Topics in CHM 1046

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Topics in CHM 1046

• Intermolecular forces (IMF)
• Themodynamics
• Chemical Kinetics
• Chemical Equilibrium
• Combination of above

Unit 11 Intermolecular Forces, Liquids Solids
(T1) Unit 12 Properties of Solutions (T1)
Unit 13 Thermochemistry (T1) Unit 14 Chemical
Thermodynamics (T2)
Unit 15 Chemical 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)
Unit 21 Electrochemistry (T4)
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Unit 16Chemical Equilibrium Gases
Heterogeneous
CHM 1046 General Chemistry and Qualitative
Analysis

• Dr. Jorge L. Alonso
• Miami-Dade College Kendall Campus
• Miami, FL

• Textbook Reference
• Chapter 17
• Module 5 ( Appendix 2)

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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 ? B
?
(red-brown gas)
(clear gas)
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!
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The Equilibrium Constant (Keq)

• Therefore, at equilibrium
• Ratef Rater
• kf N2O4 kr NO22
• Rewriting this, it becomes

Keq
The Equilibrium Constant is the ratio of the
rate constants at a particular temperature.
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The Equilibrium Constant (K)

• To generalize this expression, consider the
  reaction

• The equilibrium expression for this reaction
  would be

where X concentration of each chemical in
moles/ liter (M).
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What Does the Value of K Mean?
10 x
Kc 10 10
1

• If K gtgt 1, the reaction is product-favored
  product predominates at equilibrium.

10 x
Kc 1 0.1
10

• If K ltlt 1, the reaction is reactant-favored
  reactant predominates at equilibrium.

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The Equilibrium Constant for Gases
Partial press (75 torrs) (150 torrs)
(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

where Kp equilibrium constant for gases, and
Px partial pressure of each gas.
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Relationship between Kc and Kp

• ideal gas law

PV nRT

Plugging this into the expression for Kp
?
Kp Kc (RT)?n
Where
?n (moles of gaseous product) - (moles of
gaseous reactant)
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Equilibrium Can Be Reached from Either Direction
_at_ 100C
KC

• The ratio of NO22 to N2O4 remains constant
  at this temperature no matter what the initial
  concentrations of NO2 and N2O4 are.

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Equilibrium Can Be Reached from Either Direction
N2O4 (g)
2 NO2 (g)
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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.

1 0.212
4.72 at 100?C
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Manipulating Equilibrium Constants

• 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.

RULE 2 Coefficient Rule
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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)
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Homogeneous vs.HeterogeneousEquilibria
Reactions involving only gasses or only solutions
NO22 N2O4
Kc 0.212 at 100?C
Reactions involving different phases (s, l, g,
aq) of matter
CO2 (g) CaO(s)
CaCO3 (s)
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The Concentrations of Solids and Liquids Are
Essentially Constant
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 massand both of these are constants
at constant temperature.
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What are the Equilibrium Expressions for these
Heterogeneous Equilibria?
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How does the concentration of CO2 differ in the
two bell jars?

• As long as some CaCO3 or CaO remain in the
  system, the amount of CO2 above the solid will
  remain the same.

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Equilibrium Calculations
(1) Calculating Kc from experimental data i
e
(2) Calculating Equilibrium Concentrations e
when Kc i are known
ICE Tables
I Ai Bi 0 0
C - change - change change change
E Ae Be Ce De
Initial
Change
Equilibrium
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(1) Calculating Kc from experimental data i
e

• 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

H2, M I2, M ? 2 HI, M
Initially 1.000 x 10-3 2.000 x 10-3 0
Change -9.35 x 10-4 -9.35 x 10-4 1.87 x 10-3
At Equilibrium 6.5 x 10-5 1.065 x 10-3 1.87 x 10-3
HI Increases by 1.87 x 10-3 M
Stoichiometry tells us H2 and I2 decrease by
half as much
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and, therefore, the equilibrium constant
H2, M I2, M ? HI, M
Initially 1.000 x 10-3 2.000 x 10-3 0
Change -9.35 x 10-4 -9.35 x 10-4 1.87 x 10-3
At Equilibrium 6.5 x 10-5 1.065 x 10-3 1.87 x 10-3
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Calculating Kc from experimental data i e
_at_ 100C
- 2x
x
? x
- 2x
x
- 2x
x
2x
- x
I 0.0 0.0200
C x - 2x
E 0.0014 0.0200-2x
Kc

• The ratio of NO22 to N2O4 remains constant
  at this temperature no matter what the initial
  concentrations of NO2 and N2O4 are.

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Practice Problem
2006B Q5


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Equilibrium Calculations
(1) Calculating Kc from experimental data i
e
H2, M I2, M ? HI, M
Initially 1.000 x 10-3 2.000 x 10-3 0
Change
At Equilibrium 1.87 x 10-3
(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
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(2) Calculating Equilibrium Concentrations e
when K i are known
I Ai Bi 0 0
C - ax - bx cx dx
E Ai- ax Bi- bx cx dx
Solve for x by using the quadratic equation
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Calculating Equilibrium Concentrations when K
i are known
Problem for the reaction H2(g) I2(g)
2HI(g) , K 54.3 _at_ 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 0.0224 2x
Deciding if x is negligible
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Problem for the reaction H2(g) I2(g)
2HI(g) , K 54.3 _at_ 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.
Deciding if x is negligible

1. Take the largest Ai 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 AI
4. If these calc values are significant compared to
   K, then x is not negligible and you must use the
   quadratic to solve for x.

YOU MUST USE QUADRATIC!
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Calculating Equilibrium Concentrations when K
i are known
H2i 0.00623
E 0.00632- x 0.00414- x 0.0224 2x
I2i 0.00414
First answer is physically impossible, since
conc. of H2 and I2 would be more than original
conc.
x calc without quadratic
H2e 0.00623 0.00156) 0.00467 M
I2e (0.00414 0.00156) 0.00258 M
HIe (0.0242 2 x 0.00156) 0.0255 M
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Equilibrium Calculations
(1) Calculating Kc from experimental data i
e
H2, M I2, M ? 2 HI, M
Initially 1.000 x 10-3 2.000 x 10-3 0
Change
At Equilibrium 1.87 x 10-3
(2) Calculating Equilibrium Concentrations
when K i are known
H2(g) I2(g) 2HI(g)
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I 0.00623 0.00414 0.0224
C - x - x 2x
E 0.00632- x 0.00414- x 0.0224 2x
Deciding if x is negligible
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Le Châteliers Principle
Equi.Intro1
Equi.Intro2

• 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.

KC
The effect of (1) Concentration
Henri Louis Le Chatelier 1850-1936
?
(clear gas)
(red-brown gas)
The effect of (2) Pressure Volume
(3) Temperature
L.C.PressVol
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Le Châteliers Principle
The effect of (3) Temperature
HEAT
?
?H -58.0 kJ
(clear gas)
(red-brown gas)
Why does heat favor the formation of NO2 over
N2O4?
Le ChâteliersTemp
Molecular Explanation
Keq at different Temp
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The Effect of Temperature Changes
Add heat (?), which way will the equlibrium be
shifted?
?
?H
(pink)
HEAT
(blue)
?
CoCl
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Briggs-Rauscher reaction
ClockReaction
IO3- 2H2O2 CH2(COOH)2 H ? ICH(COOH)2 2O2
3H2O
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Le Châteliers 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 _at_ 250 C
NH3 HNO3 ? NH4NO3 (explosives fertilizers)
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Le Châteliers Principle the Haber Process
Shift equilibrium to the right by increasing
press conc. of N2 H2 and by removing NH3
from system
Fe
N2 (g) 3H2 (g) 2NH3 (g)

• This apparatus helps push the equilibrium to
  the right by removing the ammonia (NH3) from the
  system as a liquid

Removed by liquefaction
Subs b. pt.
NH3 -33C
N2 -196C
H2 -253C
Fe
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Increase the rate of both the forward and
reverse reactions.
The Effect of Catalysts on Equilibrium
How does the addition of a catalyst affect the
equilibrium between subs. A and B?
catalyst

• Equilibrium is achieved faster, but the
  equilibrium composition remains unaltered.

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Equilibrium Constant (K) vs. Reaction Quotient
(Q)
At Equilibrium
At Non-Equilibrium

• 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.

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If Q K, the system is at equilibrium.
If Q lt K, there is too much reactant, and the
equilibrium shifts to the right.
If Q lt K?
If Q gt K, there is too much product and the
equilibrium shifts to the left.
If Q gt K?
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Kc, Q and Spontaneity (?G)
?G -
Reverse Process ?G
?G -
Reverse Process ?G
_at_ Equilibrium K Q, ?G 0
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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 _at_ 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

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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
K e??G?/RT
Rearranging
?
?G? - RT ln K
- 2.303 RT log K
?G? K Product formation
gt 1 Product-favored _at_ Equilibrium
0 1 Products Reactants _at_ Equilibrium (RARE)
lt 1 Reactant-favored _at_ Equilibrium
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Free Energy (?G), the Equilibrium Constant
(?K),and Temperature
How would T ?or ? affect K?
Low Temp.
?H -58.0 kJ
High Temp.
(clear gas)
(red-brown gas)
_at_ T ?, K ?
?G -
?G? - RT ln K
_at_ T ?, K ?
?G
Low Temperature
High Temperature
?G -
?G
Q K
Products only
Reactants only
Equilibrium mixture
?G, Q T?NO2 ? N2O4HighT
?G, Q T? NO2 ? N2O4LowT
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2003B Q1
2 HI?H2 I2
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2003B Q1
Kp Kc(RT)?n
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2000A Q1
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2000A Q1
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2004B Q1
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2007B Q1
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2000 1
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2003 B
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2004 B
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2006 A
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2006 (B)
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2007 (B)
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