# Chem 106 Chapter 16: Principles of Reactivity: Chemical Equilibria by unm8F6

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```									Chapter 16 Principles of Reactivity: Chemical Equilibria

1
I. Equilibrium
A. Equilibrium expressions
B. Manipulating equilibrium expressions
C. Magnitude of K
D. The reaction quotient
II. Applications of Equilibrium Constants
A. Calculating equilibrium constants
B. Calculating equilibrium concentrations
III. Le Châtelier's Principle
A. Change in reactant or product quantity
B. Change in pressure or volume
C. Change in temperature
D. Effect of catalysts

2
I. Equilibrium

e.g., H2O + CO2    H2CO3
Dynamic equilibrium

Rateforward = Ratereverse

[H2O]

conc.
[CO2]

[H2CO3]

time
3
I. Equilibrium
General: A            B
[B]
Equilibrium lies on side of product.
Rateforward = Ratereverse
conc.

kf[A] = kr[B]
kf [B]
[A]           1      kf > kr
k r [ A]
time

[A]
Equilibrium lies on side of reactant.
Rateforward = Ratereverse
conc.

kf[A] = kr[B]
kf [B]
[B]           1      kf < kr
k r [ A]
time                                                     4
I. Equilibrium
A. Equilibrium expressions

For the reaction, aA + bB         cC + dD

At equilibrium:   Rateforward = Ratereverse

kf[A]a[B]b = kr[C]c[D]d

kf        [C] c [D]d
 Kc                     equilibrium expression
kr        [A]a [B] b
concentration
equilibrium constant
(at a given temperature)

5
I. Equilibrium
A. Equilibrium expressions

Also, for the reaction, aA + bB     cC + dD
c d
pC p D
Kp  a b
pA pB
pressure

e.g., 2N2O5(g)       4 NO2(g) + O2(g)

What are Kc and Kp?

6
I. Equilibrium
A. Equilibrium expressions

Heterogeneous equilibria:

1. insoluble solids are not included in Kc or Kp

e.g., AgCl(s) + 2NH3(aq)               Ag(NH3)2+(aq) + Cl–(aq)

[Ag(NH 3 )  ][Cl  ]
K c'                 2
[AgCl ( s )][ NH 3 ]2
constant ( density)

[Ag(NH 3 )  ][Cl  ]
 K c' [AgCl ( s )]    Kc             2
[ NH 3 ]2

7
I. Equilibrium
A. Equilibrium expressions

Heterogeneous equilibria:

2. solids, liquids are not included in Kp

e.g., H2O(g) + SO3(g)             H2SO4(l)
pH 2SO 4
'
Kp                            constant (pvap at T)
pH 2O pSO 3

'
Kp                  1
               Kp 
pH 2SO 4          pH 2O pSO 3

8
I. Equilibrium
A. Equilibrium expressions
e.g., Write the equilibrium expression for each of the following reactions.

Cu(OH)2(s)         Cu2+(aq) + 2 OH–(aq)      (Kc)

Ni(s) + 2Ag+(aq)        Ni2+(aq) + 2 Ag(s)     (Kc)

2 NH3(g) + CO2(g) + H2O(g)          (NH4)2CO3(s)       (Kp)

9
I. Equilibrium
B. Manipulating equilibrium expressions
1. reversing the equation
[HCl]2
H2 + Cl2        2HCl        Kc =
[H2][Cl2]

[H2][Cl2]
2HCl          H2 + Cl2      Kc´ =
[HCl]2

1
 Kc´ =
Kc

When reversing a chemical equation, take the inverse
of the equilibrium constant.

10
I. Equilibrium
B. Manipulating equilibrium expressions
2. multiplying equations by some factor
[HCl]4
2H2 + 2Cl2        4HCl     Kc´´ =
[H2]2[Cl2]2

 Kc´´ = (Kc)2

[HCl]
½H2 + ½Cl2         HCl     Kc´´´ =
[H2]½[Cl2]½

 Kc´´´ = (Kc)½

When multiplying a chemical equation by some factor,
raise the equilibrium constant to that power.

11
I. Equilibrium
B. Manipulating equilibrium expressions

[NO][SO3]
SO2 + NO2           NO + SO3             Kc1 = [SO ][NO ]
2     2
[H2SO4]
SO3 + H2O           H2SO4                Kc2 =
[SO3][H2O]

[NO][H2SO4]
SO2 + NO2 + H2O             NO + H2SO4   Kc3 =
[SO2][NO2][H2O]

[NO][SO3]    [H2SO4]    [NO][H2SO4]
Kc3 = Kc1 · Kc2 =             ·          =
[SO2][NO2] [SO3][H2O] [SO2][NO2][H2O]

When adding equations, multiply the equilibrium constants.

12
I. Equilibrium
B. Manipulating equilibrium expressions
e.g., Given that for the reaction, N2(g) + 3H2(g)       2NH3(g), Kc = 3.5 x 108,
what is Kc for the reaction NH3(g)        1/2 N2(g) + 3/2 H2(g)?

e.g., Given the following:
H2O(g) + CO(g)         H2(g) + CO2(g) Kp = 1.6
FeO(s) + CO(g)         Fe(s) + CO2(g) Kp = 0.67
What is Kp for the reaction Fe(s) + H2O(g)    FeO(s) + H2(g)

13
I. Equilibrium
C. Magnitude of K

2SO2 + O2         2SO3     K = 3 x 1024

Equilibrium lies very far to the right; reaction “goes to completion.”

2HF          H 2 + F2      K = 1 x 10–13

Equilibrium lies very far to the left; reaction “doesn’t go.”

OH
H2O + H2C=O              H C OH            K1
H
“Equilibrium mixture” of reactant and product.

14
I. Equilibrium
D. The reaction quotient, Q

For aA + bB         cC + dD

[C]c [D]d
Qc                 not necessarily at equilibrium
[A]a [B]b

If Q > K,   reaction shifts left to achieve equilibrium

If Q < K,   reaction shifts right to achieve equilibrium

If Q = K,   reaction is at equilibrium

15
I. Equilibrium
D. The reaction quotient, Q

e.g., N2 + O2       2NO      Kc = 4.0 x 10–4 at 2000 K

If [N2] = 0.50 M, [O2] = 0.25 M, and [NO] = 4.2 x 10–3 M,

Is the system at equilibrium? Which way will it shift?

16
II. Applications of Equilibrium Constants
A. Calculating equilibrium constants

e.g., 2SO2 + O2      2SO3
1.00 mol of SO2 and 1.00 mol of O2 were placed in a 1.00-L flask at 1000ºC.
When equilibrium was attained, it was found that the concentration of SO3
was 0.925 M. What is Kc?

[SO2]       [O2]           [SO3]
[ ]i
D[ ]
[ ]eq

17
II. Applications of Equilibrium Constants
A. Calculating equilibrium constants

e.g., CO + Cl2       COCl2
A mixture of 0.102 M CO and 0.00609 M Cl2 was placed in a flask. When
the reaction had come to equilibrium at 325 °C, the concentration of Cl2 was
found to be 0.00301 M. What is Kc for the reaction?

18
II. Applications of Equilibrium Constants
B. Calculating equilibrium concentrations
1. SO2 + NO2        NO + SO3         Kc = 85.0
0.50 mol of SO2 and 0.50 mol of NO2 are placed in a 10.0-L flask and
allowed to come to equilibrium. What are the equilibrium concentrations
of all species?

[SO2]    [NO2]      [NO]    [SO3]
[ ]i
D[ ]
[ ]eq

19
II. Applications of Equilibrium Constants
B. Calculating equilibrium concentrations
2. SO2 + NO2        NO + SO3         Kc = 85.0
Same as problem 1, but this time start with products: 0.050 M NO and
0.050 M SO3. What are the equilibrium concentrations of all species?

[SO2] [NO2] [NO]      [SO3]
[ ]i
D[ ]
[ ]eq

20
II. Applications of Equilibrium Constants
B. Calculating equilibrium concentrations
3. COCl2       CO + Cl2         Kc = 2.2 x 10-10
When 0.20 mol of COCl2 is placed in a 10.0-L flask, what are the
equilibrium concentrations of all species?
[COCl2]   [CO]    [Cl2]
[ ]i
D[ ]

21
II. Applications of Equilibrium Constants
B. Calculating equilibrium concentrations
4. 2CO2        2CO + O2             Kc = 6.4 x 10-7 at 2000ºC
Starting with 0.010 M CO2, what are the [ ]eq’s of all species?
[CO2]    [CO]     [O2]
[ ]i
D[ ]

22
II. Applications of Equilibrium Constants
B. Calculating equilibrium concentrations
5. CO + H2    H2CO      Kc = 4.6 x 109
0.15 0.20 M
[CO]   [H2]    [H2CO]
[ ]i
D[ ]

23
II. Applications of Equilibrium Constants
B. Calculating equilibrium concentrations
5. (cont.) CO + H2       H2CO   Kc = 4.6 x 109
0.15 0.20 M

[CO]   [H2]     [H2CO]
[ ]i
[ ]f
D[ ]

24
II. Applications of Equilibrium Constants
B. Calculating equilibrium concentrations
6. 2SO2 + O2   2SO3         Kp = 1.1 x 1012
1.0 1.0 atm
Carry reaction to completion:

pSO2     pO2      pSO3
pi
pf
Dp

25
II. Applications of Equilibrium Constants
B. Calculating equilibrium concentrations
e.g., For the decomposition of gaseous water at 500ºC,
2H2O(g)        2H2(g) + O2(g), Kc = 6.0 x 10–28.
If 0.20 mol of H2O is placed in a 5.0-L container and allowed to come
to equilibrium at 500ºC, what will be the concentrations of all species?

26
II. Applications of Equilibrium Constants
B. Calculating equilibrium concentrations
e.g., For the reaction, CO(g) + Cl2(g)      COCl2(g), Kc = 4.6 x 109. If
0.15 mol of CO and 0.30 mol of Cl2 are placed in a 1.0-L flask and allowed
to react, what would be the concentration of each species at equilibrium?

27
III. Le Châtelier’s Principle

“Any change imposed on a system at equilibrium will cause the system
to shift in such a way to re-establish equilibrium.”

Reactants             Products

28
III. Le Châtelier’s Principle
A. Change in quantity of reactant or product

A        B at equilibrium

increase [B]
[B]
K=             reaction shifts to left
[A]

 A reaction will shift: away from a substance added,
toward a substance removed.

e.g., 3H2(g) + N2(g)              2NH3(g)
2
pNH3                   Increase p H 2 : reaction shifts ______
Kp     3
pH 2 p N 2              Increase p NH3 : reaction shifts ______
Decrease p N 2 : reaction shifts ______

29
III. Le Châtelier’s Principle
A. Change in quantity of reactant or product
CH3
e.g., CH3   CH2    CH2     CH3         CH3    CH CH3        Kc = 2.5
butane (B)                   isobutane (I)
At equilibrium, [B] = 0.20 M and [I] = 0.50 M. If 0.10 mol of butane is added
and the system returns to equilibrium,
a. Which way will the reaction shift to re-establish equilibrium?
b. What will be the new [ ]eq’s?
c. Is there be more or less butane after the reaction reaches equilibrium?

30
III. Le Châtelier’s Principle
B. Change in pressure or volume
pC
A(g) + B(g)       C(g)     Kp = p p
A B

 Decrease V:     reaction shifts to side with
(increase PT)    fewer moles of gas
Increase V:    reaction shifts to side with
(decrease PT)   more moles of gas

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

Increase V: reaction shifts ________
Increase PT: reaction shifts ________

31
III. Le Châtelier’s Principle
C. Change in temperature

2NO2(g)       N2O4(g) + heat       DH = -57.2 kJ/mol
“product”

 DH < 0: increasing T shifts equilibrium left (K gets smaller)
decreasing T shifts equilibrium right (K gets larger)

DH > 0: opposite effect

e.g., 2NOBr(g)       2NO(g) + Br2(g)     DH = +16.1 kJ/mol

Increase T: reaction shifts ________
Decrease T: reaction shifts ________

32
III. Le Châtelier’s Principle
D. Effect of catalysts
 No effect on position of equilibrium
(neither a reactant nor a product of the reaction)
- increases rate at which equilibrium is established

e.g., 2N2O(g) + O2(g)       4 NO (g)           DH = 199 kJ/mol

Will the amount of NO at equilibrium be greater, less, or the same if we:
•remove O2?      __________
•increase V?     __________
•raise T?        __________
33
III. Le Châtelier’s Principle

e.g., CaCO3(s)     CaO(s) + CO2(g)         DH = 278 kJ/mol

How will the amount of CO2 change if we: