Ch. 13: Chemical Equilibrium

13.4: Heterogeneous Equilibria

Heterogeneous Equilibria

• involve more than one phase

• position of heterogeneous equilibria does NOT depend on amounts of:– pure solids– pure liquids

• because their concentrations stay constant (since they are PURE)

Heterogeneous Equilibria• do not include liquids or solids in equilibrium

expression• only include gases and solutions (aq)

Example 1

• 2H2O(l) ⇄ 2H2(g) + O2(g)

• 2H2O(g) ⇄ 2H2(g) + O2(g)

][][ 22

2 OHK

22

22

2

][

][][

OH

OHK

Ch. 13: Chemical Equilibrium

13.5/6: Applications of Equilibrium Constant (K)

Equilibrium Constant

• if we know the value of K, we can predict:– tendency of a reaction to occur– if a set of concentrations could be at equilibrium– equilibrium position, given initial concentrations

Equilibrium Constant

• If you start a reaction with only reactants:– concentration of reactants will decrease by a

certain amount– concentration of products will increase by a same

amount

Example 2

• The following reaction has a K of 16. You are starting reaction with 9 O3 molecules and 12 CO molecules.

• Find the amount of each species at equilibrium.

O3(g) + CO(g) CO2(g) + O2(g)

Example 2

16NN

NN

][CO][O]][CO[O

KCOO

COO

3

22

3

22

OO33(g) + CO(g) (g) + CO(g) O O22(g) + (g) + COCO22(g)(g)

InitialInitial II

ChangeChange CC

EquilibriumEquilibrium EE

Example 2

16]][[

]][[

3

22

3

22

COO

COO

NN

NN

COO

COOK

OO33(g) + CO(g) (g) + CO(g) O O22(g) + (g) + COCO22(g)(g)

InitialInitial II 99 1212 00 00

ChangeChange CC -x-x -x-x +x+x +x+x

EquilibriumEquilibrium EE 9-x9-x 12-x12-x xx xx

Example 2

1621108

16)12()9(

)()(2

2

xx

x

xx

xxK

172833616 22 xxx

1728336150 2 xx

)5761125(30 2 xx)8)(725(30 xx

14.4 OR 8x

Example 2

OO33(g) + CO(g) (g) + CO(g) O O22(g) + (g) + COCO22(g)(g)

II 99 1212 00 00

CC -x-x -x-x +x+x +x+x

EE 9-(8) = 19-(8) = 1 12-(8) 12-(8) =4=4

x = 8x = 8 x = 8x = 8

Extent of a Reaction• If K >>1– mostly products– goes essentially to

completion– lies far to right

• If K<< 1:– mostly reactants– reaction is negligible– lies far to left

• size of K and time needed to reach equilibrium are NOT related

• time required is determined by reaction rate (Ea)

Reaction Quotient• Q: equal to equilibrium expression but does not

have to be at equilibrium• used to tell if a reaction is at equilibrium or not• relationship between Q and K tells which way the

reaction will shift– Q=K: at equilibrium, no shift– Q > K: too large, forms reactants, shift to left– Q < K: too small, forms products, shift to right

Example 3

• For the synthesis of ammonia at 500°C, the equilibrium constant is 6.0 x 10-2. Predict the direction the system will shift to reach equilibrium in the following case:

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

23

22

23 100.6

]][[

][ HN

NHK

Example 3

• [NH3]0 = 1.0x10-3 M,

[N2]0=1.0x10-5 M

[H2]0=2.0x10-3 M

Q > K so forms reactants, shifts to left

7335

23

103.1]100.2][100.1[

]100.1[

Q

Example 4

• In the gas phase, dinitrogen tetroxide decomposes to gaseous nitrogen dioxide:

N2O4(g) ⇄ 2NO2(g)• Consider an experiment in which gaseous N2O4 was

placed in a flask and allowed to reach equilibrium at a T where KP = 0.133. At equilibrium, the pressure of N2O4 was found to be 2.71 atm.

• Calculate the equilibrium pressure of NO2.

Example 4

133.042

2

2

ON

NOP P

PK

600.0360.0

360.0)71.2)(133.0(

2

422

2

NO

ONPNO

P

PKP

Example 5• At a certain temperature a 1.00 L flask initially

contained 0.298 mol PCl3(g) and 8.70x10-3 mol PCl5(g). After the system had reached equilibrium, 2.00x10-3 mol Cl2(g) was found in the flask.

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

• Calculate the equilibrium concentrations of all the species and the value of K.

Example 5 PClPCl55(g) (g) PCl PCl33(g) + Cl(g) + Cl22(g)(g)

II 8.70x108.70x10-3-3 0.2980.298 00

CC -x-x +x+x +x+x

EE 8.70x108.70x10-3-3-x =-x =

(8.70-2.00) x10(8.70-2.00) x10-3-3 ==

6.70x106.70x10-3-3

0.298+0.298+x =x =

0.298+2.00x100.298+2.00x10-3 -3

==

0.3000.300

xx = = 2.00x102.00x10--

33

23-

-3

1096.86.70x10

)0x100.300)(2.0( K

Approximations

• If K is very small, we can assume that the change (x) is going to be negligible

• can be used to cancel out when adding or subtracting from a “normal” sized number

• to simplify algebra

32

2

2

2

4)0.1(

)2)((

)20.1(

)2)((K x

xx

x

xx

0

Example 6

• At 35°C, K=1.6x10-5 for the reaction2NOCl(g) ⇄ 2NO(g) + Cl2(g)

• Calculate the concentration of all species at equilibrium for the following mixtures

– 2.0 mol NOCl in 2.0 L flask– 1.0 mol NOCl and 1.0 mol NO in 1.0 L flask– 2.0 mol NOCl and 1.0 mol Cl2 in 1.0 L flask

Example 6• 2.0 mol NOCl in 2.0 L flask

• [NOCl]=1.0 - (2 x 0.016)=0.97 M = 1.0 M, [NO]=0.032 M, [Cl2]=0.016 M

2NOCl(g) 2NOCl(g) 2NO(g) + Cl 2NO(g) + Cl22(g)(g)

II 1.01.0 00 00

CC -2x-2x +2x+2x +x+x

EE 1.0-2x1.0-2x 2x2x xx

016.0,4)0.1(

)2)((

)20.1(

)2)((1.6x10 3

2

2

2

25-

xx

xx

x

xx

Example 6• 1.0 mol NOCl and 1.0 mol NO in 1.0 L flask

• [NOCl]=1.0 - (2x1.6x10-5)=0.999968 M = 1.0 M, [NO]=1.0 +(2x1.6x10-5)= 1.0 M, [Cl2]=1.6x10-5M

2NOCl(g) 2NOCl(g) 2NO(g) + Cl 2NO(g) + Cl22(g)(g)

II 1.01.0 1.01.0 00

CC -2x-2x +2x+2x +x+x

EE 1.0-2x1.0-2x 1.0+2x1.0+2x xx

xx

x

xx

2

2

2

25-

)0.1(

)0.1)((

)20.1(

)20.1)((1.6x10

Example 6• 2.0 mol NOCl and 1.0 mol Cl2 in 1.0 L flask

• [NOCl]=2.0 - (2x4.0x10-3)=1.992 M = 2.0 M, [Cl2]=1.0+4.0x10-3=1.0 M, [NO]=0.0080 M,

2NOCl(g) 2NOCl(g) 2NO(g) + Cl 2NO(g) + Cl22(g)(g)

II 2.02.0 00 1.01.0

CC -2x-2x +2x+2x +x+x

EE 2.0-2x2.0-2x 2x2x 1.0+x1.0+x

0040.0,)0.2(

)2)(0.1(

)20.2(

)2)(0.1(1.6x10 2

2

2

2

25-

xxx

x

xx