essential question: what mechanisms affect the rates of...
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UNIT 9: KINETICS &
EQUILIBRIUM
Essential Question: What mechanisms affect the rates of
reactions and equilibrium?
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What is Kinetics?
¨ Kinetics is the branch of chemistry that explains the rates of chemical reactions
¨ Collision Theory: in order for a reaction to occur, reactant particles MUST collide ¤ based on…
n spatial orientation n energy of colliding particles
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Factors Affecting Rates of Reaction
Nature of the Reactants ¨ Since reactions involve the breaking of
existing bonds and formation of new ones… ¨ IONIC bonds are faster to react than
COVALENT bonds ¤ (since covalent require more energy to break
the higher # of bonds) ¨ Conclusion: The more IONIC the bond, the
faster the reaction rate
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Factors Affecting Rates of Reaction
Concentration ¨ Ideally, if there’s MORE of a reactant
available/at our disposal, then…. ¨ based on the kinetic molecular theory (phys.
behavior of matter...) the MORE reactant available, the MORE product able to form
¨ Conclusion: The higher the CONCENTRATION, the faster the reaction
rate
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Factors Affecting Rates of Reaction
Surface Area ¨ What will react faster à a finely divided
powder or a lump of the same mass ¨ *sugar cube v. sugar powder
¨ Since the powder has a larger surface area exposed, there are MORE chances for the reactant particles to collide ¨ Conclusion: The more SURFACE AREA, the
faster the reaction rate
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Factors Affecting Rates of Reaction
Pressure (of gases) ¨ When pressure increases, what happens to
it’s solubility, and therefore it’s concentration? ¨ (think back to solutions) à the higher the
pressure, the more soluble the gas, the higher the gaseous concentration ¨ Conclusion: The higher the PRESSURE of a
gas, the faster the reaction rate
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Factors Affecting Rates of Reaction
Catalyst ¨ Catalysts are substances that increase the
rate of reaction by providing a different & EASIER pathway for a reaction
¨ THEY REMAIN UNCHANGED BY THE REACTION after completion *think baking pan
¨ Conclusion: In the presence of a CATALYST, the reaction rate increases
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Factors Affecting Rates of Reaction
Temperature
¨ Temperature increases energy of particles (Kinetic Molecular Theory)
¨ Temperature increases motion of particles (KMT) ¨ Conclusion: The higher the TEMPERATURE,
the faster the reaction rate
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CONCLUSIONS: Rates of Reaction
Factor: 1) Nature of reactants 2) Concentration 3) Surface Area 4) Pressure 5) Catalyst 6) Temperature
Increases Rate: à ionic MORE v. covalent à with ñ concentration à with ñ Surface Area à with ñ Pressure (GAS) à presence of Catalyst à with ñ Temperature
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¨ Illustrate the changes in potential energy that occur during a chemical reaction.
¨ REACTION COORDINATE is the horizontal axis ¤ shows DIRECTION or
progress of reaction
¨ Potential because as reactant particles approach each other, KINETIC energy is converted into POTENTIAL energy
Potential Energy Diagrams
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Reactants Activated Complex Products PE of Reactants Activation Energy PE of the Activated Complex Heat of Reaction PE of Products
1 à 2 à 3 à A à B à C à D à E à
Potential Energy Diagrams
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¨ Activated Complex ¤ temporary,
intermediate; highest PE of system before reaction COMPLETES
¨ Activation Energy ¤ Amount of energy
needed to form the activated complex from the reactants
Potential Energy Diagrams
¨ Heat of Reaction (ΔH) ¤ Difference between PE of
Reactants & Products
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Potential Energy Diagrams
1 à 2 à 3 à
PE of Reactants PE of Activated Complex PE of Products
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Potential Energy Diagrams
4 à 5 à 6 à
Activation Energy (Forward) HEAT of Reaction (ΔH) Activation Energy (Reverse)
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FORWARD REACTION Act. En. NO catalyst Act. En. WITH catalyst
4 à 7 à
P. Energy Diagrams with CATALYSTS
6 à 8 à
REVERSE REACTION Act. En. NO catalyst Act. En. WITH catalyst
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¨ Major difference in both curves?? ¤ Activation Energy needed
¨ Major constant in both curves?? ¤ Heat of Reaction (ΔH) = #5
P. Energy Diagrams with CATALYSTS
Comparing Catalyzed &
Un-Catalyzed Reactions
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Endothermic V. Exothermic
Characteristics ¨ Heat of Reaction (ΔH) =
+ (POSITIVE) ¨ curve starts at a LOWER
P.E. (reactants) & ends at a HIGHER P.E. (products)
¨ ABOSORBED energy, aka reactant “+ heat” (kJ)
¨ see table “I” *ENDOTHERMIC
Reaction*
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Endothermic V. Exothermic
Characteristics ¨ Heat of Reaction (ΔH) =
- (NEGATIVE) ¨ curve starts at a HIGHER
P.E. (reactants) & ends at a LOWER P.E. (products)
¨ RELEASED energy, aka product “+ heat” (kJ)
¨ see table “I” *EXOTHERMIC
Reaction*
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¨ A chemical reaction in a state of equilibrium is said to have both the forward and the reverse reactions occurring at the same time
¨ RATES are equal, not reactant/product quantities!
What is EQUILIBRIUM?
¨ CAN ONLY OCCUR in a system in which neither the reactants nor the products can leave the system
¨ à CLOSED SYSTEM
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¨ Occurs during PHYSICAL processes à dissolving, change of state
Physical Equilibrium ~ Phases
¨ solid/liquid: water & ice exist at same time!
¨ H2O (s) H2O (l) ¨ rate of melting equal
to rate of freezing
¨ liquid/gas: water & vapor exist at same time!
¨ H2O (l) H2O (g) ¨ evaporation rate equal to
rate of condensation
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¨ Occurs when a solution is saturated! (Gas or Liquid)
Physical Equilibrium ~ Solutions
¨ saturated: no more solute can “dissolve”
¨ C12H22O11 (s) C12H22O11 (aq) ¨ process of dissolving STILL taking
place while recrystallization occurs
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¨ Reactants are mixed to FORM products (which DON’T exist…yet)
¨ THEN, the concentrations of reactants DECREASE while producing (or INCREASING) products
Ex) CH4 (g) + H2O (g) à 3H2 (g) + CO (g) decreasing à increasing
Chemical Equilibrium
¨ THEN: rate of REVERSE reaction will now increase, until RATES of BOTH reactions become EQUAL = EQUILIBRIUM
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¨ ANY change in temp, concentration, or pressure on an equilibrium system is called a stress
¨ Le Châtelier’s Principle explains how a system at equilibrium responds to relieve any stress on it!
Concentration Changes: (of rctnts/pdts) STAYS constant
CH4 (g) + H2O (g) 3H2 (g) + CO (g)
¨ What will happen if we INCREASE CH4?? ¤ reaction will go TO THE RIGHT (forward), USE the higher
concentration of CH4 available, and create more Products (H2 & CO)
¤ THEN: reaction will keep “oscillating” direction (forward & backward) until it reaches Equilibrium!
Le Châtelier’s Principle
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4NH3 (g) + 5O2 (g) 4NO (g) + 6H2O (g) + Heat **VARYING CONCENTRATIONS**
Le Châtelier’s Principle
STRESS EFFECT SYSTEM SHIFT EFFECT EFFECT EFFECT
- NH3
+ NH3
Increase
Decrease
Increase Increase Increase
Decrease Decrease Decrease
- O2
+ O2
Decrease
Increase
+ NO + H2O + heat
- NO - H2O - heat
à
ß
AWAY from stress
TOWARD the stress
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Temperature Changes: N2 (g) + 3H2 (g) 2HN3 (g) + heat
¨ What will happen if we INCREASE heat?? ¤ reaction will go TO THE LEFT (reverse), because
HEAT is a product of the reaction & a change in temp. is essentially a change in the concentration of that product
¤ RESULT: the ENDOTHERMIC reverse reaction is favored (in this example) over the EXOTHERMIC forward reaction
Le Châtelier’s Principle
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4NH3 (g) + 5O2 (g) 4NO (g) + 6H2O (g) + Heat **VARYING TEMPERATURES**
Le Châtelier’s Principle
EFFECT EFFECT SYSTEM SHIFT EFFECT EFFECT STRESS
- NH3
+ NH3
Increase
Decrease
Decrease Decrease Increase
Increase Increase Decrease
+ O2
- O2
Increase
Decrease
- NO - H2O + heat
+ NO + H2O - heat
ß
à
AWAY from stress
TOWARD the stress
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Pressure Changes: ONLY affect gases-not liquids/solids CO2 (g) CO2 (aq)
¨ What will happen if we INCREASE pressure? ¤ concentration of gaseous CO2 increases *think
solubility rules/curve* ¤ MOVE AWAY from added stress (aka to the
RIGHT) ¨ What will happen if we DECREASE pressure??
¤ reaction shifts TO THE LEFT (toward gaseous CO2) to make MORE gas **Think SODA BOTTLE: decrease pressure (when open
bottle), dissolved gas becomes BUBBLES of gaseous CO2
Le Châtelier’s Principle
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N2 (g) + 3H2 (g) 2NH3 (g)
¨ What will happen if we INCREASE pressure NOW? ¤ concentration of ALL gases increases ¤ Reaction direction will be favored TOWARD the
side with FEWER # of gas molecules
¨ Conclusions ¤ left side (reactants) = 4 gas molecules ¤ right side (products) = 2 gas molecules ¤ Therefore: increase in pressure will favor reaction
towards products, or > amount of NH3 formed
Le Châtelier’s Principle
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N2 (g) + 3H2 (g) 2NH3 (g) ¨ What will happen if we DECREASE pressure?
¤ Reaction direction will be favored TOWARD the side with GREATER # of gas molecules
¨ Conclusions: decrease in pressure will favor reaction towards reactants, or > amount of N2 & H2 formed and reduce amount of NH3
What if a rxn has the same # gas molecules on both sides??? ¨ NO EFFECT! What if a rxn has a catalyst???
¨ changes rate of both forward/reverse rxns EQUALLY
Le Châtelier’s Principle
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¨ Exothermic reactions move toward lower energy state ¤ energy contained in the reactants is RELEASED ¤ the products have less P.E. than the reactants
¨ Exothermic reactions (lower enthalpy) are more likely than endothermic because less activation energy necessary
Enthalpy
¨ The tendency in nature to change to a state of LOWER energy
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¨ The tendency in nature to change to a state of greater CHAOS…DISORDER… RANDOMNESS
¨ the greater the disorder, the higher the Entropy ¨ systems will often go from conditions of >order (low
entropy) to conditions of > disorder (high entropy) ¤ phase changes (solid à liquid à gas) ¤ compounds v. elements (High # of molecules =
greater entropy) Low Slight High
Entropy
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¨ Mathematical expression that shows the relationship of reactants and products in a system at equilibrium
¨ Keq = equilibrium constant
How to write the FORMULA: Write the equilibrium expression for the equilibrium system of 0.5M nitrogen (N2), 0.3M hydrogen (H2), and 1.5M ammonia (NH3). 1. Write a balanced equation for the system à N2 (g) + 3H2 (g) 2NH3 (g) + heat
The Equilibrium Expression
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2. Place products as factors in numerator and reactants as factors in denominator of a fraction
à NH3 H2 x N2
3. Place a square bracket [ ] around each formula. à [NH3] = this means molar concentration (M)
[H2] [N2]
4. Write the coefficient of each substance as a POWER of its concentration, then label Keq
Keq = [NH3]2
[H2]3 [N2]
The Equilibrium Expression
= [0.5]2
[0.3]3 [1.5] Keq = 6.17
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¨ SPECIFIC for a specific temperature ¨ therefore: changes in concentration &
catalysts will NOT change the value of Keq
¨ Keq is LARGE when numerator > denominator ¤ MORE products than reactants = products favored
¨ Keq is SMALL when denominator > numerator
¤ LESS products than reactants = reactants favored
The Equilibrium Expression