ch. 13: chemical equilibrium
DESCRIPTION
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). - PowerPoint PPT PresentationTRANSCRIPT
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