UNIT 9: KINETICS &

EQUILIBRIUM

Essential Question: What mechanisms affect the rates of

reactions and equilibrium?

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

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

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

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

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

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

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

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

¨  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

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

¨ 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

Potential Energy Diagrams

1 à 2 à 3 à

PE of Reactants PE of Activated Complex PE of Products

Potential Energy Diagrams

4 à 5 à 6 à

Activation Energy (Forward) HEAT of Reaction (ΔH) Activation Energy (Reverse)

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

¨  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

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*

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*

¨  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

¨  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

¨  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

¨  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

¨  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

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

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

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

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

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

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

¨  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

¨  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

¨  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

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

¨ 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