decreasing the volume shifts the equilibrium

toward the side with fewer moles.

p605

LeChâtelier’s Principle

The Effect of a Change in Concentration p605

As expected, Q is less than K because the concentration ofN2 was increased. The system will shift to come to theequilibrium position. Rather than do the calculations, wesimply summarizes the results.

p605

Figure 13.8

(a) The initial equilibrium mixture of N2, H2, and NH3. (b)The new equilibrium position for the system containingmore N2 (due to addition of N2), less H2, and more than in(a).

P606Ex 13.13 Using Le Chatelier’s Principle IArsenic can be extracted from its ores by first reacting the

ore with oxygen (called roasting) to form solid As4O6,

which is then reduced using carbon: )(6)()(6)( 464 gCOgAssCsOAs

Predict the direction of shift of the equilibrium position

in response to each of the following changes in

conditions. a. Addition of carbon monoxide

b. Addition or removal of carbon ortetraarsenic hexoxide (As4O6)

c. Removal of gaseous arsenic (As4)

Solution:p607

The effect of a Change in Pressurep607

The volume of the container is changed p607

The volume of a gas is directly proportional tothe number of moles of gas present.

(a) (b) (c)

p608

)(4)(6)( 324 lPClgClsP

)()()( 523 gPClgClgPCl

P608Ex 13.14 Using Le Chatelier’s Principle II

Predict the shift in equilibrium position that will occur for

each of the following processes when the volume is reduced.

(a) The preparation of liquid phosphorus trichloride by the

reaction.

(b) The preparation of gaseous phosphorus pentachloride

according to the equation.

(c) The reaction of phosphorus trichloride with ammonia:

)(3)()()(3)( 3233 gHClgNHPgNHgPCl

Solution: p609

The Effect of a Change in Temperature p609

Shifting the N2O4(g) → 2NO2(g) equilibrium by changingthe temperature. (a) At 100 ℃ the flask is definitely red-dishbrown due to a large amount of NO2 present. (b) at 0℃ theequilibrium is shifted toward colorless N2O4(g).

p610

For an endothermic reaction:

As T Cause the equilibrium to shift to theright and the value of K to increase.

p610

)(2)()( 22 gNOgOgN

P610Ex 13.16 Using Le Chatelier’s Principle II

For each of the following reactions, predict how the value

of K changes as the temperature is increased.

a. ΔHo = 181 KJ

)(2)()(2 322 gSOgOgSO b. ΔHo = - 198 kJ

Solution: