Topics in CHM 1046
1. Intermolecular forces (IMF)
Unit 11 Intermolecular Forces, Liquids & Solids (T1)
Unit 12 Properties of Solutions (T1)
2. Themodynamics
Unit 13 Thermochemistry (T1)
Unit 14 Chemical Thermodynamics (T2)
3. Chemical Kinetics Unit 15 Chemical
4. Chemical Equilibrium
Kinetics (T2)
Unit 16 Chem. Equilibrium Gases & Heterogeneous (T3)
Unit 17.Acid Base Equilibria (T3)
Unit 18 Acid Base Equilibria Buffers & Hydrolysis (T3)
Unit 19 Acid Base Equilibria Titrations (T4)
Unit 20 Aqueous Equilibria Solubility Product (T4)Equilibrium
Combination of above:
Unit 21 Electrochemistry (T4)
CHM 1046: General Chemistry and
Qualitative Analysis
Unit 16
Chemical Equilibrium:
Gases & Heterogeneous
Dr. Jorge L. Alonso
Miami-Dade College –
Kendall Campus
Miami, FL
Textbook Reference:
•Chapter # 17
Equilibrium
•Module # 5 (& Appendix
2)
The Concept of Equilibrium
{Non-Equilibrium Reactions}: proceed in one direction, A  B
2 NO (g) + O2 (g)
2 NO2 (g)
(clear gases)
(red-brown gas)
{Equilibrium Reactions}: proceed in both directions, A

N2O4 (g)
2 NO2 (g)
(clear gas)
(red-brown
gas)
↔B
2X
Chemical equilibrium occurs when a
reaction and its reverse reaction proceed
at the same rate.
This does not mean chemicals are found
in same concentration!
Equilibrium
The Equilibrium Constant (Keq)
N2O4 (g)
2 NO2 (g)
• Therefore, at equilibrium
Ratef = Rater
kf [N2O4] = kr [NO2]2
• Rewriting this, it becomes
kf
Keq =
kr
[NO2]2
=
[N2O4]
The Equilibrium Constant is the ratio of the
Equilibrium
rate constants at a particular temperature.
The Equilibrium Constant (K)
• To generalize this expression, consider
the reaction
aA(aq) + bB(aq)
cC(aq) + dD(aq)
• The equilibrium expression for this
reaction would be
[C]c[D]d
Kc =
[A]a[B]b
where [X] = concentration of each chemical
in moles/ liter (M).
Equilibrium
What Does the Value of K Mean?
aA + bB
10 x
[C]c[D]d
Kc = [A]a[B]b
cC + dD
Kc =
10 = 10
1
• If K >> 1, the reaction is product-favored; product
predominates at equilibrium.
aA + bB
[C]c[D]d
Kc = [A]a[B]b
10 x
cC + dD
Kc = 1 = 0.1
10
Equilibrium
• If K << 1, the reaction is reactant-favored; reactant
predominates at equilibrium.
The Equilibrium Constant for Gases
aA(g) + bB(g)
Partial press: (75 torrs) (150 torrs)
cC(g) + dD(g)
[C]c[D]d
Kc =
[A]a[B]b
(300 torrs) (600 torrs) = total (1125 torrs)
Because pressure is proportional to
concentration for gases in a closed
system, the equilibrium expression can
also be written
(PC)c (PD)d
Kp =
(PA)a (PB)b
where Kp = equilibrium constant for gases, and
Px = partial pressure of each gas.
Equilibrium
Relationship between Kc and Kp
• ideal gas law:
PV = nRT
n
P=
RT
V
[C]c[D]d
Kc =
[A]a[B]b
Kp =
(PC)c
(PA)a
(PD)d
(PB)b
(
=
(
n
V
n
V
)c
(
RT
a (
)
RT
n
V
n
V
RT )d
b
)
RT
Plugging this into the expression for Kp
Where
Kp = Kc (RT)n
☺
Equilibrium
n = (moles of gaseous product) − (moles of gaseous reactant)
Equilibrium Can Be Reached from
Either Direction
@ 100°C
KC
[NO2]2
=
[N2O4]
2
2
[0.0172] [0.0243] [0.0310]2 [0.0310]2
Kc 



 0.212
[0.0014] [0.0028]
[0.0045] [0.0045]
The ratio of [NO2]2 to [N2O4] remains constant at this temperature noEquilibrium
matter
what the initial concentrations of NO2 and N2O4 are.
Equilibrium Can Be Reached from
Either Direction
N2O4 (g)
2
2 NO2 (g)
2
2
[ NO2 ] [0.0172 ] [0.0243] [0.0310 ]2 Equilibrium
Kc 



 0.212
[ N 2 O 4 ] [0.0014 ] [0.0028]
[0.0045]
Manipulating Equilibrium Constants
RULE #1: Reciprocal Rule
The equilibrium constant of a reverse reaction is
the reciprocal of the equilibrium constant of the
forward reaction.
N2O4 (g)
2 NO2(g)
2 NO2 (g)
N2O4(g)
[NO2]2
K =
= 0.212 at 100C
[N2O4]
I
[N2O4]
K =
[NO2]2
II
1
=
0.212
Equilibrium
= 4.72 at 100C
Manipulating Equilibrium Constants
RULE #2: Coefficient Rule
If the coefficients of a chemical equation are
changed (increased or decreased) by a factor
n, then the value of K is raised to that power.
N2O4 (g)
2 NO2 (g)
2 N2O4 (g)
4 NO2 (g)
½N2O4(g)
NO2(g)
2
[NO
]
2
KI =
= 0.212 at 100C
[N2O4]
KIII
[NO2]4
2 at 100C
=
=
(0.212)
[N2O4]2
KIV =
Equilibrium
[ NO2 ]2
 0.212 at 100C
[N 2O4 ]
Manipulating Equilibrium Constants
RULE #3: Multiple Equilibria Rule
When two or more equations are added to
obtain a final resultant equation, the equilibrium
constant for the resultant equation is product of the
constants of the added equations.
2 A(aq) + B(aq) ↔ 4 C(aq)
4 C(aq) + E(aq) ↔ 2 F(aq)
2 A(aq) + B(aq) + E ↔ 2 F(aq)
4
[
C
]
KV 
[ A ]2 [ B]
K
K VII
VI
[ F]2

[C]4 [ E]
[ F]2
Equilibrium

[ A ]2 [ B][ E ]
Homogeneous
Reactions involving only gasses or only solutions
[NO2]2
= 0.212 at 100C
N2O4 (g)
2 NO2(g) Kc =
[N2O4]
vs.
Heterogeneous
Reactions involving different phases (s, l, g, aq) of matter
CaCO3 (s)
CO2 (g) + CaO(s)
Equilibria
KC  [CO2 ]
Equilibrium
The Concentrations of Solids and
Liquids Are Essentially Constant
PbCl2 (s)
Pb2+ (aq) + 2 Cl−(aq)
Write the equilibrium expression!
Kc = [Pb2+] [Cl−]2
The concentrations of solids and liquids do not appear in
equilibrium expressions because their concentration does not
change.
Concentration of both solids and liquids can be obtained by
dividing the density of the substance by its molar mass—and
both of these are constants at constant temperature.
g
D
Concentration ( M ) 
 L 
MM MM
g
MM  m ol
L
L
Equilibrium
What are the Equilibrium
Expressions for these
Heterogeneous Equilibria?

P 

2
Kp
CO2
PCO 2
 
K p  PCO2
2
[Zn ]
Kc 
[Cu 2 ]
Equilibrium
CaCO3 (s)
CO2 (g) + CaO(s)
 
K p  PCO2
How does the concentration of CO2 differ in the two bell jars?
Equilibrium
As long as some CaCO3 or CaO remain in the system, the
amount of CO2 above the solid will remain the same.
Equilibrium Calculations
(1)
Calculating Kc from experimental data: [ ]i & [ ]e
(2)
Calculating Equilibrium Concentrations [ ]e
when Kc & [ ]i are known
ICE Tables:
Initial
[A] + [B]
[C] + [D]
[A]i
[B]i
0
0
Change
I
C
change
change
+
change
+
change
Equilibrium
E
[A]e
[B]e
[C]e
[D]e
Kc 
d
[C ]cequi [ D]equi
a
[ A]equi
[ B]bequi

d
[C ]cequi [ D]equi
Equilibrium
a
[[ A]i  change]equi
[[B]i  change]bequi
(1) Calculating Kc from experimental data: [ ]i & [ ]e
[HI]2
H2 (g) + I2 (g)
2 HI (g) Kc =
[H2] [I2]
A closed system initially containing 1.000 x 10−3 M H2 and 2.000 x 10−3 M
I2 at 448C is allowed to reach equilibrium. Analysis of the equilibrium
mixture shows that the concentration of HI is 1.87 x 10−3 M. Calculate Kc at
448C for the reaction taking place, which is
Initially
Change
At Equilibrium
[I2], M 
[H2], M +
1.000 x 10-3 2.000 x 10-3
-9.35 x 10-4
6.5 x 10-5
-9.35 x 10-4
1.065 x 10-3
2 [HI], M
0
+1.87 x 10-3
1.87 x 10-3
[HI] Increases by 1.87 x 10-3 M
Equilibrium
Stoichiometry tells us [H2] and [I2] decrease by half as much
…and, therefore, the equilibrium constant
[I2], M 
[H2], M +
1.000 x 10-3 2.000 x 10-3
-9.35 x 10-4
-9.35 x 10-4
Initially
Change
6.5 x 10-5
At Equilibrium
1.065 x 10-3
[HI], M
0
+1.87 x 10-3
1.87 x 10-3
[HI]2
Kc =
[H2] [I2]
=
(1.87 x 10-3)2
(6.5 x 10-5)(1.065 x 10-3)
Equilibrium
= 51
Calculating Kc from experimental data: [ ]i & [ ]e
@
100°C
+x
+x
+x
-x
N2O4 (g)
- 2x
- 2x
- 2x
+ 2x
x
2 NO2 (g)
I
0.0
0.0200
C
+x
- 2x
E
0.0014
0.0200-2x
[NO2]2
Kc = [N O ]
2 4
([0.02]i  2 x) 2 ([0.03]i  2 x) 2 ([0.04]i  2 x) 2 ([.00]i  2 x) 2
Kc 



 0.212
[0.0014]
[0.0028]
[0.0045]
[0.0045]
Equilibrium
The ratio of [NO2]2 to [N2O4] remains constant at this temperature no matter
what the initial concentrations of NO2 and N2O4 are.
2006B Q5
Practice Problem
{
}
Equilibrium
Equilibrium
Equilibrium
Equilibrium Calculations
(1) Calculating Kc from experimental data: [ ]i & [ ]e
[H2], M
Initially
+
1.000 x 10-3
[I2], M 
2.000 x 10-3
[HI], M
0
Change
1.87 x 10-3
At Equilibrium
(2) Calculating Equilibrium Concentrations
when K & [ ]i are known
H2(g) +
I2(g)
↔ 2HI(g)
I
0.00623
0.00414
0.0224
C
-x
-x
+ 2x
E
0.00632- x 0.00414- x
0.0224 + 2x
K  51 =
[HI]2e
[H 2 ]eEquilibrium
[I 2 ]e
(2) Calculating Equilibrium
Concentrations [ ]e when K & [ ]i are known
aA + bB
I
C
E
K
[A]i
[B]i
0
0
- ax
- bx
+ cx
+ dx
[A]i- ax
[B]i- bx
cx
dx
[C ]cequi [ D]eequi
a
equi
[ A]
cC + dD
b
equi
[ B]
(cx) c  (dx) d

a
b
([A]i  ax)  ([B]i  bx)
Solve for x by using the quadratic equation:
ax  bx  c  0
2
 b  b Equilibrium
 4ac
x 
2a
2
Calculating Equilibrium Concentrations
when K & [ ]i are known
Problem: for the reaction H2(g) + I2(g)
2HI(g) , K= 54.3 @
430ºC. Suppose that the initial concentrations of H2, I2, and HI
are 0.00623 M, 0.00414 M, and 0.0224 M, respectively.
Calculate the concentrations of these species at equilibrium.
H2(g) + I2(g)
2HI(g)
I
0.00623
0.00414
0.0224
C
-x
-x
+ 2x
E
0.00632- x 0.00414- x
K
[HI]2e
[H 2 ]e [I 2 ]e
0.0224 + 2x
ax2  bx  c  0
(0.0224 2 x) 2
54.3 
([0.00632]- x)  ([0.00414]- x)
Deciding if x is negligible:
54.3(2.58 x 1015  0.0104x  x 2 )  5.02 x 10-4  0.0896x  4 x 2
50.3x  0.654x  8.98 x 10  0
2
-4
 b  b 2  4ac
x 
2a
Equilibrium
Problem: for the reaction H2(g) + I2(g)
2HI(g) , K= 54.3 @
430ºC. Suppose that the initial concentrations of H2, I2, and HI
are 0.00623 M, 0.00414 M, and 0.0224 M, respectively.
Calculate the concentrations of these species at equilibrium.
(0.0224 2 x) 2
54.3 
([0.00632]- x)  ([0.00414]- x)
ax2  bx  c  0
50.3x 2  0.654x  8.98 x 10-4  0
YOU MUST USE QUADRATIC!
50.3x 2  8.98 x 10-4

50.3
50.3
x 2  1.79 x 10-5
x  0.00423
Deciding if x is negligible:
[A] i
 K  [A] i x 100
100
0.000224  54.3  2.24
1. Take the largest [A]i and multiply it by
100 and also divide it by 100.
2. Compare the answers of above to
the value of K
3. If comparison are not significant as
compared to K, then x is negligible
and you can delete the x that is
subtracted from [A]I
4. If these calc values are significant
compared to K, then x is notEquilibrium
negligible and you must use the
quadratic to solve for x.
Calculating Equilibrium Concentrations
when K & [ ]i are known
ax2  bx  c  0
50.3x 2  0.654x  8.98 x 10-4  0
 b  b  4ac
x 
2a
2
0.654  (0.654) 2  4(50.3)(8.98 x 10-4 )
x 
2(50.3)
[H2]i = 0.00623
[I2]i = 0.00414
x  0.0114 M
E
or
x  0.00156 M
0.00632- x 0.00414- x
0.0224 + 2x
First answer is physically impossible, since conc. of H2 and I2
would be more than original conc. x calc without quadratic:
x  0.00423 M
[H2]e = 0.00623 – 0.00156) = 0.00467 M
Equilibrium
[I2]e = (0.00414 – 0.00156) = 0.00258 M
[HI]e = (0.0242 + 2 x 0.00156) = 0.0255 M
Equilibrium Calculations
(1) Calculating Kc from experimental data: [ ]i & [ ]e
[H2], M
+
1.000 x 10-3
Initially
[I2], M 
2 [HI], M
2.000 x 10-3
0
Change
1.87 x 10-3
At Equilibrium
[ HI ]2
K
[ H 2 ]e [ I 2 ]e
(2) Calculating Equilibrium Concentrations
when K & [ ]i are known
H2(g) + I2(g)
2HI(g)
[HI]2e
K  51 =
I
0.00623
0.00414
0.0224
e
C
E
-x
-x
+ 2x
0.00632- x 0.00414- x
ax2  bx  c  0 x 
[H 2 ]e [I 2 ]e
0.0224 + 2x
 b  b  4ac
2a
2
Deciding if x is negligible:
[A] i
100
Equilibrium
 K
 [A] i x 100
Le Châtelier’s Principle
“If a system at equilibrium is disturbed
by a change in temperature, pressure, or the
concentration of one of the components, the
system will shift its equilibrium position so
as to counteract the effect of the
disturbance.”
[N2O4]
KC = [NO ]2
2
The effect of (1) Concentration
2 NO2 (g)

(red-brown gas)
{Equi.Intro1}
{Equi.Intro2}
Henri Louis Le Chatelier
1850-1936
N2O4 (g)
(clear gas)
Equilibrium
The effect of (2) Pressure & Volume
{L.C.&PressVol}
(3) Temperature
Le Châtelier’s Principle
The effect of (3) Temperature
2 NO2 (g)
(red-brown gas)

N2O4 (g) + HEAT
(clear gas)
H = -58.0 kJ
Why does heat
favor the formation
of NO2 over N2O4?
{Le Châtelier’s&Temp}
{Molecular Explanation}
Keq at different Temp:
Equilibrium
The Effect of Temperature Changes
Add heat (), which way will the equlibrium be shifted?
Co(H2O)62+(aq) + 4 Cl(aq)
(pink)
+ HEAT

CoCl4 (aq) + 6 H2O (l)
(blue)
H= +

{CoCl}
Equilibrium
Briggs-Rauscher reaction
{ClockReaction}
Equilibrium
IO3− + 2H2O2 + CH2(COOH)2 + H+ → ICH(COOH)2 + 2O2 + 3H2O
Le Châtelier’s Principle & the Haber Process
It was not until the early 20th century that this method was
developed to harness the atmospheric abundance of nitrogen
to create ammonia, which can then be oxidized to make the
nitrates and nitrites essential for the production of fertilizers and
ammunitions.
∆H = -92.4 kJ/n
@ 250 C
Equilibrium
NH3 + HNO3  NH4NO3 (explosives & fertilizers)
Le Châtelier’s Principle & the Haber Process
N2 (g) + 3H2 (g)
Fe
2NH3 (g)
Shift equilibrium to the right
by increasing press & conc. of
N2 + H2 and by removing NH3
from system
This apparatus helps
push the equilibrium to the
right by removing the
ammonia (NH3) from the
system as a liquid
Removed by liquefaction
Fe
Subs
NH3
N2
H2
b. pt.
-33°C
-196°C
Equilibrium
-253°C
The Effect of Catalysts on Equilibrium
How does the
addition of a
catalyst affect
the equilibrium
between subs. A
and B?
catalyst
• Increase the rate of both the forward and reverse reactions.
• Equilibrium is achieved faster, but the equilibrium Equilibrium
composition remains unaltered.
Equilibrium Constant (K)
vs. Reaction Quotient (Q)
aA(aq) + bB(aq)
At Equilibrium:
K=
[C]c[D]d
[A]a[B]b
cC(aq) + dD(aq)
At Non-Equilibrium:
[C]ic[D]id
Q=
[A]ia[B]ib
• To calculate Q, one substitutes the initial
concentrations on reactants and products into the
equilibrium expression.
• Q gives the same ratio the equilibrium expression
gives, but for a system that is not at equilibrium.
Equilibrium
aA(aq) + bB(aq)
[C]c[D]d
K=
[A]a[B]b
cC(aq) + dD(aq)
[C] c[D] d
Q=
[A] a[B] b
i
i
i
i
If Q = K, the system is at equilibrium.
If Q < K? If Q < K, there is too much reactant, and the
equilibrium shifts to the right.
If Q > K? If Q > K, there is too much product and the
equilibrium shifts to the left.
Equilibrium
Kc, Q and Spontaneity (ΔG)
[NH3]2eq
KC =
[N2]eq[H2]eq
2
[NH
]
3 i
Q=
[N2]i[H2]i
N2(g) + 3H2(g)
2NH3(g)
ΔG= Reverse Process
ΔG = +
ΔG= Reverse Process
ΔG = +
@ Equilibrium
K = Q, ΔG = 0
Equilibrium
Free Energy (G) & Reaction Quotient
(Q) under non-Standard Conditions
☺ G = G + RT ln Q = G + RT 2.303 log Q
Non-standard
conditions
Non-standard conditions
G = - (spontaneity), implies COMPLETE conversion of all reactants to →
products @ standard conditions (25ºC, 1 atm, 1M).
G implies DIFFERENT CONCENTRATIONS of both reactants ↔ products
and under non-standard conditions of concentration, temperature and
pressure.
•
If [Reactants]↑, then G = -
•
If [Products]↑, then G = +
•
If at Equilibrium , then G = 0
[ Products ]p
Q
[ Reactants ]r
Equilibrium
Free Energy (G) & Equilibrium Constant (K)
at Equilibrium
At equilibrium, G = 0 and Q = K so……
G = G + RT ln Q
0 = G + RT ln K
Rearranging ……
G = - RT ln K
G
–
0
+
K
>1
=1
<1
K = eG/RT
= - 2.303 RT log K
☺
Product formation
Product-favored @ Equilibrium
[Products] = [Reactants] @ Equilibrium
Reactant-favored @ Equilibrium
(RARE)
Equilibrium
Free Energy (ΔG), the Equilibrium
Constant (ΔK),and Temperature
2 NO2 (g)
Low Temp.
High Temp.
N2O4 (g)
(clear gas)
(red-brown gas)
G = - RT ln K
Low Temperature:
How would T ↓or ↑ affect K?
H = -58.0 kJ
[ Products ]p @ T ↓, K ↑ G°= K
[ Reactants ]r @ T ↑, K ↓ G°= +
High Temperature:
G°= -
G°= +
Q=K
Reactants
only
Equilibrium
mixture
Products
only
{G, Q & T↓ NO2 → N2O4}LowT
Equilibrium
{G, Q & T↑NO2 ← N2O4}HighT
2003B Q1
2 HIH2 + I2
Equilibrium
2003B Q1
Kp = Kc(RT)Δn
Equilibrium
2000A Q1
Equilibrium
2000A Q1
Equilibrium
Equilibrium
Equilibrium
2004B Q1
Equilibrium
Equilibrium
2007B Q1
Equilibrium
2000 1
Equilibrium
Equilibrium
Equilibrium
2003 B
Equilibrium
Equilibrium
2004 B
Equilibrium
Equilibrium
2006 A
Equilibrium
Equilibrium
2006 (B)
Equilibrium
Equilibrium
Equilibrium
Equilibrium
Equilibrium
Equilibrium
Equilibrium
2007 (B)
Equilibrium
Equilibrium