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Unit II - Equilibrium - Chapter 13
2.1. Characterize chemical reactions in terms of reversibility and
relative concentrations of reactants and products.
[Readings 13.1 Problems 18]
Demonstration:
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Demonstration:
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Fe+3 + SCN- --> FeSCN+2
Fe+3 + SCN- --> FeSCN+2
Fe+3 + SCN- --> FeSCN+2
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The relative amounts of reactants and products in a chemical reaction are effected by
concentration changes.
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Fe+3 + SCN- --> FeSCN+2
Fe+3 + SCN- --> FeSCN+2
Fe+3 + SCN- --> FeSCN+2
All reactants and products are present during the reaction.
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Observations on the chemical reaction:
Fe+3 + SCN- --> FeSCN+2
All reactants and products are present during the reaction.
Reactions are reversible.
The relative amounts of reactants and products in a chemical reaction are effected by
concentration changes.
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AgNO3(aq) + NaC2H3O2(aq) --> AgC2H3O2(s)+ NaNO3(aq)
Ag+ + NO3- + Na+ + C2H3O2- --> AgC2H3O2(s) + Na+ + NO3Net Ionic Equation
Ag+(aq) + C2H3O2-(aq) --> AgC2H3O2(s)
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Ag+ + C2H3O2- --> AgC2H3O2(s)
Initial
.002 moles .002 moles
Final (expected)
0 moles
0 moles
Final
(experimental)
0 moles
.002 moles
.122 g = .000730 moles
Where did the rest go?
.00127 moles .00127 moles
.000730 moles
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AgC2H3O2(s) -->Ag+ + C2H3O2Initial
.002 moles
0 moles
0 moles
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Final
.000730 moles
.00127 moles .00127 moles
----------------------------------------------------------------------------------
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Initial
.02 moles
Final
.01873 moles
0 moles
0 moles
.00127 moles .00127 moles
AgC2H3O2(s) <==> Ag+(aq) + C2H3O2-(aq)
AgC2H3O2(s) <==> Ag+(aq) + C2H3O2-(aq)
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[Ag+] vs. [C2H3O2- ]
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2.2. Determine equilibrium expressions for homogeneous and
heterogeneous chemical reactions from stoichiometry.
[Readings 13.2, 13.3, & 13.4 Problems 1, 2, 26, 28, 34, 36, 38, &
46]
aA + bB <==> cC + dD
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[Ag+] vs. [C2H3O2-] ---------- [Ag+] vs. 1/[ C2H3O2 -]
[Ag+] vs. [C2H3O2-] ---------- [Ag+] vs. 1/[ C2H3O2 -]
[Ag+] vs. [C2H3O2-] ---------- [Ag+] vs. 1/[ C2H3O2 -]
In heterogeneous equilibria, pure liquids or solids are not included.
In aqueous equilibria, pure water is not included.
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Superscripts?
A + B <==> 2C
A + B <==> C + C
K = [C] [C] / [A] [B]
K = [C]2 / [A] [B]
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Example
• Reaction
N2 (g) + 3 H2 (g) –––> 2 NH 3 (g)
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Example
• Reaction
N2 (g) + 3 H2 (g) –––> 2 NH 3 (g)
K=?
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Example
• Reaction
N2 (g) + 3 H2 (g) –––> 2 NH 3 (g)
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Example
• Reaction
N2 (g) + 3 H2 (g) –––> 2 NH 3 (g)
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For rxn:
2 NH3 (g) –––> N2 (g) + 3 H2 (g)
K’ = ?
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Example
• Reaction
N2 (g) + 3 H2 (g) –––> 2 NH 3 (g)
For rxn:
2 NH3 (g) –––> N2 (g) + 3 H2 (g)
K’ = 1/K
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Pure solids or liquids?
What is the [AgC2H3O2] ?
= moles / Liter
{moles = weight / MW}
= weight / Lx MW
= density / MW = constant for pure substance
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Water not included?
What is the concentration of water in pure water?
1000g / L (1 mole / 18 g) = 55.6 M
What is the concentration of water in 0.1 M NaCl solution? (5.9 g)
994g / L (1 mole / 18 g) = 55.2 M
Therefore [H2O] in solutions is a constant.
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aA + bB <==> cC + dD
In heterogeneous equilibria, pure liquids or solids are not included.
In aqueous equilibria, pure water is not included.
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Example
Write equilibruim expressions for the following reactions: (in a 1 liter container)
A. 1/8 S8 (s) + O2 (g) –––> SO2 (g)
B. NH3 (aq) + H2O (l) –––> NH 4+ (aq) + OH- (aq)
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Gas Phase Reactions
Often partial pressures are used instead of concentrations for gas phase reactions.
Rxn:
N2O4 (g) –––2 NO2 (g)
What is the eauilibrium constant?
Solve for Kp
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Gas Phase Reactions
Often partial pressures are used instead of concentrations for gas phase reactions.
Rxn:
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N2O4 (g) –––2 NO2 (g)
What is the eauilibrium constant?
Solve for Kp
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K is temperature dependent:
K Values for Related Chemical Equations
(1) A <=> B
K1= x
(2) B <=> A
K2= K1-1 = x-1
(1) A <=> B
K1= x
(3) 2A <=> 2B
K3= K12 = x2
2.3. Determine the stoichiometric relationship between initial and
equilibrium concentrations of reactants and products.
[Readings 13.5 Problems]
2.4. Determine values for K from equilibrium concentrations of
reactants and products in a chemical reaction.
[Readings 13.5 Problems 80, ]
H2O + HC7H5O2 <=> H3O+ + C7H5O2Initial
Equilibrium
Change
Equilibrium
0.10 M
0M
0M
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2.57 x 10-3
?
-2.57 x 10-3
9.74 x 10-2
+2.57 x 10-3
+2.57 x 10-3
2.57 x 10-3
2.57 x 10-3
Initial ----- amount based on experiment
Change ----- amount based on stoichiometry
Equilibrium ----- amount based on equilibrium
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2.5. Determine the equilibrium concentrations of reactants and
products of a chemical reaction from initial concentrations and
values of K. [Readings 13.5 Problems 10, 11, 25, 48, 52, 54, 56, &
58]
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H2O + HC7H5O2 <=> H3O+ + C7H5O2If 61 g of HC7H5O2 is dissolved in 1 L of solution, what is [H3O+] at equilibrium? Kc = 6.3x10-5
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2.6. Determine if equilibrium has been reached in a chemical
reaction; determine the direction the reaction will shift if
equilibrium has not been reached.
[Readings 13.5 Problems ex 13.6 & prob. 13.7]
rR <==> pP
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K = [P]ep / [R]er
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Q = [P]ip / [R]ir
2.7. Use Le Châtelier’s Principle to predict the direction a
reaction at equilibrium will shift as a result of changes in
concentration, pressure/volume, and temperature as it
approaches a new equilibrium. [Readings 13.6, 13.7, 13.8, & 13.9
Problems 66, 68, 70, & 72]
Le Châtelier’s Principle
If a stress is put on a system in equilibrium, it will shift (adjust)
to minimize (offset, reverse) the effect of the stress.
Le Châtelier’s Principle
If a stress is put on a system in equilibrium, it will shift (adjust)
to minimize (offset, reverse) the effect of the stress.
Equilibrium:
A system is at equilibrium when the rate of the forward reaction is equal to the rate of the reverse
reaction.
This does not mean:
The amounts are equal
The reaction has stopped
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A <==> B
Stress -- changes that effect rates of forward and reverse
reactions differently
concentration
temperature
catalyst?
pressure?
Shift -- An adjustment in the concentrations of reactants and
products
Shift to right
-- increase [P], decrease [R]
Shift to left
-- increase [R], decrease [P]
Minimize
A(g) + B(g) <=> 2C(g)
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0
C
-1
-1
+2
K = [C]2 / [A] [B]
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E
add 19 A
C
new E
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1
4
2
K=1
20
4
2
Q
-2
-2
+4
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K=1
N2 + 3 H2 <==> 2 NH3
N2 + 3 H2 <==> 2 NH3
Demo:
2 NO(g) + O2(g) --> 2 NO2(g)
Initial
20 mL 10 mL
Predicted
20 mL
Actual
15 mL
2 NO2(g) <==> N2O4(g) +
brown
colorless
Change Temperature
2 NO2(g) <==> N2O4(g) +
Increase T -- More Heat -- P constant
Equilibrium Shifts to Reduce Heat Color Deepens
K = [N2O4] / [NO2]2 , K decreases
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Therefore K is T dependent
Change Pressure (by decreasing V)
2 NO2(g) <==> N2O4(g) +
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PT
Kp
Equilibrium
20 atm
20 atm
40 atm
.05
P x 2 (1/2 V)
40 atm
40 atm
80 atm
.025
Change
-10 atm
+5 atm
New
30 atm
Equilibrium
45 atm
75 atm
.05
Self Test
X(g) + H2O(l) <==> Q (s) + Y(g)
Change
increase Px
increase [Y]
remove 1/2 Q
increase PT
increase T
Shift
H = +150 kJ
Change in K
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catalyst
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