Gases, Liquids and Solids Bettelheim, Brown, Campbell and Farrell Chapter 5.
Reaction Rates and Chemical Equilibria Bettelheim, Brown, Campbell and Farrell Chapter 7.
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Transcript of Reaction Rates and Chemical Equilibria Bettelheim, Brown, Campbell and Farrell Chapter 7.
![Page 1: Reaction Rates and Chemical Equilibria Bettelheim, Brown, Campbell and Farrell Chapter 7.](https://reader033.fdocuments.us/reader033/viewer/2022061607/56649e925503460f94b9875e/html5/thumbnails/1.jpg)
Reaction Rates and Chemical Equilibria
Bettelheim, Brown, Campbell and Farrell
Chapter 7
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Chemical Kinetics
• Chemical kinetics:Chemical kinetics: the study of the rates of chemical reactions
Example, the concentration might increase from 0 to 0.12 mol/L over a 30 minute time period
CH3-Cl I- CH3-I Cl-+ +Chloro-methane
Iodo-methane
30 min(0.12 mol CH3I/L) - (0 mol CH3I/L)
=0.0040 mol CH3I/L
min
![Page 3: Reaction Rates and Chemical Equilibria Bettelheim, Brown, Campbell and Farrell Chapter 7.](https://reader033.fdocuments.us/reader033/viewer/2022061607/56649e925503460f94b9875e/html5/thumbnails/3.jpg)
Reaction Rates
• The rates of chemical reactions are affected by several factors– Molecular collisions– Activation energy– Nature of the reactants– Concentration of the reactants– Temperature– Presence of a catalyst
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Molecular Collisions
– Two species, A and B (molecules or ions), must collide in order to react
– Most collisions do not result in a reaction– A collision that does result in a reaction is
called an effective collisioneffective collision– Effective collisions must have
• Enough energy to reach the activation energy• Correct orientation of A and B at the time of
collision
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Molecular Collisions• EEaa Activation energy: Activation energy: the minimum energy
required for a reaction to take place– Most reactions involve breaking covalent
bonds initially– Energy is required to break covalent bonds– Energy comes from the collision between A
and B
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– If the collision energy is large, there is sufficient energy to break the necessary bonds, and reaction takes place
– If the collision energy is too small, no reaction occurs
![Page 7: Reaction Rates and Chemical Equilibria Bettelheim, Brown, Campbell and Farrell Chapter 7.](https://reader033.fdocuments.us/reader033/viewer/2022061607/56649e925503460f94b9875e/html5/thumbnails/7.jpg)
Molecular Collisions• Orientation at the time of collision
– Colliding particles must be properly oriented for bond breaking and bond making
– In reaction between H2O and HCl, the oxygen of H2O must collide with the H of HCl so that the new O-H bond can form and the H-Cl bond can break
+ +
H2O + HCl H3O+ Cl-+
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Energy Diagrams
Exothermic
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Energy Diagrams
Endothermic
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Energy Diagrams
• Transition state:Transition state: maximum on an energy diagram
+ +
H2O + HCl H3O+ Cl-+
+ -
transition state
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Factors Affecting Rate
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Factors Affecting Rate• Catalyst:Catalyst: a substance that increases the rate of a
chemical reaction without itself being used up
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Reversible Reactions
• Equilibrium:Equilibrium: a dynamic state in which the rate of the forward reaction is equal to the rate of the reverse reaction– No change in concentration of either reactants
or products– Reaction is still taking place but the rates of
the two reactions are equal
forward reactionreversereaction
CO(g) + H2O(g) CO2(g) + H2(g)
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Equilibrium Constant, K
• For the general reaction
– the equilibrium constant isaA + bB cC + dD
K =[C]c[D]d
[A]a[B]b
CO(g) +H2O(g) CO2(g) +H2(g)
[CO2][H2]
[CO][H2O]K =
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Equilibrium Constant ExpressionKeq
aA + bB → cC + dD reactants products
Keq = [C]c [D]d
[A]a [B]b
For equilibrium constant expressions, we use coefficients as the powers.
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Equilibrium Constants
• Problem:Problem: when H2 and I2 react at 427°C, the following equilibrium is reached
– Equilibrium concentrations are [I2] = 0.42 mol/L, [H2] = 0.025 mol/L, and [HI] = 0.76 mol/L. Using these values, calculate the value of K
– Solution:Solution:
I2(g) +H2(g) 2HI(g)
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Equilibrium Constants
• Problem:Problem: when H2 and I2 react at 427°C, the following equilibrium is reached
– Equilibrium concentrations are [I2] = 0.42 mol/L, [H2] = 0.025 mol/L, and [HI] = 0.76 mol/L. Using these values, calculate the value of K
– Solution:Solution:
I2(g) +H2(g) 2HI(g)
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[HI]2
[I2][H2]K = = (0.76 M)2
(0.42 M) x (0.025 M)= 55
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Coefficients from different sources
Equilibrium constants:
coefficients are the powers in equation
Reaction Rate equations:
coefficients are NOT the powers
(must be determined experimentally)
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What does Keq mean?At equilibrium:
Keq > 100: mostly products present
Keq < 0.01: mostly reactants present
0.01 < Keq < 100: significant amounts of both products and reactants present
Keq value changes at different temperatures
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Equilibrium and Rates
• There is no relationship between a reaction rate and the value of K– Reaction rate depends on the activation
energy of the forward and reverse reactions; these rates determine how fast equilibrium is reached but not its position
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• LeChatelier’s Principle:LeChatelier’s Principle: when a stress is applied to a chemical system at equilibrium, the position of the equilibrium shifts in the direction to relieve the applied stress
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Le Chatelier’s Principle
When system at equilibrium is “disturbed” or “stressed” the system moves in the direction that relieves the stress
Types of “stress”:
Add or remove reactant
Add or remove a product
Change temperature
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Ways to stress system
Increase the concentration of a chemical
Decrease the concentration of a chemical
Heat the system (add heat to the system)
Cool the system (remove heat energy)
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Le Chatelier’s Principle
2 O2 + N2 ↔ 2 NO2
colorless colorless red/brown
Add O2: Turns Redder
Concentration increases on left
System makes more product to relieve stress
System “shifts” to right
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Le Chatelier’s Principle
2 O2 + N2 ↔ 2 NO2
colorless colorless red/brown
Remove O2: Turns less red
Concentration decreases on left
Some product changes to reactants to relieve stress
System “shifts” to left
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• Can treat heat as if it were a reactant or product for LeChatelier
A + B ↔ C + heat