Chemical Equilibrium

Chapter 17Chapter 17

Equilibrium vs. Kinetics

Kinetics:Kinetics: speed of a reaction or processspeed of a reaction or process

how fast?how fast?

Equilibrium:Equilibrium: extent of reaction or processextent of reaction or process

how much?how much?

Chemical Equilibrium

Reactant and product concentrations remain constantReactant and product concentrations remain constant

Molecular level:Molecular level: rapid activity (dynamic)rapid activity (dynamic)

Macroscopic level:Macroscopic level: unchangingunchanging

At equilibrium:At equilibrium: raterateforwardforward = rate= ratereversereverse

Does not limit timeDoes not limit time

Law of Mass Action

If:If:

Then:Then:

And:And:

reverseforward rate rate

nm ]k[products ]reactants[k

K]reactants[k

]k[products

rate

ratem

n

reverse

forward

K = equilibrium constant

m, n = coefficients in balanced equation

Law of Mass Action

For:For:

Equilibrium Equilibrium expression:expression:

mD lC kB jA

kj

ml

c[B][A]

[D][C] K

Value of K Favors

K < 0.01 Reactants

K > 100 Products

0.01 < K < 100 Neither

Direction of Reaction and Q

For:For:

Equilibrium Equilibrium expression:expression:

Reaction Reaction Quotient:Quotient:

mD lC kB jA

kj

ml

c[B][A]

[D][C] K

tat time, [B][A]

[D][C] Q

kj

ml

c

]O[N

][NO K

42

22

c

N2O4(g) 2 NO2(g)

Practice

1.1. Writing expression for KWriting expression for K

2. 2. Q vs. K and reaction directionQ vs. K and reaction direction

3.3. K for a multistep processK for a multistep process

4.4. K for reaction “multiples”K for reaction “multiples”

Write K

The decomposition of dinitrogen pentoxide:

N2O5(g) NO2(g) + O2(g)

The combustion of propane gas:

C3H8(g) + O2(g) CO2(g) + H2O(g)

Write K

The decomposition of dinitrogen pentoxide:

2 N2O5(g) 4 NO2(g) + O2(g)

The combustion of propane gas:

C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(g)

252

24

2c

]O[N

][O][NO K

5283

42

32

c]][OH[C

O][H][CO K

Write K

4 NH3(g) + O2(g) 4 NO(g) + 6 H2O(g)

2 NH3(g) + 5/2 O2(g) 2 NO(g) + 3 H2O

K vs. Q

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

Kc = 0.21 at 1000C.

At a point during the reaction, [N2O4] = 0.12M and [NO2] = 0.55M.

(a) Find Q. Is the reaction at equilibrium?

(b) If not, in which direction is it progressing?

K vs. Q

N2O4(g) 2NO2(g) Kc = 0.21 at 1000C.

At a point, [N2O4] = 0.12M and [NO2] = 0.55M.

(a) Find Q. Is the reaction at equilibrium?

(b) If not, in which direction is it progressing?

5.2(.12)

(0.55)

]O[N

][NO Q

2

42

22

c

left Right toso K Q cc

K for multistep reactions

Nitrogen dioxide, a toxic pollutant that contributes to photochemical smog, can develop in combustion engines from N2 and O2.

(1) N2 + O2 2NO Kc1 = 4.3 x 10-25

(2) 2NO + O2 2NO2 Kc2 = 6.4 x 109

(a) Show that Qc for the overall reaction is the same as the product of Qcs of the individual reactions.

(b) Calculate Kc for the overall reaction.

K for multistep reactions(1) N2 + O2 2NO Kc1 = 4.3 x 10-

25

(2) 2NO + O2 2NO2 Kc2 = 6.4 x 109

N2 + 2 O2 2 NO2

(a) Qc

(b) Kc,overall

222

22

22

22

22

2

c2c1c]][O[N

][NO

][O[NO]

][NO

]][O[N

[NO]QQ Q

15925c2c1c 108.2106.4 103.4K KK

K for multistep reactions

For the following

(1) Br2 2 Br

(2) Br + H2 HBr + H

(3) H + Br HBr

(a) Write the overall balanced reaction.

(b) Write out the individual expressions for Qc and show that their product is equivalent to the overall Qc.

Multiples of K

For the ammonia reaction:

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

Kc is 2.4x10-3 at 1000K.

Find K for the following:

(a) 1/3 N2+ H2 2/3 NH3

(b) NH3 1/2 N2 + 3/2 H2

Multiples of K

N2(g) + 3H2(g) 2NH3(g), Kc = 2.4x10-3

(a) 1/3 N2+ H2 2/3 NH3

(b) NH3 1/2 N2 + 3/2 H2

13.0104.2 KK 31

331

cac,

.20104.2 KK 21

321

cbc,

MultiplesN2(g) + O2(g) 2 NO(g) Kc = 1 x 10-30

Write the expression for Q and determine its value for ½ N2(g) + ½ O2(g) 2 NO(g)

H2(g) + Cl2(g) 2 HCl(g) Kc = 7.6 x 108

Write the expression for Q and determine its value for 2/3 HCl(g) 1/3 H2(g) + 1/3 Cl2(g)

Heterogeneous Equilibrium

PURE solids and liquids do not appear in expression for K (or Q).

2CO K

Kc vs. Kp

For:For: mD lC kB jA n

p K(RT)K

molK

atmL 0.08206 R

gas moles # in change

)nn( n reactantsproducts

Kp and Kc

For the ammonia reaction,

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

Kc = 2.4x10-3

Find Kp at 1000 K.

Kp and Kc

N2(g) + 3H2(g) 2NH3(g), Kc = 2.4x10-3

Find Kp at 1000 K.

2)4()2( n

)nn( n reactantsproducts

7p

23p

2c

Δncp

103.56 K

1000) )(0.08206104.2( K

(RT)K(RT) K K

Kp and Kc

For the following reaction,

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

Kc = 1.67 at 500 K

Find Kp at 500 K.

K vs. Q

For the reaction: CH4(g) + Cl2 CH3Cl(g) + HCl

Kp = 1.6x104 at 1500 K.

At a point during the reaction,

PCH4 = 0.13 atm, PCl2= 0.035 atm,

PCH3Cl = 0.24 atm, and PHCl = 0.47 atm.

(a) Find Q. Is the reaction at equilibrium?

(b) If not, in which direction is it progressing?

Problems

1.1. Given equilibrium concentrations or pressures, Given equilibrium concentrations or pressures, find K or Q.find K or Q.

2. Given K and initial conditions (conc’s or P’s), 2. Given K and initial conditions (conc’s or P’s), find equilibrium quantities (conc’s or P’s).find equilibrium quantities (conc’s or P’s).

Le Châtelier’s Principle

. . . if a change is imposed on a . . . if a change is imposed on a system at equilibrium, the system at equilibrium, the position of the equilibrium will position of the equilibrium will shift in a direction that tends to shift in a direction that tends to reduce that change.reduce that change.

Le Châtelier’s Principle

1.1. ConcentrationConcentration

2.2. TemperatureTemperature

3.3. PressurePressure

4.4. VolumeVolume

5.5. Catalysts*Catalysts*

13_315

H2

NH3N2

Time

Concentration

Equilibrium