Chapter 15 Chemical Equilibrium. Copyright McGraw-Hill 2009 Double arrows () denote an equilibrium...

Chapter 15

Chemical Equilibrium

Copyright McGraw-Hill 2009

Double arrows ( ) denote an equilibrium reaction.

15.1 The Concept of EquilibriumMost chemical reactions are reversible.

reversible reaction = a reaction that proceeds simultaneously in both directions

Examples:

)(NO 2 )(ON 242 gg

)(NH 2 )(H 3 )(N 322 ggg

)OH(CH )(H 2 )CO( 32 ggg


Equilibrium)(NO 2 )(ON 242 gg Consider the reaction

At equilibrium,

the forward reaction: N2O4(g) 2 NO2(g), and

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

proceed at equal rates.

Chemical equilibria are dynamic, not static – the reactions do not stop.


EquilibriumLet’s use 2 experiments to study the reaction

each starting with a different reactant(s).

)(NO 2 )(ON 242 gg

Exp #2

pure NO2

Exp #1

pure N2O4


EquilibriumExperiment #1 )(NO 2 )(ON 242 gg


EquilibriumExperiment #2 )(NO 2 )(ON 242 gg


Equilibrium)(NO 2 )(ON 242 gg

Are the equilibrium pressures of NO2 and N2O4

related? Are they predictable?


15.2 The Equilibrium Constant

rate forward ratereverse

At equilibrium,

2f 2 4 eq r 2 eq[N O ] [NO ]k k

)(NO 2 )(ON 242 gg

or2

2 eqfc

r 2 4 eq

[NO ]

[N O ]

kK

k

where Kc is the equilibrium constant


The Equilibrium Constant

This constant value is termed the equilibrium constant, Kc, for this reaction at 25°C.


The Equilibrium ConstantFor the NO2 / N2O4 system:

equilibrium constant expression

equilibrium constant

2

4

[NO ]

O ]K

2

2

0.143 at 25 C[N

)(NO 2 )(ON 242 gg

Note: at 100°C, K = 6.45


The Equilibrium Constantreaction quotient = Qc = the value of the

“equilibrium constant expression” under any conditions.

)(COCl )(Cl )CO( 22 ggg For,

2 eqc

eq 2 eq

[COCl ]

[CO] [Cl ]K 2

c2

[COCl ]

[CO][Cl ]Q

Q > K reverse reaction favored

Q = K equilibrium present

Q < K forward reaction favored


The Equilibrium Constant

The Law of Mass Action:Cato Maximilian Guldberg & Peter Waage, Forhandlinger: Videnskabs-Selskabet i Christiana 1864, 35.

K c d

c a b

[C] [D]

[A] [B]

For a reaction: D C B A dcba

C D

A BP

P PK

P P

c d

a b For gases:

For solutions: [ ] = mol/L

P in atm


The Equilibrium ConstantNote:

• The equilibrium constant expression has products in the numerator, reactants in the denominator.

• Reaction coefficients become exponents.

• Equilibrium constants are temperature dependent.

• Equilibrium constants do not have units. (pg. 622)

• If K >>> 1, products favored (reaction goes nearly to completion).

• If K <<< 1, reactants favored (reaction hardly proceeds).


15.3 Equilibrium Expressionshomogeneous equilibria = equilibria in which all

reactants and products are in the same phase.

• [CaO] and [CaCO3] are solids.

• Pure solids and liquids are omitted from equilibrium constant expressions.

)(CO )CaO( )(CaCO 23 gss Ex:

The equilibrium constant expression is,

K = [CO2]

heterogeneous equilibria = equilibria in which all reactants and products are not in the same phase.


Exercise: Write the expressions for Kp for the

following reactions:)O(H 2 )O(N )(NONH (a) 2234 ggs

)(CuCl )(Cl )Cu( (b) 22 sgs

Solution:

2 2

2

N O H OPK P P (a)

2Cl

1PK P

(b)


Equilibrium ExpressionsA. Reverse Equations

[1] )(NO 2 )(ON 242 gg For,

Conclusion:

C 1 6.99 at 25

0.143

C

PK

P 2

2 4

NO1

N O

2

0.143 at 25

[2] )(ON )(NO 2 422 gg For,

2 4

2

N O

NO

PK

P2 2


Equilibrium ExpressionsB. Coefficient Changes

[1] )(NO 2 )(ON 242 gg For,

Conclusion:

C 0.143 0.378 at 25

For, [3] )(NO )(ON 21

242 gg

2

2 4

NO1/2

N O

PK

P3

C

PK

P 2

2 4

NO1

N O

2

0.143 at 25


Equilibrium ExpressionsC. Reaction Sum (related to Hess’ Law)

[1] )(NO 2 )(ON 242 gg For,

For, [4] )(O )NO( 2 )(NO 2 22 ggg

2

2

NO O

NO

P PK

P

2

4 2

Add [1] + [4],

[5] )(O )NO( 2 )(ON 242 ggg

2

2 4

NO O

N O

P PK

P

2

5

42

2

ON

NO1 P

P K

2

K K 1 4


Equilibrium Expressions


Exercise: At 500ºC, KP = 2.5 1010 for,

)(SO 2 )(O )(SO 2 322 ggg

Compute KP for each of the following:

(a) At 500ºC, which is more stable, SO2 or SO3?

(g)O 21

(g)SO (g)SO (d) 223

(g)SO (g)O (g)SO (b) 322

21

(g)SO 3 (g)O (g)SO 3 (c) 322 23


15.4 Using Equilibrium Expressions to Solve Problems

Q > K reverse reaction favored

Q = K equilibrium present

Q < K forward reaction favored

Predicting the direction of a reaction

Compare the computed value of Q to K


Exercise #1: At 448°C, K = 51 for the reaction,

Predict the direction the reaction will proceed, if at

448°C the pressures of HI, H2, and I2 are 1.3, 2.1

and 1.7 atm, respectively.Solution:

2 2

2HI

H I

P

QP P

0.47 )7.1()1.2(

)3.1(

2

0.47 < 51 system not at equilibrium

Numerator must increase and denominator must decrease.

Consequently the reaction must shift to the right.


Exercise #2: At 1130°C, K = 2.59 102 for

At equilibrium, PH2S = 0.557 atm and PH2 = 0.173 atm,

calculate PS2 at 1130°C.

Solution:

2 2

2

2

H S 22

H S

2.59 10P P

KP

PS2 = 0.268 atm

2

2S 2

2

(0.173) 2.59 10

(0.557)

P


Exercise #3: K = 82.2 at 25°C for,

Initially, PI2 = PCl2

= 2.00 atm and PICl = 0.00 atm.

What are the equilibrium pressures of I2, Cl2, and ICl?Solution:

Initial 2.00 atm 2.00 atm 0.00 atmChange x x +2xEquilibrium (2.00 – x) (2.00 – x) 2x

2 2

2ICl

I Cl

P

KP P

perfect square

2(2 ) 82.2

(2.00 )(2.00 )xx x


(2 ) 9.066

(2.00 )xx

square root

2 x = 18.132 – 9.066 x

11.066 x = 18.132

x = 18.132 / 11.066 = 1.639

PI2 = PCl2

= 2.00 – x = 2.00 – 1.639 = 0.36 atm

PICl = 2x = (2)(1.639) = 3.28 atm

2(2 ) 82.2

(2.00 )(2.00 )xx x

Exercise #3: (cont.)


Exercise #4: At 1280°C, Kc = 1.1 103 for

Initially, [Br2] = 6.3 102 M and [Br] = 1.2 102

M. What are the equilibrium concentrations of Br2

and Br at 1280°C?

Initial 6.3 102 M 1.2 102 MChange -x +2x

Equilibrium (6.3 102) - x (1.2 102) + 2x

Solution:

2 2 23

c 22

[Br] [(1.2 10 ) 2 ] 1.1 10

[Br ] (6.3 10 )

xK

x

4 x2 + 0.0491x + (7.47 105) = 0


4 x2 + 0.0491x + (7.47 10-5) = 0

quadratic equation: a x2 + b x + c = 02 4

2

b b acx

a solution:

x = 1.779 103 and 1.050 102 Q: Two answers? Both negative? What’s happening?

Equilibrium Conc. x = 1.779 103 1.050

102

[Br2] = (6.3 102) – x = 0.0648 M 0.0735

M

[Br] = (1.2 102) + 2x = 0.00844 M 0.00900 M

impossible[Br2] = 6.5 102 M

[Br] = 8.4 103 M


Exercise #5: A pure NO2 sample reacts at 1000 K,

KP is 158. If at 1000 K the equilibrium partial

pressure of O2 is 0.25 atm, what are the equilibrium

partial pressures of NO and NO2.

)(O )NO( 2 )(NO 2 22 ggg

Solution:

Initial ? 0 atm0 atm

Change Equilibrium

0.25 atm

+0.25+0.50+0.50 atm

0.50

2

2 2

22

NO O

2 2

NO NO

(0.50) (0.25) 158P

P PK

P P

rearrange and solve

PNO2


2

2

NO

(0.50) (0.25) 158

P

2

2P

2

2

NO(0.50) (0.25)

158

P 2

4NO 3.956 10 0.01989


= 3.956 104

PNO2= 0.020 atm

PNO = 0.50 atmsee ICE table


Exercise #6: The total pressure of an equilibrium mixture of N2O4 and NO2 at 25°C is 1.30 atm. For

the reaction:

KP = 0.143 at 25°C. Calculate the equilibrium

partial pressures of N2O4 and NO2.

)(NO 2 )(ON 242 gg

2

2 4

2

NO

N O

0.143P

PK

P

PNO2 + PN2O4

= 1.30 atm

two equations and two unknowns – BINGO!


PNO22 + 0.143 PNO2

0.1859 = 0

PN2O4 = 1.30 atm - PNO2

2

2

2

NO

NO

0.143(1.30 )

P

P

2

2 4

2

NO

N O

0.143P

PK

P PNO2

+ PN2O4 = 1.30 atm


Use the quadratic formula,

PNO2 = +0.366 atm and 0.509 atm

PN2O4 = 1.30 atm - PNO2

= 1.30 0.366 = 0.934 atm

PN2O4 = 0.93 atm


15.5 Factors That Affect Chemical Equilibrium

“If an equilibrium system variable is changed, the equilibrium will shift in the direction (right or left) that tends to reduce the change.”

Example: N2, H2, and NH3 are at equilibrium in a

container at 500°C.

(continued on next 5 slides)

Le Châtelier’s Principle

kJ 92 H )(NH 2 )(H 3 )(N rxn322

ggg


Case I: Change: N2 is added

Shift: ???to the right

Q: Why?

Ans: [N2] has increased. Which direction

will decrease [N2]?

)(NH 2 )(H 3 )(N 322 ggg

N2 decreases

N2 increases

right

left


Case II: Change: compress the system

Shift: ???to the right

Q: Why?

Ans: Total pressure has increased. Which direction will decrease the total pressure? Recall: P n

)(NH 2 )(H 3 )(N 322 ggg

N2 H2

NH3

(4 moles gas) (2 moles gas)

less gasless pressure

more gasmore pressure


Case III: Change: increase the temperature

Shift: ???to the left

Q: Why?

Ans: Temperature has increased. Which direction decreases the

temperature?

Recall, the reaction is exothermic.

Δ 92 kJH rxn

endothermic heat absorbed

right

left

exothermic heat evolved


Case IV: Change: add helium at constant volume

Shift: ???none

Q: Why?

Ans: Helium is not a reactant or product. Adding helium (at constant V) does not change PN2, PH2 or PNH3. Hence the

equilibrium will not shift.


Case V: Change: add helium at constant total pressure

Shift: ???to the left

Q: Why?

Ans: If the total pressure is constant, PN2 +

PH2 + PNH3 must decrease. Which

direction increases this sum?Recall: P n

)(NH 2 )(H 3 )(N 322 ggg (4 moles gas) (2 moles gas)

less gasless pressure

more gasmore pressure


Exercise: Hydrogen (used in ammonia production) is produced by the endothermic reaction,

)(H3 )(CO )(OH )(CH 224 gggg 750C

Ni

Assuming the reaction is initially at equilibrium, indicate the direction of the shift (L, R, none) if

(a) H2O(g) is removed.

(b) The temperature is increased.

(c) The quantity of Ni catalyst is increased.

(d) An inert gas (e.g., He) is added.

(e) H2(g) is removed.

(f) The volume of the container is tripled.

Left

Right

None

None

Right

Right

Chapter 15 Chemical Equilibrium. Copyright McGraw-Hill 2009 Double arrows () denote an equilibrium...

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