Unit 7: Solutions, Kinetics, and Equilibrium

Unit 7: Solutions, Kinetics, and Equilibrium

I. What is a solution? Heterogeneous Mixture – Homogenous Mixture – Solutions are examples of _______________ mixtures Solution: Solute: Solvent: Substances in which particles are NOT uniform (example - soil) Substances in which particles are uniform (example: aqueous (aq) solutions) HOMOGENOUS Homogenous mixture of substances in the same physical state Substance that is being dissolved Salt Water (solute) Substance that dissolves the solute (solvent)

Check for Understanding
What type of bonding is going on in NaCl? What type of compound is NaCl? What type of bonding is going on in H2O? What type of compound is going on in H2O? What type of molecule is H2O? When NaCl is added to H2O, what happens to the Na+ and Cl- ions with respect to the H2O molecules? ionic ionic Polar covalent molecular polar Positive Na ions are attracted to the negative pole (O) in the water molecule Negative Cl ions are attracted to the positive pole (H’s) in the water molecule

I. What is a solution? POLAR IONIC + and - the polar solvent
Dissolving and Dissociation: Salt (NaCl) and Water NaCl (s)  Na+ (aq) + Cl- (aq) Dissolving occurs when the solvent is ______________________ Dissociation is when ____________ compounds’ ions get pulled apart by the ________________ end of _____________: POLAR IONIC + and - the polar solvent DISSOLVING ANIMATION

I. What is a solution? Atmosphere (O2, N2, CO2) Antifreeze
Types of Solutions: - Gas with gas: Liquid with liquid: Gas with Liquid: Solid with liquid: Sold with Solid: Atmosphere (O2, N2, CO2) Antifreeze Soda (CO2 (g) in H2O (l)) Salt water (NaCl (s) in H2O (l)) Alloys: Mixture of metals (brass – copper and zinc)

II. Solubility Factors Amount of solute that will dissolve in solvent
Nature of Solute and Solvent (video demo): - Amount of solute that will dissolve in solvent “likes dissolves likes” Solution Type Nonpolar Solvent Polar Solvent Nonpolar Polar Ionic Soluble insoluble insoluble soluble insoluble soluble

II. Solubility Factors 2. Temperature:
Solids: ________ temperature, _________ solubility (more solid will dissolve) Gases: ________ temperature, _________ solubility 3. Pressure: Only effects gases in liquids Gases: ________ pressure, ___________ solubility - increase increase decrease increase increase increase

III. Factors affecting rate of solution
increase increase 1. Size of particles: (_______ surface area, ________ rate of solution) 2. 3. Amount already dissolved: (_______ dissolved, ________ rate) 4. Temperature: Solids and Liquids: _________ temperature, _______ rate - Gases: __________ temperature, ___________ rate - Stirring – more surface area brought into contact more slower increase increase increase decrease

IV. Expressing concentrations
A. Saturated: - Solution that contains the maximum amount of solute at a given temperature Saturated solutions are in equilibrium (rate of dissolving = rate of recrystalization)

IV. Expressing concentrations
B. Unsaturated: Solution is holding ______________________ If more solute is added, ____________ C. Supersaturated: Solution is holding ____________________ Unstable – ____________________________________ LESS solute than maximum it will dissolve more solute than maximum Addition of more solute will bring it back to a saturated solution Supersaturated solution video

V. Solubility Curves (Found on Table ___ )
G Table G Shows: mass of solute that will dissolve in 100 g (or 100 mL) of water at different temperatures Dilute: __________ amount of solute ______________ Concentrated: __________ amount of solute ____________ What is a salt? SMALL DISSOLVED LARGE DISSOLVED AN IONIC COMPOUND (POSITIVE METAL ION WITH NEGATIVE NONMETAL OR PAI)

G How to use the chart: Saturated will be ______________ Unsaturated will be ____________ Supersaturated will be ___________ ON THE LINE BELOW THE LINE ABOVE THE LINE

G Table G Practice Questions 1. A solution contains 14 g of KCl in 100. g of water at 40 oC. What is the minimum amount of KCl that must be added to make this saturation solution? 2. How many grams of the compound KCl must be dissolved in 200. g of water to make a saturated solution at 60oC? 3. Identify the following as being saturated, unsaturated, or supersaturated: A) 20 oC and 20 g of KNO3 B) 40 oC and 20 g of KClO3 C) 90 oC and 10 g of NH3 D) 50 oC and 55 g of NH4Cl 4. Ninety grams of NaNO3 is added to 100. G of H2O at 0oC. With constant stirring, to what temperature must the solution be raised to produce a saturated solution with no solid remaining? 39 g – 14 g - = 25 g 45 g X 2 = 90 g unsaturated Supersaturated saturated supersaturated Approx. 22oC

VI. How to determine if a compound is soluble (Found on Table ___ )
Ionic compounds can or cannot be soluble (dissolve in water). If it is soluble the phase is ___________, if it is insoluble the phase is ___________ Use Table _____ to identify the phase: Examples: 1) PbCO3______ 2) KNO3 _______ 3) Li2S _______ aqueous (aq) solid (s) F aq s aq

VII. Concentration Molarity: Equation (See Ref. Tabs.) Tells how much solute is dissolved in a given amount of solvent # of moles of solute in 1 L of solution Molarity = moles of solute liters of solution M = mol L

VII. Concentration Examples: What is molarity of a solution that contains 4.0 mol of NaOH in L of solution? 2. Calculate the molarity of 2.0 moles of HCl dissolved in 500. mL solution. mol = 4.0 mol NaOH V = 0.50 L M = ? mol = L 4.0 mol NaOH = 8.0 mol/L (M) Molarity (M) = 0.50 L mol = 2.0 mol HCl V = 500. mL = L M = ? mol = L 2.0 mol HCl M = = M 0.500 L

VII. Concentration Step 1: GFM = 1(40g/mol) + 2(14g/mol) +
Molarity from Grams: If you are given grams instead of mole, follow the steps: Step 1: Convert grams to moles by finding the GFM Step 2: Use the molarity equation Example 1: What is the molarity of a solution containing 82.0 g of Ca(NO3)2 in 2.0 L of solution? Step 1: GFM = 1(40g/mol) + 2(14g/mol) + 6(16g/mol) = 164 g/mol 1 mol 82.0 g x = mol 164 g 0.500 mol Step 2: Molarity = mol = L = 0.25 M 2.0 L

VII. Concentration Step 1: GFM = 2(23g/mol) + 1(12g/mol) +
Molarity from Grams: If you are given grams instead of mole, follow the steps: Step 1: Convert grams to moles by finding the GFM Step 2: Use the molarity equation Example 2: There is a L solution with 53 g of Na2CO3 completely dissolved. What is the molarity of the solution? Step 1: GFM = 2(23g/mol) + 1(12g/mol) + 3(16g/mol) = 106 g/mol 1 mol = 0.50 mol 53 g x 106 g 0.50 mol Step 2: Molarity = mol = L = 2.0 M 0.25 L

VII. Concentration M = mol/L  mol = (M)(L) = (0.250 M) (2.00L)
How to make a solution: What mass of sodium carbonate is required to prepare 2.00 L of M Na2CO3 solution? Step 1: Find out how many moles are needed Step 2: Convert moles to grams M = mol/L  mol = (M)(L) = (0.250 M) (2.00L) = mol Na2CO3 GFM = 2(23g/mol) + 1(12g/mol) + 3(16g/mol) = 106 g/mol 106 g Na2CO3 0.500 mol Na2CO3 x = 53.0 g Na2CO3 1 mol Na2CO3

VII. Concentration Parts Per Million (PPM): Equation (See Ref. Tabs.) Ratio between mass of solute and total mass of solution PPM = mass of solute* 1,000,000 mass of solution

VII. Concentration PPM = mass of solute x 1,000,000 mass of solution
Example 1: Approximately g of oxygen can be dissolved in 100. mL of water at 20oC. Express this in terms of parts per million. Example 2: 2.5 grams of a groundwater solution is found to contain 5.4 x 10-6 grams of the Cu+2 ion. What is the concentration of the copper ion in ppm? 100. mL H2O = 100. g H2O PPM = mass of solute x 1,000,000 mass of solution = g O2 x 106 (100.g g) = 43 PPM PPM = mass of solute x 106 mass of solution = 5.4 x10-6 g Cu2+ ions x 106 (2.5 g) = 2.2 PPM

VIII. Colligative properties of solutions
Properties of solvent that change when a solute is added They depend on the concentration of the solute in solution: _____ ___concentration, _____ _____change Electrolytes: CaCl2 (s)  Nonelectrolytes: C6H12O6 (s)  greater greater Solutions that conduct electricity due to the free-moving ions (mostly ionic compounds) Ca +2 (aq) + 2 Cl -1 (aq) Substances formed from covalent bonding do not dissociate into ions upon entering water C6H12O6(aq)

Boiling Point Elevation: (animation) - The Higher the concentration of a solution, the higher the boiling point of the solution - The more ionic particles in a solution, the higher the boiling point of the solution MgCl2 (s) Mg2+ (aq) 2Cl- (aq) (3 particles – higher BP) NaCl (s) Na+ (aq)+ Cl- (aq) (2 particles) In 2 cups of water

Freezing Point Depression: Example: Salt is added to roads in the winter time - The higher the concentration of a solution, the lower the freezing point of the solution - The more ionic particles in a solution, the lower the freezing point of the solution MgCl2 (s) Mg2+ (aq) 2Cl - (aq) (3 particles – lower FP) NaCl (s) Na+ (aq)+ Cl - (aq) (2 particles)

Which type of solute will have the greatest effect on MP and BP, ionic or covalent? Why? - Ionic, they break apart when in water releasing more particles NaCl  Na + (aq) + Cl – (aq) (2 particles) C6H12O6  C6H12O6 (aq) (1 particle)

CHECK FOR UNDERSTANDING
1) Which of the following solutions will boil at the highest temperature? a) 100 g NaCl in 1000 g of water b) 100 g NaCl in 500 g water c) 100 g NaCl in 250 g of water d) 100 g NaCl in 125 g of water The answer is D because it has the highest concentration: Concentration = mass of solute/mass of solvent. The less solvent there is, the higher the concentration of the solution there will be. As the concentration of solution increases, the boiling point increases.

2) Which solution has the highest boiling point? a) 1.0 M KNO3 b) 2.0 M KNO3 c) 3.0 M C6H12O6 d) 2.0 M Ca(NO3)2 The answer is D because it has the most amount of particles when dissolved: 1.0 M KNO3  1 particle of K - and 1 particle of NO3 - = 2 total particles 3.0 M C6H12O6  3 particles of C6H12O6 = 3 total particles 2.0 M KNO3  2 (1 particles of K - and 1 particles of NO3 - ) = 4 total particles 2.0 M Ca(NO3)2  2 (1 particles of Ca – 2 and 2 particles of NO3 - ) = 6 total particles

1) Which of the following solutions will freeze at the lowest temperature? a) 100 g NaCl in 150 g of water b) 100 g NaCl in 600 g water c) 100 g NaCl in 125 g of water d) 100 g NaCl in 250 g of water The answer is C because it has the highest concentration: Concentration = mass of solute/mass of solvent. The more solvent there is, the lower the concentration of the solution there will be. As the concentration of the solution decreases, the freezing point increases.

2) Which of the following solutions will freeze at the lowest temperature? a) 100 g C6H12O6 in 500 g of water b) 100 g AlCl3 in 500 g of water c) 100 g KBr in 500 g of water d) 100 g MgF2 in 500 g of water The answer is B because AlCl3 breaks into 4 particles, the most of any of the choices. The more particles that a solute ionizes into, the higher the boiling point will be.

Part B: kinetics I. Kinetics
Collision Theory: - Study of how chemical reactions occur and reaction rates - It is collisions between reactants that allow a reaction to occur - In order for a reaction to occur, there has to be the correct amount of energy and orientation of the collisions How to Speed Up Chemical Reactions (and get a date) Video Clip

Ii. Six factors affecting rate of a reaction
1) Nature of the Reactants: 2) Concentration: 3) Surface Area: - Ionic reactants, especially aqueous, react very quickly (easier to break apart ions) Covalent reactants take a long time (hard to break apart the atoms that are sharing e-) - If the concentration of the reactants increases, then the reaction will occur faster - If the surface area of the reactants increases, then the reaction will occur faster

1) Why would putting batteries in the refrigerator make them last longer? Decrease in temperature, decrease in the amount of collisions. 2) At STP, which 4.0 g zinc sample will react fastest with dilute hydrochloric acid? Lump C) bar Powdered D) sheet metal B – powdered – more surface area 3) Which substancee would react fastest with hydrochloric acid? C6H12O6 C) MgCl2 H2O D) SF2 C – ionic compound

Ii. Six factors affecting rate of a reaction
4) Pressure (ONLY EFFECTS FOR GASEOUS REACTANTS) 5) Presence of Catalyst: 6) Temperature: If the pressure of the reactants INCREASES, then the reaction will occur FASTER A substance that increases the rate of a reaction by creating a different pathway. It is NOT used in the reaction equation, so it is NOT a reactant or product. It appears above the reaction - If the temperature of the reactants INCREASES, then the reaction will occur FASTER

IIi. Rates and equilibrium
Example: NaCl (aq) Na+ (aq) + Cl – (aq) Not all reactions can exist at equilibrium ( _________________ ). Some reactions ___________________________ which means all the reactants get _____________ and ___________________ go backwards Rates can refer to how fast the forward reaction occurs or how fast the reverse reaction occurs. Forward = backward go to completion used up does not

IIi. Rates and equilibrium
A reaction will ____________________ if it is not a closed system. Another reason it may not have a reverse rxn is if it forms: ______________________________ Example: Pb(NO3)2 (aq) + 2 KI (aq)  PbI2 (s) + 2 K(NO3) (aq) go to completion A precipitate (insoluble solid)

IV. Energy changes in chemical reactions
absorbed Endothermic Reactions: heat is _______________ from the surroundings and the temperature of the surroundings ___________ General Equation: Using Table _______: Example: N2(g) + O2 (g)  2 NO (g) ∆ H (kJ) = ________ Can be rewritten as: decreases A + B + Heat/energy  C + D I kJ N2(g) + O2(g) kJ  2 NO (g) Excess stored energy in the products makes them UNSTABLE and more reactive

IV. Energy changes in chemical reactions
released Exothermic Reactions: heat is _______________ INTO the surroundings and the temperature of the surroundings ___________ General Equation: Using Table _______: Example: 2 C (s) + 3H2 (g)  C2H6 (g) ∆ H (kJ) = ________ Can be rewritten as: increases A + B  C + D + Heat/energy I kJ 2 C (s) + 3H2 (g)  C2H6 (g) kJ The products have less stored energy making them more STABLE

V. Potential Energy Diagrams (VIDEO) (ANIMATION)
Endothermic Reactions: Energy (heat) is _________ by the reactants, so the net amount of potential energy ________________ absorbed increases A + B +50 kJ C Reactants PE (60 kJ) A) Potential Energy (PE) of the Reactants: amount of energy the reactants have

A + B +50 kJ C Activation Energy (70 kJ) Reactants PE (60 kJ) B) Activation energy: amount of energy it takes to BREAK EXISTING BONDS and get the reaction STARTED

V. Potential Energy Diagrams
A + B +50 kJ C Activation Energy(70 kJ) PE of AC (130 kJ) Reactants PE (60 kJ) C) PE of Activation Complex: amount of energy activated complex has

A + B +50 kJ C Activation Energy (70 kJ) PE of AC (130 kJ) PE of Products (110 kJ) Reactants PE (60 kJ) D) PE of Products: amount of energy the products have

A + B +50 kJ C Activation Energy (70 kJ) + ∆ H for ENDO (+50 kJ) PE of AC (130 kJ) Products PE (110 kJ) Reactants PE (60kJ) E) Heat of Reaction (∆H) : the NET CHANGE IN ENERGY for the reaction (Difference between Products and Reactants)

A + B +50 kJ C Reverse Activation Energy (20 kJ) Activation Energy (70kJ) + ∆ H for ENDO (+50 kJ) PE of AC (130 kJ) Products PE (110kJ) Reactants PE (60kJ) F) Activation Energy of Reverse Reaction: amount of energy it takes to get THE REVERSE REACTION GOING

A + B +50 kJ C Reverse Activation Energy (20 kJ) Activation Energy (70kJ) + ∆ H for ENDO (+50 kJ) - ∆ H for EXO (-50 kJ) PE of AC (130 kJ) Products PE (110kJ) Reactants PE (60kJ) G) Heat of Reaction (∆H) of Reverse Reaction : the NET CHANGE IN ENERGY for the REVERSE reaction

V. Potential Energy Diagrams (VIDEO) (ANIMATION)
RELEASED Exothermic Reactions: Energy (heat) is _________ by the reactants, so the net amount of potential energy ________________ DECREASES X + Y Z + 40 kJ Reverse Activation Energy (120 kJ) Activation Energy (80kJ) - ∆ H for EXO (- 40 kJ) + ∆ H for ENDO (+40 kJ) PE of AC (150kJ) Products PE (30 kJ) Reactants PE (70kJ)

Effects of a Catalyst with Potential Energy Diagrams: Catalysts often _______________a chemical reaction by providing a ___________ with a ____________________. However, the heat of reaction (∆ H) remains_________________ SPEED UP NEW PATHWAY LOWER ACTIVATION ENERGY THE SAME AE without AE with WITH CATALYST

Part C: Equilibrium I. types of equilibrium
- Occurs when rate of forward reaction equals the rate of reverse reaction - Does NOT mean there is the same amount of reactants and products. However, the CONCENTRATION of them REMAINS CONSTANT “CON CON and REqual” (CONcentration CONstant and REaction Equal)

Phase Equilibrium: Example: H2O (s)  H2O (l) or H2O (l)  H2O (g) - Occurs when rate of the forward phase equals the rate of the reverse phase

B. Solution Equilibrium: Example: NaCl (s)  Na + (aq) + Cl – (aq) (animation) When sodium chloride is first placed in the water, the salt dissolves. As the concentration of dissolved ions increases, some of those dissolved ions will rejoin each other and form a precipitate, which is almost immediately re-dissolved. Eventually, all of the water molecules will be engaged in holding ions apart and no more salt can go into solution until some ions come out of solution as precipitate. The rate of dissolving equals the rate of precipitating. This is called a saturated solution. If you place a salt cube in a beaker of saturated sodium chloride solution, the cube will change shape over time as ions precipitate onto the cube from solution as salt from the cube dissolves. This demonstrates very well the dynamicity of systems at equilibrium. - Formed when a saturated solution has the rate of dissolving equal to the rate of precipitating

C. Chemical Equilibrium: Example: N2 (g) + 3 H2 (g)  2 NH3 (g) + heat - Reached when rate of the forward reaction is equal to the rate of the reverse reaction and the concentration of each substance remains constant - Forward Reaction: N2 + 3H2  2NH3 + heat - Reverse Reaction: 2NH3 + heat  N2 + 3H2

Part C: Equilibrium II. Le Chatelier’s Principle
Stress – any change in _________________ , _____________________, or ________________ on an equilibrium system is called a stress. LeChatelier’s Principle – if stress is applied to a equilibrium system, the system will _______________________ A. Stress #1: Concentration Changes: concentration temperature pressure Shift to relieve the stress - Reaction will always shift away from something that is added and towards something that is taken away (Concentration [ ] = M)

Example: 4 NH3 (g) + 5 O2 (g)  4 NO (g) + 6 H2O (g) + Heat Increase [NH3] results in: Shifts to the __________________, ____ [O2], _____ [NO], _____ [H2O] _____ heat Less [NH3] results in: Shifts to the __________, ____ [O2], _____ [NO], _____ [H2O,]_____ heat + --- + + + RIGHT ↓ ↑ ↑ ↑ --- + --- --- --- LEFT ↑ ↓ ↓ ↓

B. Stress #2: Temperature Changes: Adding Heat (warming): When heat is added, the _______________ reaction is always ______ Reaction shifts ____________________________________ Example: 4 NH3 (g) + 5 O2 (g)  4 NO (g) + 6 H2O (g) + Heat Increase Heat results in: Shifts to the __________, _____ [NH3], _____ [O2], ___ [NO], _____ [H2O] ENDOTHERMIC FAVORED AWAY FROM WHAT IS ADDED --- + + + --- LEFT ↑ ↑ ↓ ↓

B. Stress #2: Temperature Changes: Removing Heat (cooling): When heat is removed, the ____________ reaction is always ______ Reaction shifts ____________________________________ Example: 4 NH3 (g) + 5 O2 (g)  4 NO (g) + 6 H2O (g) + Heat Decrease Heat results in: Shifts to the __________, _____ [NH3] _____ [O2], _____ [NO], ___ [H2O] EXOTHERMIC FAVORED TOWARD WHERE HEAT IS REMOVED + + --- --- --- RIGHT ↓ ↓ ↑ ↑

C. Stress #3: Pressure Changes: Increasing Pressure (decreasing volume): Reaction shifts ____________________________________ Example: N2 (g) + 3 H2 (g)  2 NH3 (g) Increase Pressure results in: Shifts to the __________, _____ [N2], _____ [H2], _____ [NH3] TO THE SIDE WITH LESS GAS MOLES 4 total 2 total RIGHT ↓ ↓ ↑

C. Stress #3: Pressure Changes: DECREASING Pressure (INCREASING volume): Reaction shifts ____________________________________ Example: N2 (g) + 3 H2 (g)  2 NH3 (g) DECREASE Pressure results in: Shifts to the __________, _____ [N2], __ [H2,], _____ [NH3] Important Note: If the same number of gaseous reactant and products molecules, pressure changes have _________________ on system TO THE SIDE WITH MORE GAS MOLES 4 total 2 total LEFT ↓ ↑ ↑ NO EFFECT

D. Effect of a Catalyst: Changes rate of _______________ the forward and reverse reaction BOTH EQUALLY Does NOT change any equilibrium concentrations

1. H2(g) + I2 (g) + 53 kJ  HI (g) What happens if: A) Increase Temperature: Shifts to the _______, ______ HI, _____ H2, _____ I2 B) Increase [H2 (g)]: Shifts to the _______, ______ HI, _____ I2 C) Increase [HI (g)]: Shifts to the _______, _____ H2, _____ I2 D) Decrease Pressure: + right --- ---- right + --- left + + No change. Equal number of moles of gas on both sides

PCl5 (g) + heat  PCl3 (g) + Cl2(g) What happens if: A) Adding Cl2: Shifts to the _______, ______ PCl5, _____ PCl3 B) Increasing Pressure: Shifts to the ____, ____ PCl5, _____ PCl3, _____ Cl2 C) Lowering Temperature: Shifts to the ____, ___ PCl5, _____ PCl3, ____ Cl2 D) Removing PCl3: Shifts to the _______, ______ PCl5, _____ Cl2 + --- left left + --- ---- left + ---- ---- left --- +

Part C: Equilibrium VI. Entropy
state of greater randomness or disorder Nature tends to want to change to _________________________________ Entropy: ( _______________ disorder = ______________ entropy) Systems in nature tend to undergo changes toward _____________________ and __________________________________ Measure of disorder or randomness of a system MORE MORE Lower energy (enthalpy) Higher entropy (disorder)

Physical Examples of Changes in Entropy: Entropy INCREASES and is favored Entropy DECREASES and is not favored

Chemical Examples of Changes in Entropy: 1 EXOTHERMIC (less energy) favored and entropy INCREASES 1 ENDOTHERMIC (more energy) not favored and entropy DECREASES

Check for understanding
Complete Chart A on pg 22 in Work Packet #2 Check for understanding Increases Favored decreases unfavored Increases Favored decreases unfavored decreases unfavored decreases unfavored

Unit 7: Solutions, Kinetics, and Equilibrium

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