11:11 pm unit 1 kinetics and equilibrium chemistry 3202
TRANSCRIPT
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Unit 1Kinetics and
EquilibriumChemistry 3202
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Part 1: Reaction Kinetics (Chp. 12) Reaction Kinetics is the study of the
rate of a chemical reaction Qualitative:
Reactions may be described as being FAST or SLOW Fast – burning, explosions, precipitation Slow – rusting, fermentation
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Reaction Kinetics
Quantitative:
The rate of a reaction measures how fast products are formed or how fast reactants are consumed
Rate = Change in quantity
Change in time
POSSIBLE UNITS ??
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Measuring Reaction Rate The method used to determine
reaction rate will depend on the reaction being studied. (p. 466)
Methods:1. monitor pH if there is an acid or
base in the equation2. record gas volume or changes in
pressure if there is a gas in the reaction
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Measuring Reaction Rate
Methods: (cont’d)3. record changes in mass if solids are
present4. monitor absorption of light if there is a
color change5. changes in electrical conductivity
indicate changes in ion concentration
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MC: What could we use to measure the rate of this reaction?
Cu(s) + 2 AgNO3(aq) 2 Ag(s) + Cu(NO3)2(aq)
a) pressure c) gas volume
b) pH d) mass
Answer: d) because a solid is present
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MC: What could we use to measure the rate of this reaction?
SO3(g) + H2O(l) H2SO4(aq)
a) pressure c) gas volumeb) pH d) massAnswers: a) and c) because a gas is present b) because an acid is being produced
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What determines RATE??
All chemical reactions are bond breaking/bond forming events
The rate of a reaction depends on how quickly bonds are broken and how rapidly new bonds form.
KMT and Collision Theory are used to explain reaction rates.
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Kinetic Molecular Theory (KMT)
Matter is made of particles (atoms, ions, or molecules) in continuous motion
An increase in temperature: increases the speed of particles reduces the forces of attraction between
particles
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Kinetic Molecular Theory (KMT) KMT is supported by:
Diffusion – particles of a gas spread to fill their container (‘perfume in a room’)
- solids dissolve uniformly in liquids over time.
Pressure – a balloon remains inflated because gas particles are continuously hitting the sides of the balloon
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Energy Distribution of Particles
# of
par
ticle
s
25 °C
200 °C
Kinetic Energy
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Collision Theory
reactant particles must collide with one another for a chemical reaction to occur
particles must collide with proper orientation
collisions must have enough intensity to break old bonds and allow new bonds to form
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Collision Theory
to increase reaction rate you must increase the number of successful collisions between reactant molecules
VIDEO (VHS): Reaction Rates LASERDISK: 3 VIDEO CLIPS
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Factors Affecting Reaction Rate
1. Concentration
– an increase in the concentration of a reactant usually increases the rate of a chemical reaction
- the rate increases because there are:
- more particles resulting in
- more collisions between particles &
- more successful collisions.
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Factors Affecting Reaction Rate
2. Temperature
- an increase in the temperature increases the rate of a chemical reaction
- the higher temperature results in:
- more collisions between particles
- more intense collisions
NOTE: A temperature increase of 10 ºC usually causes reaction rate to double.
more successful collisions
faster rate
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Factors Affecting Reaction Rate3. Nature of Reactants– compounds with fewer bonds to break
will react more rapidly than compounds with many bonds
eg. propane (C3H8) burns faster than candlewax (C25H52) because it has fewer bonds
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Factors Affecting Reaction Rate3. Nature of Reactants
- compounds with weak bonds react more rapidly than compounds with strong bonds
– ions will react more rapidly than atoms and molecules
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Factors Affecting Reaction Rate4. Surface Area- crushing a solid to produce a powder, or
changing a substance to the gas phase, exposes more particles for collision
- if more particles are available for collision there will be:- more collisions- more successful collisions
faster ratefaster rate
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Factors Affecting Reaction Rate5. Catalysts
- a catalyst increases the reaction rate by providing a different reaction pathway or mechanism with a lower activation energy
- a catalyst IS NOT consumed by a chemical reaction.
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# of
par
ticle
s
Kinetic Energy
Ea withoutcatalyst
Ea withcatalyst
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Potential Energy Diagrams (p. 473)
PE diagrams show changes in potential energy (stored chemical energy) during chemical reactions
Exothermic reactions release more energy than they absorb (eg. burning)
Endothermic reactions absorb more energy than they release
(eg. photosynthesis)
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NEW SLIDE
ΔH written in the equation and outside the equation (Thermochemical Equation)
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Potential Energy Diagrams
∆H represents the heat of reaction or enthalpy of reaction
∆H is the difference between the PE of the reactants and the PE of the products
the minimum energy needed for a chemical reaction to occur is the activation energy
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Potential Energy Diagrams
the activated complex for a reaction is a temporary, unstable, intermediate species that quickly decomposes to products
eg. H2 + I2 → H2I2 → 2 HI
activated complex
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Reaction Progress
PE
Reactants
Products
∆H
ENDOTHERMIC
(positive)
site of the activated complex
activation energy (Ea forward)
Eareverse
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Reaction Progress
PE Reactants
Products
∆H
EXOTHERMIC
(negative)Ea forward
site of AC
Ea reverse
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Formula: (OPTIONAL)
Eaforward - Eareverse = ΔH
This formula is NOT necessary if you prefer using the PE diagram.
Animation
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Ea fwd Ea rev ΔH Endothermic or
Exothermic
25 -30
50 20
150 250
65 28
Sketch a PE diagram for each reaction
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Reaction Progress
PE
∆HEafwd
CO2 + H2O
C6H12O6 + O2
PhotosynthesisEarev
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Reaction Progress
PE
∆H
Eafwd
CO2 + H2O
C6H12O6 + O2
Respiration
Earev
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p. 474
∆H
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∆H
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∆H
p. 475
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Reaction Progress
PE
no catalyst
catalyzed
Effect of a catalyst
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Reaction Progress
PE
no catalystEXOTHERMIC
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Sample problem: p. 475 Questions:
p. 476; #’s 1 – 4
p. 484; #’s 1 – 4
p. 486; #’s 1,2, 4, 6, & 7
p. 468; # 4
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Reaction Mechanisms (pp. 477 – 485)reaction mechanism – the steps that
occur in a chemical reactionelementary reaction - each step in a reaction mechanismreaction intermediate – a molecule, atom or ion formed in one step and consumed in a later step NOTE: reaction intermediates are NOT included in the overall equation
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Reaction Mechanisms
eg. #1
Step #1 NO(g) + O2(g) NO3(g)
Step #2 NO3(g) + NO(g) 2 NO2(g)
Overall Equation:
2 NO(g) + O2(g) 2 NO2(g)
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HBr + O2 → HOOBrfast
HOOBr + HBr → 2 HOBrslow
2 HOBr + 2 HBr → 2 H2O + 2 Br2
fast
p. 478 #’s 5 – 8
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Reaction Mechanisms
rate-determining step (RDS)
- the slowest step in a reaction mechanism
- to increase the rate of a reaction you must speed up the RDS
- increasing the concentration of a reactant will increase the rate ONLY IF the reactant is in the RDS
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Reaction Mechanisms
PE diagrams
- every step in a reaction mechanism has an activation energy which can be drawn on a PE diagram
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Reaction Progress
PE
Reaction Mechanisms
3-step mechanism
#1
#2
#3RDS ??
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Reaction Mechanisms
eg:
Step #1 H2CO2 + H+ H2CO2H+ fast
Step#2 H2CO2H+ HCO+ + H2O slow
Step #3 HCO+ CO + H+ fast
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Reaction Mechanisms
eg:
Overall H2CO2 H2O + CO
Omit H+ - catalyst
Omit H2CO2H+ & HCO+ - reaction intermediates
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Reaction Mechanisms
Reaction Progress
PE
H2CO2 + H+CO + H+
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p. 476; #’s 1 – 4
p. 484; #’s 1 – 4
p. 486; #’s 1,2, 4, 6, & 7
p. 468; # 4
p. 478; #’s 5 - 8 p. 484 #’s 5 – 9 p. 485 #’s 10, 12 p. 486 #’s 8, 10, 11 p. 487 #’s 14, 17 p. 829 #’s 128,129, 131, 132
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p. 829 # 128
Step 1 H2(g) + NO(g) → H2O(g) +
N(g)
Step 2
Step 3 H2(g) + O(g) → H2O(g)
2H2(g) + 2NO(g) → N2(g) + 2H2O(g)
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Part 2: Chemical EquilibriumEquilibrium
A balancing Act!Text Ch 13: p 488 - 541
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Part 2: Chemical Equilibrium
All reactions we have done have shown reactants being converted 100% to products
Many reactions are reversible with some products being converted back to reactants
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Part 2: Chemical Equilibrium
Dynamic equilibrium occurs when 2 opposing processes occur at the same rate
A chemical equilibrium occurs when two opposing chemical reactions occur at equal rates.
Demo: p. 491
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Types of Equilibria
1. Phase EquilibriaAn equilibrium may be established between different phases of a compound in a sealed container
eg.
H2O(l) in a sealed container
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Types of Equilibria
Initially: H2O(l) changes to H2O(g)
H2O(l) → H2O(g)
Gradually: H2O(g) changes to H2O(l)
H2O(l) ← H2O(g)
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Types of Equilibria
Using equilibrium notation:
H2O(l) ⇌H2O(g)
Temperature changes?
Closed vs. open system?
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Types of Equilibria
2. Solubility Equilibria occur in saturated solutions when NaCl(s) is placed in water, the
initial rate of dissolving is fast
NaCl(s) NaCl(aq)
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Types of Equilibria
as more solid dissolves, the rate of dissolving slows and recrystallization begins.
eg. NaCl(s) NaCl(aq)
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Types of Equilibria
when the solution is saturated there are NO VISIBLE CHANGES
At equilibrium, the RATE of dissolving and the RATE of recrystallization are EQUAL.
eg. NaCl(s) ⇌ NaCl(aq)
equilibrium
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Types of Equilibria
Temperature change?? Open vs. Closed ??
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Types of Equilibria
3. Chemical Equilibrium Chemical reactions that are reversible
usually result in chemical equilibrium
eg. NO2 gas changing to N2O4
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Types of Equilibria
Initially the forward rate is high
eg. 2 NO2(g) N2O4(g)
as more product forms, the reverse reaction begins and increases in rate.
eg. 2 NO2(g) N2O4(g)
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Types of Equilibria eventually the forward rate slows and
the reverse rate increases such that the FORWARD AND REVERSE RATES ARE EQUAL
eg. 2 NO2(g) ⇌ N2O4(g)
http://www.chm.davidson.edu/ronutt/che115/EquKin/EquKin.htm
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Conditions for Equilibrium (p. 492)
1. Macroscopic properties are constant
ie. NO OBSERVABLE CHANGE
2. Forward and reverse rates must be equal
3. A CLOSED SYSTEM is required for equilibrium
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Conditions for Equilibrium4. Equilibrium may occur from either
direction
eg. 2 NO2(g) ⇌ N2O4(g)
OR
N2O4(g) ⇌ 2 NO2(g)
p. 493; #’s 1 – 6Kinetics & Equilibrium #4
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Shifts in Equilibrium
Equilibrium occurs when the forward rate equals the reverse rate.
Changes in concentration, temperature and pressure/volume can cause the forward or reverse reaction rate to change.
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Shifts in Equilibrium
Eventually, a new equilibrium will be established with different reactant and product concentrations
Le Châtelier’s Principle is used to predict changes in concentrations when a stress is applied to a system at equilibrium.
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Le Châtelier’s Principle (p.520)When a stress is applied to a system at equilibrium, the system will adjust or shift to relieve the stress.
A change(stress) is applied to a system at equilibrium
Forward or reverse rate will change New equilibrium established
WHAT???
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Le Châtelier’s Principle1. Changes in concentration
An increase in concentration on one side of an equation favors or drives the reaction to the opposite side.
eg. What will happen if CO is added to this system at equilibrium?
CO(g) + 2 H2(g) ⇌ CH3OH(g)
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Le Châtelier’s Principle- the system will ‘shift’ to the right to use
the CO and produce more CH3OH
- some H2 will be used
[CH3OH] will increase
[H2] will decrease[CO] ‘spikes’ and then drops to a value higher than it was before the change
?? possible graph ??
forward rate increases;reverse rate catches up
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mol/L
time
CO
CH3OH
H2
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Le Châtelier’s Principle
An equilibrium shifts away from a substance that increases in concentration or toward a
substance that decreases in concentration.
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Le Châtelier’s PrincipleWHY???
- rates at equilibrium ???
- increasing the [CO] & forward rate ??
- reverse rate??
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Le Châtelier’s Principle- [H2] ??
- new equilibrium concentrations ??
** NO CHANGE in Keq **
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Le Châtelier’s Principle
decrease [CO]
increase [H2]
decrease [CH3OH]
CO(g) + 2 H2(g) ⇌ CH3OH(g)
What happens if we:
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Le Châtelier’s Principle
IMPORTANT NOTE:
- adding a solid or liquid does not change molar concentration
- changing the amount of a solid or liquid in an equilibrium will NOT cause a shift
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Le Châtelier’s Principle
eg.
CaCO3(s) ⇌ CaO(s) + CO2(g)
add CaCO3(s) ??
add CO2(g) ??
What happens if we:
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Le Châtelier’s Principle (ΔH & equations)2. Temperature
- Raising the temperature of an exothermic equilibrium favors the formation of reactants.
eg. CO(g) + 2 H2(g) ⇌ CH3OH(g) + 65 kJ
What happens if we increase temperature in this equilibrium?
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Le Châtelier’s Principle- Raising the temperature of an endothermic equilibrium favors formation of products.
eg. CaCO3(s) + heat ⇌ CaO(s) + CO2(g)
What happens if we increase temperature in this equilibrium?
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Le Châtelier’s PrincipleOR
- Raising the temperature shifts the equilibrium away from the energy term.
- Decreasing the temperature shifts the equilibrium toward the energy term.
CO(g) + 2 H2(g) ⇌ CH3OH(g) + 65 kJ
CaCO3(s) + heat ⇌ CaO(s) + CO2(g)
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Le Châtelier’s Principleeg. How would an increase in
temperature affect these equilibria?
N2(g) + 3 H2(g) ⇌ 2 NH3(g) + heat
2 SO3(g) ⇌ 2 SO2(g) + O2(g) ΔH = +197 kJ
2 SO3(g) + 197 kJ ⇌ 2 SO2(g) + O2(g)
**A change in temperature changes Keq**
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Le Châtelier’s Principle
3. Pressure/Volume An increase in pressure of a system at
equilibrium has the same effect as a decrease in the volume of the system.
(inverse relationship)
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Le Châtelier’s Principle Increasing the pressure of a system at
equilibrium by reducing volume causes the equilibrium to shift in the direction that reduces pressure
ie. shift to the side with fewer molecules of GAS!!
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Le Châtelier’s Principleeg. How would an increase in pressure -
caused by a decrease in volume - affect these equilibria?
N2(g) + 3 H2(g) ⇌ 2 NH3(g)
4 NH3(g) + 5 O2(g) ⇌ 4 NO(g) + 6 H2O(g)
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Le Châtelier’s Principle
eg. Why would a change in pressure NOT affect the following equilibria?
H2(g) + I2(g) ⇌ 2 HI(g)
2 Ag(s) + Zn2+(aq) ⇌ 2 Ag+
(aq) + Zn(s)
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Le Châtelier’s Principle
pp. 529, 530 #’s 33 – 37
(omit 36 for now)
p. 533 #’s 1 – 3, 5
Answers on p. 537
Lab: Perturbing Equilibrium
Animation
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Le Châtelier’s Principle
4. Catalyst- Does NOT cause a shift in equilibrium- Increases BOTH rates equally so
equilibrium is reached faster
5. Surface Area - same as a catalyst
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Equilibrium Constant (Keq)For any system at equilibrium, there is a
mathematical relationship between reactant and product concentrations
(Guldberg and Waage, 1864)
P. 494
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Equilibrium Constant (K)
See p. 495 eg. N2O4(g) ⇌ 2 NO2(g)
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Equilibrium Constant (K)
For the general equilibrium below:
aP + bQ ⇌ cR + dS
Keq = [R]c [S]d
[P]a [Q]b
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Equilibrium Constant (K)
- a, b, c, & d are coefficients used to balance the equation
- P, Q, R, & S are the reactants and products
- Kc is sometimes used instead of Keq when units are molar concentration
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Equilibrium Constant (K)
eg. Write the expression for Keq for:
2 SO2(g) + O2(g) ⇌ 2 SO3(g)
2 HCl(g) ⇌ H2(g) + Cl2(g)
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Equilibrium Constant (K)
NOTE!!
Solids or liquids ARE NOT included in the Keq or Kc expression because their concentration is constant
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Equilibrium Constant (K)eg. Write the expression for Kc for:
2 NaCl(s) + H2SO4(aq) ⇌ 2 HCl(g) + Na2SO4(aq)
CaCO3(s) ⇌ CaO(s) + CO2(g)
p. 497; #’s 1 - 5
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Interpreting K
Large vs small K values?
K much larger than 1 - products favoured
K much smaller than 1 - reactants favoured
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Changing K (p. 497)
For a given system at equilibrium, the value of the equilibrium constant depends only on temperature
ie. For any equilibrium, the only way to change the actual value of K is to change the temperature
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Calculations with K
Types of K calculations:
1. Given equilibrium concentrations, find K
2. Find a missing concentration given K and other concentrations.
3. Given initial concentrations and equilibrium data, find K (ICE tables)
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1.Given equilibrium concentrations, find K
eg. Calculate Kc for this equilibrium using the equilibrium concentrations given:
H2(g) + I2(g) ⇌ 2 HI(g)
[H2] = 0.22 mol/L
[I2] = 0.30 mol/L
[HI] = 1.56 mol/L
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Given the equilibrium concentrations;
[CO(g)] = 0.105 mol/L,
[H2(g)] = 0.250 mol/L
[CH3OH(g)] = 0.00261 mol/L,
What is the value of Keq for the equilibrium below?
CO(g) + 2 H2(g) ⇌ CH3OH(g)
(A) 0.0994
(B) 0.398
(C) 2.51
(D) 10.0
Be Careful!!The other answers are possible if you mess
up the calculation
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Given the equilibrium concentrations below, what is the value of Keq for
N2(g) + O2(g) ⇌ 2 NO(g) ?
[N2(g)] = 0.10 mol/L
[O2(g)] = 0.20 mol/L
[NO(g)] = 0.0030 mol/L
(A) 2.2×10−4
(B) 4.5×10−4
(C) 1.5×10−1
(D) 3.0×10−1
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Find Keq for the equilibrium below:
4 NH3(g) + 5 O2(g) ⇌ 4 NO(g) + 6 H2O(g)
[NH3(g) ] = 0.100 mol/L
[O2(g) ] = 0.200 mol/L
[ NO(g) ] = 0.300 mol/L
[ H2O(g) ] = 0.250 mol/L
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2. Find a missing concentration . . .
eg. Find the [HI] in the equilibrium below if Kc = 36.9, [H2] = 0.125 mol/L and [I2] = 2.56 mol/L.
H2(g) + I2(g) ⇌ 2 HI(g)
p. 499; #’s 6 – 9
p. 538; # 6 Which reaction is faster - forward
or reverse?
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3. ICE tables
Keq may be calculated given initial concentrations and at least one equilibrium concentration
Using the Initial concentration and the Change in concentration we can find the missing Equilibrium concentrations and calculate Keq
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ICE tables
eg. 4.30 mol of NH3 was placed in a 1.00 L closed container to establish this equilibrium:
2 NH3(g) ⇌N2(g) + 3 H2(g)
Calculate Kc if the equilibrium concentration of H2(g) = 0.500 mol/L
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ICE tables
2 [NH3] [N2] 3 [H2]
I
C
E
4.30 mol/L 0 0
-2x +x +3x
4.30 - 2x x 3x
0.500 mol/L
2x x 3x
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[H2] = 0.500 mol/L3x = 0.500 x = 0.167
[N2] = x
= 0.167 mol/L
[NH3] = 4.30 - 2x
= 3.97 mol/L
K = (0.500)3 x (0.167)
(3.97)2
= 0.00132
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ICE tableseg. 2.00 mol of N2, 4.00 mol of H2 and
3.00 mol of NH3 were allowed to come to equilibrium in a 1.00 L container
2 NH3(g) ⇌N2(g) + 3 H2(g)
Calculate Kc if the equilibrium concentration of NH3(g) = 3.50 mol/L
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ICE tables
2 [NH3] [N2] 3 [H2]
I
C
E
3.00 2.00 4.00
2x x 3x
3.00 + 2x 2.00 - x 4.00 - 3x
+2x -x -3x
= 3.50
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[H2] = 4.00 – 3x
= 3.25
[N2] = 2.00 - x
= 1.75 mol/L
[NH3] = 3.50 mol/L
3.00 + 2x = 3.50
x = 0.25
K = (3.25)3 x (1.75)
(3.50)2
= 4.90
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ICE tableseg. The oxidation of ammonia occurs
according to the following expression:
4 NH3(g) + 5 O2(g) ⇌4 NO(g) + 6 H2O(g)
0.800 mol of each chemical were placed in a 1.00 L container and there was 0.450 mol of NH3 at equilibrium. Calculate the equilibrium concentrations and Kc
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4 NH3(g) 5 O2(g) 4 NO(g) 6 H2O(g)
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[NH3] = 0.450 mol/L
0.800 – 4x = 0.450
x = 0.0875
[O2] = 0.3625 mol/L
[NO] = 1.15 mol/L
[H2O] = 1.325 mol/L
K = 37000
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ICE tables
eg. 2.30 mol of NH3 was placed in a 2.00 L closed container to establish this equilibrium:
2 NH3(g) ⇌N2(g) + 3 H2(g)
Calculate Kc if 25 % of the NH3(g) reacts.
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[NH3] = 2.30 mol/2.00 L = 1.15 mol/L
2 [NH3] [N2] 3 [H2]
I
C
E
1.15 0 0
2x x 3x
1.15 - 2x x 3x
-2x +x +3x
= (0.75 x 1.15) = 0.8625 mol/L
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[NH3] = 0.8625
1.15 - 2x = 0.8625
x = 0.1438
[N2] = x
= 0.1438 mol/L
[H2] = 3x
= 0.4314 mol/L
K = 0.0155
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ICE tables
eg. 0.800 mol of NH3 was placed in a 1.00 L closed container to establish this equilibrium:
2 NH3(g) ⇌N2(g) + 3 H2(g)
Calculate Kc if 70% of the NH3 reacts.
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Le Châtelier’s Principle
Chapter Review
p. 535 #’s 2, 3, 5-7
p. 536 #’s 13, 15
p. 538 #’s 7, 8, & 10 (MC)
p. 539 #’s 15, 17, 19, 20, 22
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ICE tables
Assignment
test
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ΔH & equations The energy term may be included in a
chemical equation, or written as ΔH to the right of the equation.
eg. CO(g) + 2 H2(g) ⇌ CH3OH(g) + 65 kJ
OR
CO(g) + 2 H2(g) ⇌ CH3OH(g) ΔH = - 65 kJ
EXOTHERMIC
Energy is PRODUCED
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ΔH & equations
eg. N2(g) + O2(g) + 90 kJ ⇌ 2 NO(g)
OR
N2(g) + O2(g) ⇌ 2 NO(g) ΔH = + 90
kJ
Energy is REQUIRED ENDOTHERMIC
Back to Temperature
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06:36 PM
ΔH & equations The energy term may be included in a
chemical equation, or written as ΔH to the right of the equation.
eg. CO(g) + 2 H2(g) ⇌ CH3OH(g) + 65 kJ
OR
CO(g) + 2 H2(g) ⇌ CH3OH(g) ΔH = - 65 kJ