TOPIC 9: KINETICS AND EQUILIBRIUM

TOPIC 9: KINETICS AND EQUILIBRIUM
Part 1 – Kinetics Part 2 – General Equilibrium Part 3 – Solution Equilibrium Part 4 – Thermodynamics

PART 1 – AIMS What is kinetics?
What factors affect the rate of chemical reactions? How can we classify energy in chemical reactions? How can we interpret Potential Energy Diagrams?

PART 2 – AIMS What is equilibrium? What is the equilibrium expression?
How can we calculate K and Kp? What are external factors that affect equilibrium?

PART 3 - AIMS How can we determine and use solubility product constants Ksp? How can we estimate salt solubility from Ksp? How can we determine the formation of precipitates? What is the common ion effect?

PART 4 – AIMS What is entropy? How can we calculate ΔHformation?
How can we calculate bond energies? How can we use Gibbs Free Energy to predict spontaneity?

AIM: WHAT IS KINETICS? Kinetics: the branch of chemistry that deals with the rates of chemical reactions Collision Theory: In order for a reaction to occur, reactants must collide with each other An effective collision is when reactants come together with the correct amount of energy and in the correct position to form a product Example: Fender Bender vs. Head On Collision

AIM: What Factors Affect the Rate of a Reaction?
As the amount of effective collisions increases, the faster products are formed (reaction rate increases) How can we increase the reactions rate (increase the # of effective collisions??

FACTORS THAT AFFECT THE RATE OF CHEMICAL REACTIONS Nature of Reactants Concentration Surface Area Pressure Temperature Catalyst

NATURE OF REACTANTS: Reactions involve the breaking of old bonds and the formation of new bonds. In general: Covalent bonds are slower to react than ionic bonds Breaking more bonds requires more energy than making bonds during collisions (Table I Reactions 1-6)

CONCENTRATION: Generally: Increase concentration increase rate of reaction (especially if volume is decreased)

SURFACE AREA: Generally: the more surface area that is exposed the more chances there are for collisions (effective collisions) and will increase rate of reaction Demo: Lycopodium lump vs. lycopodium poder Sugar lump vs. granulated sugar Alka Seltzer tablet vs. ground tablet

PRESSURE: No effect on solids and liquids, only gases Increasing pressure, decreases the volume, increasing the rate of effective collisions – increases the rate of reaction Demo Example: Poston/Blocking Syringe

TEMPERATURE: Generally: Increasing temperature increases kinetic energy of molecules, leading to an increase in the amount of effective collisions – increasing the rate of reaction Demo examples: Dissolving copper sulfate crystals in cold water and hot water Glow stick in hot water and cold water – hot water will get really bright fast and then burn out increasing the temperature increases the rate of the reaction, in cold water it will stay dim but will glow longer then the one in hot water since we slowed down the rate of the reaction

CATALYST Addition of a catalyst increases the rate of the reaction by providing a different and easier pathway for the reaction

AIM: HOW CAN WE CLASSIFY ENERGY IN CHEMICAL REACTIONS?
Look at the reactions (where is E term?) CH4 + 2O CO2 + 2H2O kJ N2 + O kJ NO2 Look at the H = heat of the reaction (Table I) Potential Energy Diagrams H = PE products – PE reactants 3 WAYS TO CLASSIFY ENERGY IN CHEMICAL REACTIONS

AIM: HOW CAN WE INTERPRET POTENTIAL ENERGY DIAGRAMS?
Show how the potential energy of reactant particles changes to chemical potential energy stored in bonds PE diagrams keep track of PE changes during a chemical reaction in stages

For Example: A + B AB The forward reaction is read from left to right Compare the potential energy of the reactants to the potential energy of the products in the forward reaction (PE diagram #1) PE REACTANTS: 25 Joules PE PRODUCTS: 75Joules *Energy must have been absorbed during the reactions PE diagrams with this pattern represent endothermic reactions

Label 1ST line: H ( -) 2nd line: exothermic reaction FORWARD REACTION!!

Lets Label the Diagram PE of Reactants PE of Products H (heat of reaction) = H = PE products – PE reactants If H is positive [PE products  PE reactants] – endothermic If H is negative [PE products  PE reactants] – exothermic Table I – shows different chemical reactions and H for each one

Label the Diagram PE of Activated Complex – intermediate molecule that forms when reactants have an effective collision. It is unstable and temporary Activation Energy of forward reaction – amount of energy needed to start the reaction in the forward direction Activation Energy (with catalyst) – amount of energy needed to start the reaction if a catalyst is added * Catalysts – speed the reaction rate by lowering the activation energy needed to start a reactions (gives an alternate pathway)

For Example: A + B AB Compare the potential energy of the reactants to the potential energy of the products in the reverse reaction (PE diagram #2) PE REACTANTS: 75 Joules PE PRODUCTS: 25Joules *Energy must have been released during the reactions PE diagrams with this pattern represent exothermic reactions

1st line H (+) 2nd line Endothermic

Which diagram is endothermic? Exothermic?

TABLE I – Gives us ΔH values for reactions
PRACTICE USING TABLE I: A negative value means that the reaction is exothermic If we were to rewrite the equation with the heat of reaction shown it would looks like this 𝐂 𝐇 𝟒 𝐠 +𝟐 𝐎 𝟐 →𝐂 𝐎 𝟐 𝐠 +𝟐 𝐇 𝟐 𝐎 𝐥 +𝟖𝟗𝟎.𝟒𝐤𝐉 It is added to the right side of the equation because it is an exothermic reaction in which heat is released as a product

TABLE I – Gives us ΔH values for reactions
PRACTICE USING TABLE I: A positive value means that the reaction is endothermic If we were to rewrite the equation with the heat of reaction shown it would looks like this 𝐍 𝟐 𝐠 + 𝐎 𝟐 +𝟏𝟖𝟐.𝟔𝐤𝐉 →𝟐𝐍𝐎(𝐠) It is added to the left side of the equation because it is an endothermic reaction in which heat is absorbed as a reactant

EQUILIBRIUM AIM: What is equilibrium?
Equilibrium: when the forward and reverse reactions occur at the same rate – it is a state of balance Dynamic Equilibrium: The motion in which the interactions of reacting particles are balanced by the interaction of product particles Reversible Equilibrium: Many reactions in equilibrium are considered reversible. This is indicated by a double arrow

Physical Equilibrium: the changes that take place in chemical reactions during physical processes such as changes of state or dissolving. Phase Equilibrium: Equilibrium between phases Solid and liquid phase – during melting the rate of dissolving is equal to the rate of crystallization in a closed container (system) (H20 (s) H2O (l)) Closed system – system in which neither reactants not products can leave Equilibrium does not mean that the amounts are the same but that the RATES are the same

2. Liquid and Gas phase – during this phase the rate of evaporation is equal to the rate of condensation in a closed container (H20 (l) H2O (g))

EQUILIBRIUM AIM: How Can We Examine Various Systems at Equilibrium?
Solution Equilibrium: 1. Solid/Liquid Solution – saturated solutions are examples of solid/liquid solution equilibrium in a closed system (C12H6O11 (s) / C12H22O11 (aq) 2. Gas/Liquid Solution – in a closed system or container, there is equilibrium between the gaseous and dissolved state of the gas

The equilibrium position – whether the reaction lies far to the right or to the left depends on three main factors: 1. The initial concentrations (more collisions – faster reaction) 2. Relative energies of reactants and products (nature goes to minimum energy) 3. Degree of organization of reactants and products (nature goes to maximum disorder) 4. Significance of K K>1 means that the reaction favors the products at eq. K<1 means that the reaction favors the reactants at eq.

AIM: What is the equilibrium expression?
Mathematical expression that shows the relationship of reactants and products in a system at equilibrium is called the equilibrium expression It is a fraction with the concentrations of reactants and products expressed in moles per liter Each concentration is then raised to the power of its coefficient in a balanced equation. The expression equals a value called the equilibrium constant Keq, which remains the same for a particular reaction at a specified temperature

To write an equilibrium expression follow these steps: Write a balanced equation for the system. Place the products as factors in the numerator of a fraction and the reactants as factors in the denominator, Place a square bracket around each formula. The square bracket means molar concentration. Write the coefficient of each substance as the power of its concentration. The resulting expression is the equilibrium expression, which should be set equal to the Keq for that reaction.

Ex) Write the equilibrium expression for the equilibrium system of nitrogen, hydrogen, and ammonia. ___N2 (g) + ___H2 (g) <-> ___NH3 (g)

The equilibrium constant is a specific numerical value for a given system at a specified temperature Changes in concentrations will not cause a change in the value of Keq, nor will the addition of a catalyst. Only temperature will change the value PURE SOLIDS AND LIQUIDS ARE NOT INCLUDED IN THE EQUILIBRIUM EXPRESSION (g) and (aq) are.

AIM: How can we calculate K and Kp?

AIM: How can we calculate K and Kp?
Can you… … write an equilibrium constant expression? … tell how to find K for a summary equation? … explain what K is telling you about a reaction?

AIM: What are external factors that affect a reaction at equilibrium?
If a stress is applied to a system at equilibrium the equilibrium will shift to release the effects of the stress Stressors: Temperature Concentration Pressure

Le Chateliers Principle TEMPERATURE
Increase in temperature favors the endothermic reaction Decrease in temperature will favor the exothermic reaction * *The side you shift towards will increase and the side you shift away from will decrease** NOTE: Only temperature can disturb equilibrium and cause a change. The system will shift to establish a new equilibrium Demonstration: Equilibrium involving Cobalt (II) Chloride ions- Temperature: MATERIALS: 1. Prepare a 0.2M cobalt chloride solution but dissolving 2.6 g of CoCl2- 6H2O in 100mL of distilled water 2. Hot water bath 3. Ice water bath Cobalt equilibrium system is represented by the following equation CoCl2 * 6H2O <-> CoCl2 + 6H2O ΔH = + 50kJ (PINK) (BLUE) FORWARD ENDO REVERSE EXO PROCEDURE : Add some pink from the system to a test tube and place it into the hot water bath, or carefully warm the solution using a Bunsen burner. The solution will turn blue Place the blue test tube in the ice water bath; it will return back to its pink color EXPLANATION: - THE FORWARD REACTION IS AN ENDOTHERMIC REACTION. According LeChateliers Principle adding heat favors the endothermic direction. Thus warming the system will favor the blue cholor complex side of the reactions. When placed in the ice bath, the exothermic direction will now be favored, shifting to the pink hydrated side

Le Chateliers Principle CONCENTRATION
Increase conc. of reactants Favors FORWARD reaction Speeds up the forward direction SHIFTS RIGHT Decrease conc. of reactants Favors REVERSE reaction Speeds up the REVERSE direction SHIFTS LEFT Increase conc. of products Favors REVERSE reactions Speeds up the REVERSE direction Decrease conc. of the products Speeds up the FORWARD direction Demonstration: Equilibrium involving Cobalt (II) Chloride ions- Concentration MATERIALS: Prepare a 0.2M cobalt chloride solution but dissolving 2.6 g of CoCl2- 6H2O in 100mL of distilled water 12M HCl Cobalt equilibrium system is represented by the following equation CoCl2 * 6H2O <-> CoCl2 + 6H2O ΔH = + 50kJ (PINK) (BLUE) FORWARD ENDO REVERSE EXO PROCEDURE : Begin with 50mL of the pink hydrated form of the cobalt ion in a larger 250mL Erlenmeyer flask Carefully add HCl to the solution until the blue form of the equilibrium system appears Add distilled water to the solution in the beaker or flask. The equilibrium will shift back to the pink form EXPLANATION: When HCl is added to the pink hydrated form of the cobalt ion, the concentration of Cl- increases (H+ remains in solution as a spectator ion) The increase in Cl- causes the equilibrium to shift to the product side, and the solution turns blue Adding water will shift the equilibrium to the left, forming more pink hydrated cobalt ion. Q? What does this pink color tell us? A: have reactant in the test tube **add HCl to test tube see color change ** Q? Why did adding HCl turn the solution blue? A: More product was formed Conclusion: changing the concentration of the reactant or product will affecgt the direction of equilibrium Notes: adding HCl – increase in Cl ions

CONCENTRATION ADD AWAY (increase) TAKE TOWARDS (decrease)
Pen/pencil will go up on which ever side you shift towards Trick only some students will like/use this method --- can copy it down in their notes if they choose to

PRESSURE Need to know how many gas molecules are on the reactant side and on the product side Equal # of gas molecules – pressure will have no effect!! Increase pressure – shift from more gas molecules  towards less gas molecules Decrease pressure – shift from less  toward more

PRESSURE Ex) 4NH (g) + 5O2(g) 4NO(g) + 6H2O(g) = 9 gas molecules 4+6 = 10 gas molecules Increase pressure: shift to the left (more to less) 9 10 Decrease pressure: shift to the right (less to more) 9 10

AIM: What are the conditions at equilibrium?
At equilibrium the rate of the forward reaction is equal to the rate of the reverse reaction Concentrations are constant not equal Adding a catalyst would speed up the forward and reverse reactions to the same extent

AIM: How can we determine and use solubility product constants? Ksp
A special case of equilibrium involving dissolving Because the constant is a product of a solubility, we call it the solubility product constant Ksp

SOLUBILITY PRODUCT PROBLEMS: Given Ksp, find solubility Given solubility, find Ksp Predicting precipitation Find solubility in a solution with a common ion

Solid  Positive Ion + Negative Ion Mg(NO3)2  Mg NO3- Keq = [Mg2+ ] [NO3-]2

EX) Write out the equilibrium law expression: BaF2(S)  Ba+2(aq) + 2F-(aq) Ksp = [Ba+2][F-]2

SOLUBILITY GENERALIZATIONS (TABLE F) All nitrates are soluble All compounds of the alkali metals are soluble (Li, Na, K, etc.) All compounds of the ammonium (NH4+) are soluble

AIM: How can we estimate salt solubility from Ksp?
Need to set up RICE tables R- Reaction I – initial concentration C – Change in concentration E – Concentration at equilibrium

AIM: How can we estimate salt solubility from Ksp?
Ex) What is the solubility of silver bromide (Ksp = 5.2 x 10-13) AgBr  Ag+ + Br- Let x = the solubility (x)(x) = 5.2 x 10-13 X2 = 5.2 x 10-13 X = 7.2 x 10-7\

Ex) What is the solubility of PbI2 (Ksp = 7.1 x 10-9)

AIM: How can we determine the formation of precipitates?
Step 1: Write out the dissolving equations Step 2: Determine the most likely precipitate and write out its equation Step 3: Determine the molar concentrations and calculate the reaction quotient (Q) The reaction quotient (Q) - The product of the Ksp equation using the ion concentration before any reaction interaction If Q > Ksp then a precipitate will form

Ex) A student mixes mole Ca(NO3)2 in 2 liters of 0.10M Na2CO3 solution. Will a precipitate form?

Ex) 0.15 moles of AgNO3 is mixed with 5 liters of .02M NaCl solution. What is the most likely precipitate and will it form?

AIM: What is the common ion effect?
The common ion effect is used to describe the effect on an equilibrium involving a substance that adds an ion that is a part of the equilibrium. The common ion effect is responsible for the reduction in the solubility of an ionic precipitate when a soluble compound is combining one of the ions of the precipitate is added to the solution in equilibrium with the precipitate.

When AgNO3 is added to a salt solution of AgCl it is described as a source of a common ion, Ag+ ion. Common ion – ion that enters the solution from 2 different sources Common ion effect can be used to make an “insoluble” salt even less soluble

Ex) Calculate solubility of CaF2 in a M solution of NaF

AIM: What is enthalpy? Enthalpy (H) – flow of energy (heat exchange) at constant pressure when two systems are in contact Measure only the change in enthalpy ΔH (the difference between the potential energies of the products and reactants) Exothermic reactions are favored ΔH= - Δ H = q at constant pressure - open container

AIM: How can we calculate ΔH ?
Δ H can be calculated from several sources including: Stoichiometry Calorimetry q=mcΔT From tables of standard values Hess’s Law Bond Energies

STOICHIOMETRY:

CALORIMETRY:

TABLES:

HESS’S LAW ΔHrxn = ΣΔHf(products) – ΣΔHf(reactants)

HESS’S LAW

AIM: How can we calculate bond energies ?
Energy must be added/absorbed to BREAK bonds (endothermic) in order to overcome the attractive forces between each nuclei and the shared electrons Energy is released when bonds are FORMED (exothermic) because resulting attractive forced between the bonded atoms lowers potential energy causing a release.

**BARF** ΔH = Σ Bond Energies (broken) – Σ Bond Energies (formed)

Ex)

SUMMARY FOR ENTHALPY What does ΔH tell you about the changes in energy regarding a chemical reaction? ΔH = + reaction is endothermic and heat energy is added into the system ΔH = - reaction is exothermic and heat energy is lost from the system (Nature tends to favor the lowest possible energy state!)

AIM: What is entropy? ENTROPY ΔS: disorder or randomness of the matter and energy of a system (more disordered/dispersal is favored) Nature favors CHAOS (high entropy low energy)

AIM: What is entropy? Thermodynamically favored processes or reactions are those that involve a decrease in internal energy of the components (ΔH<0) and increase in entropy (ΔS >0) These are spontaneous or thermodynamically favored

AIM: What is entropy? ΔS is + when dispersal/disorder increases (favored) ΔS is – when dispersal/disorder decreases NOTE: Units are usually J/(molrxn • K) (not kJ!) ΔSrxn = ΣΔSf(products) – ΣΔSf(reactants)

AIM: What is entropy? Ex)Predict which has the largest increase in entropy: CO2(s)  CO2(g) H2(g) + Cl2(g)  2HCl(g) KNO3(s)  KNO3(l) C(diamond)  C(graphite)

AIM: What is entropy? Ex)

AIM: How can we use Gibbs free energy to predict spontaneity?
The calculation of Gibbs free energy, ΔG is what ultimately decides whether a reaction is thermodynamically favored or not A NEGATIVE sign on ΔG indicates that a reaction is thermodynamically favored (spontaneous) Several ways to calculate ΔG that links thermochemistry, entropy, equilibrium, and electrochemistry together!

ΔG = ΔH – TΔS

Ex)

SUMMARY Ex) If ΔG is NEGATIVE, the reaction is thermodynamically favorable If ΔG is ZERO, the reaction is at equilibrium and If ΔG is POSITIVE, the reaction is NOT thermodynamically favorable

SUMMARY Ex)

AIM: How can we determine if a reaction is spontaneous?
Spontaneous reactions: DOESN’T REQUIRE EFFORT (proceed on their own without intervention) Ex) ice melting, nuclear reactions- natural transmutation WILD FIRES- dry and hot in southern california AIM: How can we determine if a reaction is spontaneous?

2 Conditions for a reaction to be considered spontaneous 1. Tendency toward lower energy (PE) exothermic ΔH = (-) Table I * products have lower energy and more stable

2. Tendency toward randomness ΔS = (+) Entropy  measure of randomness or disorder Physical changed and entropy : Solid Liquid Gas Intermediate Entropy Low Entropy High Entropy Table I (stability) greater exothermic reaction the more stable the products are ( lower PE)

Chemical changes and Entropy: Free elements (ex. O2, Na, Fe) high entropy Compounds (ex. H2O, NH3) lower entropy ***NATURE FAVORS REACTIONS THAT HAVE LOW ENERGY AND HIGH ENTROPY*** Free Energy ΔG = Lets us know if a reaction is spontaneous ΔH  (negative, exothermic) ΔS  (positive, high entropy) **always have ΔG = (-) SPONTANEOUS Gibbs Free Energy Equation ΔG = ΔH - TΔS where ΔG = Gibbs Free Energy, in kJ ΔH = heat of reaction T = temperature, in Kelvin ΔS = entropy change (in kJ · K-1)

TOPIC 9: KINETICS AND EQUILIBRIUM

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