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Name: ________________________  

Block: ____________ 

Unit 10: Part 1: Polarity and Intermolecular Forces

Intermolecular Forces of Attraction and Phase Changes Intramolecular Bonding: attractive forces that occur between atoms WITHIN a molecule; these

are true chemical bonds, of two types: 1. ionic bond = electrostatic attraction between metal cation and non-metal anion 2. covalent bond = electron pair sharing between non-metal and non-metal Intermolecular Forces of Attraction: attractive forces that occur BETWEEN molecules

4 types of Intermolecular Forces of attraction:

o Dipole - Dipole: strongest type of intermolecular forces of attraction between 2 polar molecules with dipole moments

o Hydrogen Bonding: strong intermolecular forces of attraction between hydrogen and highly electronegative oxygen, nitrogen or fluorine of 2 different polar molecules *both molecules must have dipole-dipole forces*

o London Dispersion Forces: weak intermolecular forces of attraction between NON-polar molecules; larger mass molecules have higher London Dispersion forces.

o Van der Waal's Forces: weak, temporary intermolecular forces of attraction between molecules, assume present in all molecules - polar/non-polar

What is an Intermolecular Force? Force between molecules (weak force) Differs from an intramolecular force (strong force)

which forms Covalent Bonds

Intramolecular Forces

Intermolecular Forces

Covalent Bonds H-Bonds Dipole-dipole London Dispersion

400 kcal 12-16 kcal 2-0.5 kcal Less than 1 kcal

Notice: covalent bonds are almost 40 times the strength What creates an Intermolecular force?

• Unequal distribution of electrons • Created as a result of differences in:

• Electronegativity

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Phase change = when energy enters or leaves a compound to cause changes from solid, liquid, or gas phases; substance overcomes weak intermolecular forces of attraction 6 phase changes: 1. melting = solid - liquid 4. condensation = gas - liquid 2. freezing = liquid - solid 5. deposition = gas - solid 3. evaporation = liquid - gas 6. sublimation = solid - gas Normal melting point = temperature where substance is in equilibrium between solid and liquid

phases at standard pressure of 1 atm. Normal boiling point = temperature where substance is in equilibrium between liquid and gas

phases at standard pressure of 1 atm.

In order for a substance to move between the states of matter; for example, to turn from a solid into a liquid, which is called fusion, or from a liquid to a gas (vaporization), energy must be gained or lost. (As we move from solid to gas it is gained and from gas to a solid it is lost. Why? Molar Heat of Vaporization requires more energy to change phase from liquid to gas phase. Gas molecules have high kinetic energy and distance between gas molecules is very high, requiring more energy to overcome the intermolecular forces of attraction.) Changes in the states of matter are often shown on phase diagrams, and you will probably see at least one of two different types of phase diagrams. Let’s start with the phase diagram for water. The phase diagram for water is a graph of pressure versus temperature. Each of the lines on the graph represents an equilibrium position, at which the substance is present in two states at once. For example, anywhere along the line that separates ice and water, melting and freezing are occurring simultaneously.

The intersection of all three lines is known as the triple point (represented by a dot and a T on the figure). At this point, all three phases of matter are in equilibrium with each other. Point X represents the critical point, and at the critical point and beyond, the substance is forever in the vapor phase.

This diagram allows us to explain strange phenomena, such as why water boils at a lower temperature at higher altitudes, for example. At higher altitudes, the air pressure is lower, and this means that water can

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reach the boiling point at a lower temperature. Interestingly enough, water would boil at room temperature if the pressure was low enough!

Example What happens to water when the pressure remains constant at 1 atm but the temperature changes from -10ºC to 75ºC? Explanation Looking at the phase change diagram for water and following the dashed line at 1 atm, you can see that water would begin as a solid (ice) at 0ºC and begin melting. So from 10º C to 75º C water would be in a liquid phase until it reaches 100ºC.

The second type of phase change graph you might see on the SAT II Chemistry exam is called a heating curve. This is a graph of the change in temperature of a substance as energy is added in the form of heat. The pressure of the system is assumed to be held constant, at normal pressure (1 atm). As you can see from the graph below, at normal pressure water freezes at 0ºC and boils at 100ºC.

The plateaus on this diagram represent the points where water is being converted from one phase to another; at these stages the temperature remains constant since all the heat energy added is being used to break the attractions between the water molecules.

    

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Part 2: Solutions Vocabulary Solution - a homogeneous mixture of two or more substances in a single physical state

Solvent‐ substance that does the dissolving 

Solute‐ substance that is dissolved 

Soluble‐ when a substance is able to dissolve in another substance 

Insoluble‐ when a substance cannot dissolve in another substance 

Alloy‐ solution containing two or more metals 

Miscible‐ when two liquids can dissolve in one another in any amount 

Immiscible‐ when two liquids do not mix together  

Aqueous solution‐ a solution where the solvent is water 

Concentration‐ a solution that contains a large amount of solute 

Dilute‐ a solution that contains a little solute 

Saturated‐ a solution that contains a maximum amount of solute 

Unsaturated‐ a solution that contains below the maximum amount of solute 

Supersaturated‐ a solution that contains above the maximum amount of solvent 

Solubility‐ describes the maximum amount of solute that can be dissolved in a 

solvent at a given temperature 

The rate at which a solution is formed is affected by: 

Surface area (particle size)‐  

More surface area, smaller particles dissolves FASTER 

Temperature‐ 

Increase temperature will increase the dissolving rate 

Agitation 

Example stirring, shaking, mixing will increase the dissolving rate 

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Measuring Solution Concentrations Molarity

Molarity (M)- the number of moles of solute dissolved in each liter of solution M = mol L

Calculate the molarity of a solution formed by mixing 10.0 g of sulfuric acid (H2SO4) with enough water to make 100.0 mL of solution.

1. Calculate the moles of sulfuric acid.

10.0 g H2SO4 1 mol H2SO4 = 0.102 mol H2SO4 98.08 g H2SO4

2. Calculate liters of solution.

100.0 mL 1 L = 0.1000L

1000 mL

3. Calculate molarity. M = mol 0.102 mol H2SO4= 1.02 M L 0.1000L

What mass of sodium nitrate is needed to produce 500.0 mL of a 0.50 M solution?

  M = mol  0.50 M =  .      X = 0.25 mol NaNO3   L  0.5000L 

0.25 mol NaNO3 85.00 g = 21.25 g NaNO3 1 mol

Molality (m)- moles of solute per kilogram of solution (mol/kg) What is the molality of saltwater that contains 684 g of NaCl in 20.0 mL of water?

Step 1 convert to moles 684 g NaCl 1 mol NaCl

= 11.70 mol NaCl 58.45 g NaCl

Step 2 convert to kilograms (1 ml of water is equal to 1 g of water) 20.0 g 1 Kg

= 0.0200 kg 1000 g

Step 3 convert to molality

m = mol   11.70 mol NaCl = 585 m   kg  0.0200kg 

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Dilutions

Dilutions- many solutions come as concentrated stock solutions and must be diluted before use.

M1V1 = M2V2

What volume of a 12.0 M stock solution of hydrochloric acid is required to make 250.0 mL of a 0.10 M solution?

M1 = 12.0 M M2 = 0.10 M

V1 = ? V2 = 250.0 mL 12.0 M X = (0.10 M) (250.0 mL) 12.0 M X = (0.10 M) (250.0 mL) 12.0 M 12.0 M X = 2.08 mL You would add 2.08 mL to a volumetric flask. You would than add 247.02 mL to the flask for a final volume of 250.0 mL. You now have 250.0 mLof 0.10 M solution.

   

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Solubility - We can predict miscibility using the rule “LIKE DISSOLVES LIKE”

Polar + Polar = miscible Non-polar + Non-polar = miscible Polar + non-polar = immiscible

Polar + ionic = miscible Non-polar + ionic = immiscible

To determine polarity

Polar Rules Non-polar Rules

Hydroxyl group – OH

Organic changes CxHy

Polar solutes dissolves in polar solvent

Dissolves in non-polar substance

Asymmetric molecule w/ polar bonds Symmetric molecule w/polar bonds

Lone pairs on central atom ionic bonds

No polar bonds (look at electronegativity difference)

Polar Examples Non-polar Examples

Water, Salts, Sugar, Acids, Bases, Ammonia

Butter, oil, lard, Some Paints

 

Soap is an emulsifier:

Has a polar and non-polar end. Non-polar end dissolves in oil, Polar end dissolves

in water: water and oil appear to mix Substances that DO NOT dissolve in water are hydrophobic Substances that DO dissolve in water are hydrophilic

   

Oil 

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Solubility Charts 

The __Solubility_ of a solute dissolved in 100g of water is tested at different temperatures. 

 

The amount in grams is plotted on a graph based on 

the __Saturation_.  

Then the data points are  Connected  by a line or curve. 

The curve represents the maximum amount of a solute dissolved in 100g of water for ALL temperatures between 0°C and 100°C. 

 

Q: Why is the scale only between 0°C and 100°C? Water boils and turns to a gas at 100° C At 40°C, 100g of water can dissolve how much solute? Between 44 to 45 g

Will 50 grams dissolve in 100g of water at 75°C? Yes

For any point _below__ the solubility curve the solution is _Unsaturated_.  For any point _on__ the solubility curve the solution is _Saturated_. 

 

Is a solution with 70g of solute dissolved at 40°C saturated, unsaturated, or supersaturated? 

For any point _above__ the solubility curve the solution is _Supersaturated  

 

Solubility Questions: 

1. At 60°C, 25g of solute is dissolved in 100g of water.  What 

is the name of the solute?  KClO3 . 

2. If 50g of KCl are dissolved in 100g of water, at 80°C, is the 

solution saturated, unsaturated, supersaturated? 

                               Usaturated  

3. What is the solubility of KNO3 at 45°C in 200g of water? 

70 g = x g so 140 g of KNO3

100g 200 g 4. What is the solubility of NaCl at 99°C in 50g of water? 

40 g = x g so 20 g of NaCl

100g 50 g    

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Dilution Problems 

1) How much concentrated 18 M sulfuric acid is needed to prepare 250 mL of a 6.0 M solution? 

M1V1 = M2V2 M1 = 18 M M2 = 6.0 M

V1 = ? V2 = 250.0 mL

18 M X = (6.0 M) (250.0 mL)

18 M X = (6.0 M) (250.0 mL) 18 M 18 M

X = 83 mL  

2) How much concentrated 12 M hydrochloric acid is needed to prepare 100. mL of a 2.0 M solution? 

M1V1 = M2V2 M1 = 12 M M2 = 2.0 M

V1 = ? V2 = 100 mL

12 M X = (2.0 M) (100 mL)

12 M X = (2.0 M) (100 mL) 12 M 12 M

X = 17 mL  

Molarity Problems 

1) What is the molarity of a solution in which 58 g of NaCl are dissolved in 1.0 L of solution? 

Step 1 convert to moles 58 g NaCl 1 mol NaCl

= 1 mol NaCl 58.45 g NaCl

Step 2 convert to Liters (it is in Liters already) Step 3 convert to molality

m = mol   1 mol NaCl = 1 M     L  1 L 

 

 

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2) What is the molarity of a solution in which 10.0 g of AgNO3 is dissolved in 500. mL of solution? 

Step 1 convert to moles 10.0 g AgNO3 1 mol AgNO3 = 0.0589 mol AgNO3 169.88 g AgNO3

Step 2 convert to Liters

500.0 mL 1 L = 0.5000 L

1000 mL

Step 3 convert to molality

m = mol   0.0589 mol AgNO = 0.118 M     L  0.5 L 

 

3) How many grams of KNO3 should be used to prepare 2.00 L of a 0.500 M solution? 

  M = mol  0.500 M =  .      X = 1.00 mol KNO3   L    2.00 L 

1.00 mol KNO3 101.11 g = 101.11 g KNO3 1 mol

 

4) To what volume should 5.0 g of KCl be diluted in order to prepare a 0.25 M solution? 

5.0 g KCl 1 mol = 0.067 mol KCl

74.55 g  

  M = mol  0.25 M =  0.067 mol = 0.268 L KCL   L    X 

   

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Part 3: Colligative Properties Notes

Colligative properties- property that depends on the concentration of solute particles, not their identity

Freezing point depression (Δtf)- freezing point of a solution is lower than the freezing point of the pure solvent

Boiling point elevation (Δtb)- boiling point of a solution is higher than the boiling point of the pure solvent

Calculating Changes in Freezing or Boiling Points Boiling Point Elevation: A solution will boil at a higher temperature than the pure solvent. This is the colligative property called boiling point elevation. The more solute dissolved, the greater the effect. An equation has been developed for this behavior. It is:

ΔTb = i Kb m  

Freezing Point Depression: A solution will solidfy (freeze) at a lower temperature than the pure solvent. This is the colligative property called freezing point depression. The more solute dissolved, the greater the effect. An equation has been developed for this behavior. It is:

ΔT f = i Kf m  

ΔTb or ΔTf = change in boiling point and freezing point i= number of ions (particles)

m= molality 

kb or kf = boiling point elevation constant and freezing point depression constant

Example: If 45.0 g of glucose are dissolved in 255 g of water what will the boiling point 

and freezing point of the solution? The kb of water is 0.515 oC/m; the kf of water is 1.86 o 

C/m. (A reminder ΔT is the change in temperature not the new temperature of the 

freezing point or boiling point) 

Boiling Point          ΔTb = i Kb m 

C6H12O6 is covalent i = 1, kb = 0.515 oC/m, m = moles Kg   Convert 45.0 g of glucose to moles and 255g of water to kg 

ΔTb = i Kb m = (1) (0.515 oC/m) 0.25 moles 0.255 kg ΔTb = 0.505 oC so new boiling point is

100 oC + 0.505 oC = 100.5 oC    

i = The Number of Ions Covalent molecules do not disassociate in water so

i = 1 Ionic compounds disassociate in water so i = the number of ions (elements) in the compound.

NaCl i = 2 MgCl2 i = 3 AlCl3 i = 4

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Freezing Point          ΔTf = i Kf m 

C6H12O6 is covalent i = 1, kf = 1.82 oC/m, m = moles Kg Convert 45.0 g of glucose to moles and 255g of water to kg ΔTf = i Kf m = (1) (1.86 oC/m) 0.25 moles 0.255 kg ΔTf = 1.82 oC so new freezing point is

0 oC - 1.82 oC = - 1.8 oC   

Electrolytes and Nonelectrolytes.

‐ Electrolyte – a compound that conducts electric current when it is in an aqueous solution or in the molten state.

o Examples: Ionic Compounds ‐ Nonelectrolyte – a compound that does not conduct electric current in either an

aqueous solution or in the molten state. o Examples: Covalent Compounds

‐ For conduction to occur, the ions must be mobile.

Human body requires Na+1, K+1 and Ca+2 for bodily functions.

 

Molarity and Stoichiometry:

Ammonium chloride and calcium hydroxide react according to the following unbalanced equation:

NH4Cl (aq) + Ca(OH)2 (aq) CaCl2(aq) + NH3 (g) + H2O (l)

a. What mass of ammonium chloride is needed to make 1.5 liter of a 5.0 M ammonium chloride solution?

Calculate the moles in a 5.0 M solution of NH4Cl (aq) M = moles = 5.0 M = X x= 7.5 moles

L 1.5 L Convert Moles to grams

7.5 mol NH4Cl 53.45 g = 400 g NH4Cl

1 mol

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b. What mass of calcium hydroxide is needed to make 2.5 liters of a 5.0 M calcium hydroxide solution?

NH4Cl(aq) + Ca(OH)2(aq) CaCl2(aq) + NH3(g) + H2O(l)

Calculate the moles in a 5.0 M solution of Ca(OH)2 (aq) M = moles = 5.0 M = X x= 12.5 moles

L 2.5 L Convert Moles to grams

12.5 mol Ca(OH)2 74 g = 925 g Ca(OH)2 1 mol

After we have converted to grams then we can follow our regular stoichiometry steps to solve any stoichiometry problem.

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To make a reaction go faster: Increase temperature

- a direct relationship between kinetic energy & temperature exists - the higher the temp, the faster the molecules will go - more chaotic motion will lead to more collisions

Increase surface area of reactants - the more sites exposed to react, the more collisions can occur - a large piece of copper will react slower in acid than many small pieces

Increase concentration of reactants - more moles (therefore particles) available to react in a 6.0M solution vs. a 0.10M solution

- copper will react faster with 6.0 M acid than 0.10 M. Add a catalyst

- will speed up the reaction by lowering activation energy, Ea pathway with added catalyst......           Exothermic Endothermic Why do reactions occur at different rates? The rate of a reaction is the speed at which a chemical reaction happens. If a reaction has a low rate, that means the molecules combine at a slower speed than a reaction with a high rate. Some reactions take hundreds, maybe even thousands, of years while others can happen in less than one second. If you want to think of a very slow reaction, think about how long it takes plants and ancient fish to become fossils (carbonization). Ultimately: Molecules moving too slowly, elements electronegativity, phases (best is liquid). What prevents a reaction from occurring immediately? Energy: .Some molecules will have high energy; some low; many intermediate. Only those with energies greater than the activation energy will be able to react Why don’t all products form at the same instant? Energy

Each reaction is special. Conditions are different for each reaction. Sometimes it takes longer for molecules/atoms to arrange themselves so they can react, therefore forming new substances

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Part 4: Chemical Equilibrium A state when the concentrations of all substances in a reaction remain constant You have assumed that when reactants products, the only “thing” in a reaction

vessel (beaker) are the products and excess reactants In other words, you have assumed that all reactions proceed to completion However, most reaction are actually REVERSIBLE

ex: H2O (g) + CO (g) H2 (g) + CO2 (g)

the “reverse arrow” indicates that the reaction can occur in either direction until equilibrium occurs

No net changes occur that are visible at equilibrium, but on the molecular level, reaction still occurs

No changes in concentration occurs once equilibrium has been established

Graph concentration vs time for reaction above,

- water and carbon monoxide decrease over time, they’re the reactants, they go away - hydrogen and carbon dioxide are the products, there is no concentration of them initially, because they have not yet been formed, they increase over time - The Reaction is said to be AT EQUILIBRIUM when

Rate forward = Rate Reverse

- With kinetics, we saw that the rate of the forward reaction slows (neg sign) because the reactants are being used up, the opposite applies for the reverse reaction - Once some of the products form, they can react together to form “reactants.” Their rate of reaction starts slow (rate reverse), but increases

- When the rates equal, equilibrium has been achieved.

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The Equilibrium Constant, K or Keq

The constant, K, was derived from experimentation. For the reaction aA + bB ↔ cC + dD where lower case letters are coefficients and uppercase letters are chemical substances:

K = [C]c[D]d [A]a[B]b this is the equilibrium expression Put more simply:

K = [products]n [reactants]m MOST IMPORTANT only AQUEOUS and GAS substances can go into the

equilibrium expression

WHY? [ ] is concentration and ONLY aq and g can change [ ] Both aq and g depend on the volume in which they are contained Pure solids (s) and pure liquids (l) do not have concentration measurements

therefore, their concentration values cannot change (

Definition: homogeneous equilibria – all substances in the reaction are in the same state Definition: heterogeneous equilibria – substances are in more than one state What does the Magnitude of K mean? Larger K values indicates that the reaction really wants to (and does) occur, proceeds toward completion Smaller K values means that the reactants are favored, not many products will form

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Ex. Write the homogenous equilibrium expression for the following reactions: a. Sulfur dioxide reacts with oxygen gas to form sulfur trioxide (all gases)

Reaction: 2 SO2 (g) + O2 (g) ↔ 2 SO3 (g)

K == [SO3]2 [SO2]2[O2]1

b. Phosphorus pentachloride decomposes into chlorine and phosphorus trichloride (all

gases) Reaction:

PCl5 (g) ↔ Cl2 (g) + PCl3 (g)

K == [Cl2] [PCl3] [PCl5]

Ex. Write the heterogeneous equilibrium expression for the following reactions:

a. Sulfur trioxide gas is bubbled into water producing sulfuric acid solution Reaction:

SO3 (g) + H2O (l) ↔ H2SO4 (aq)

K == [H2SO4] [SO3]

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LeChatelier’s Principle You will predict what changes occur when a system that is at equilibrium is disturbed, and you will suggest how to re-establish equilibrium in that system. Back to EQUILIBRIUM POSITION – Dependent upon:

- Temperature - Volume of container - Concentrations of aq/g - Pressure

The position of a chemical reaction is said to

- lie to the LEFT if the reaction FAVORS the REACTANTS - lie to the RIGHT if the reaction FAVORS the PRODUCTS LeChatelier’s Principle: When a change is imposed on a system at equilibrium, the

position of equilibrium shifts in a direction to reduce the effect of that change

1. Effect of a concentration change

- K remains constant - Add a reactant/product, the system will shift AWAY from the added

reactant/product

- Remove a reactant/product, sys will shift TOWARD the added reactant/product

- System will compensate for addition or removal

Ex N2 (g) + 3H2 (g) 2NH3 (g) at equilibrium, disturb the system by: a. adding N2 b. adding NH3 c. removing NH3 Shift to Products Shift to Reactants Shift to Products Ex: CaCO3 (s) CaO (s) + CO2 (g) at equilibrium, disturb the sys by: a. removing CO2 b. adding CaO c. adding CaCO3 Shift to Products Shift to Reactants Shift to Products

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2. Effect of changing Pressure - K remains constant - add or remove gas reactant or product, same effect as concentration

- add inert gas Ptotal increase, concentration does not change, system remains at

equilibrium. 3. Effect of changing Volume of container

- both concentration and partial pressures of all sub change

- when Volume deceases, the sys responds by deceases the total number of molecules (that seems impossible, you can’t destroy matter)

- the reaction system will respond by shifting to the side with the least

number of gas molecules

- when Volume increase, shift to side with most molecules, more space available, naturally take up that space

Ex: N2 (g) + 3H2 (g) 2NH3 (g) at equilibrium, disturb the system by:

4 molecules 2 molecules a. decreasing volume of container b. increasing volume of container - shift to right, only 2 molecules - shift to L, 4 molecules Ex: CaCO3 (s) CaO (s) + CO2 (g) at equilibrium, disturb the system by: 0 molecules 1 molecule a. decreasing volume of container b. increasing volume of container - shift L, 0 gas molecules - shift R, 1 gas molecules

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4. Effect of changing Temperature

- K changes with temperature change

- equilibrium constant K is constant for a given reaction at a certain temperature; now we’re changing temp

- treat ∆H as a reaction or products being added or released as

temperature changes - how? recall: endothermic reactions means E is being absorbed into the reaction to

make it happen, + value, therefore add value to reactants side exothermic reactions means E is being released from the reaction; -

value, therefore add value to products side Ex: N2 (g) + 3H2 (g) 2NH3 (g) ∆H = -92 kJ this is an exothermic reaction (negative value) add it to the products side N2 (g) + 3H2 (g) 2NH3 (g) + 92 kJ - Disturb equilibrium by increasing T – Shift Left, away from addition

(just like adding ammonia) - Decrease T – Shift Right, toward removal

Ex: 2SO2 (g) + O2 (g) 2SO3 (g) ∆H = +198 kJ

this is an endothermic reaction (positive value), add it to the reactant side, this much energy is needed to make the reaction go

2SO2 (g) + O2 (g) + 198 kJ ↔ 2SO3 - increase temperature – shift R to the products, away from addition (just like

adding oxygen) - decrease temperature – shift L to reactants, toward removal (just like removing

oxygen)

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LeChatelier’s Problem

2NO2 (g) N2O4(g) ∆Hreaction = -58 kJ When the following changes are imposed on the reaction system above, tell in which direction the equilibrium position will shift to re-establish equilibrium:

Change Shift 1. Add N2O4 Left 2. Remove N2O4 Right 3. Add NO2 Right 4. Remove NO2 Left 5. Add Ne Right (inert gas) 6. Inc container vol Left (2 gas molecules vs. 1 gas mlc) 7. Dec container vol Right 8. Inc pressure Right 9. Dec pressure Left

10. Inc temp Left (2NO2 (g) ↔ N2O4 + 58 kJ exo reaction)

11. Dec temp Right

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Molarity Practice Problems and Solubility Curves 1. How many grams of CuSO4 are needed to prepare 100. mL of a 0.10 M solution?   M = mol  0.10 M =  .      X = 0.010 mol CuSO4   L    0.100 L 

0.0100 mol CuSO4 159.6 g = 1.60 g CuSO4 1 mol

2. What is the molarity of a solution that contains 20.45 g of sodium chloride dissolved in 700.0 mL of solution?

20.45 g NaCl 1 mol = 0.3499 mol NaCl

58.45 g   M = mol  x M =  0.3499 mol = 0.4998 M    L    0.700 L 

3. Calculate the molarity of 0.205 L of a solution that contains 156.5 g of sucrose (C12H22O11).

156.5 g C12H22O11 1 mol = 0.4573 mol NaCl

342.22 g   M = mol  x M =  0.4573 mol = 2.23 M    L    0.205 L 

4. A 0.600 L sample of a 2.5 M solution of potassium iodide contains what mass of KI?

  M = mol  2.5 M =  .      X = 1.5 mol KI   L    0.600 L 

1.5 mol KI 166 g = 249 g KI

1 mol

5. What mass of ammonium chloride would you use to prepare 85.0 mL of a 1.20 M solution of NH4Cl?

  M = mol    1.2 M =  .      X = 0.102 mol NH4Cl   L          0.085 L 

0.102 mol NH4Cl 53.5 g = 5.46 g NH4Cl

1 mol

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6. Calculate the molarity of a solution containing 106 g of naphthalene (C10H8) dissolved in 3.15 L of water.

128

106 g C10H8 1 mole C10H8 = 0.828 mol C10H8 128 g C10H8

M = mol/L 0.828 mol C10H8 / 3.15 L M = 0.263 M

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7. Which compound is LEAST soluble at 10 °C? _____KClO3________

8. How many grams of KCl can be dissolved in 100g of water at 80°C?

_________50 g_________

9. How many grams of NaCl can be dissolved in 100g of water at 90°C?

_________40 g___________

10. At 40°C, how much KNO3 can be dissolved in 100g of water? __62-63 g___

11. Which compound shows the least amount of change in solubility from 0°C-

100°C? _____NaCl_______

12. At 30°C, 90g of NaNO3 is dissolved in 100g of water. Is this solution saturated or

unsaturated? __ unsaturated __

13. At 60°C, 72g of NH4Cl is dissolved in

100g of water. Is this solution saturated or

unsaturated? _Super Saturated_

14. Which compounds show a decrease in

solubility from 0°C-100°C? ___NH3

and Ce2(SO4)3__________

15. Which compound is the most soluble at

10°C? ___________KI_________

16. Which compound (besides Ce2(SO4)3) is

the least soluble at 50°C?___KClO3__

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Molarity and Stoichiometry Worksheet 1) The calcium phosphate used in fertilizers can be made according to the following

unbalanced reaction: H3PO4 (aq) + Ca(OH)2 (s) Ca(H2PO4)2 (aq) + H2O (l)

a) Balance the reaction equation. 2H3PO4 (aq) + Ca(OH)2 (s) Ca(H2PO4)2 (aq) + 2 H2O (l)

b) If 7.50 L of 5.00 M H3PO4 react with excess Ca(OH)2, how many grams of water will be produced?

Calculate moles of H3PO4 are in solution M=mol/L 5.00 M = x/7.5 L x = 37.5 mol

2H3PO4 (aq) + Ca(OH)2 (s) Ca(H2PO4)2 (aq) + 2 H2O (l) 37.5 mol Excess ? g mol 37.5 mol H3PO4 2 mol H2O 18.02 g H2O

= 675 g H2O 2 mol H3PO4 1 mole H2O

2) A sugar solution is prepared by dissolving 3.00 grams of sucrose into 200.0 grams of

water. Calculate the molality of the solution.

3.00 g C6H12O6 1 mol = 0.0167 mol C6H12O6 180 g

m = mol/Kg 0.0167mol / 0.2000Kg m = 0.0835 m

Dilutions Worksheet (M1 V1 = M2 V2)

3) If I have 340 mL of a 0.5 M NaBr solution, what will the concentration be if I add 560 mL more water to it?

M1 V1 = M2 V2

(0.5 M) (340 mL) = M2 (560 mL)

M2 = 0.3 M

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4) If I dilute 250 mL of 0.10 M lithium acetate solution to a volume of 750 mL, what will the concentration of this solution be?

M1 V1 = M2 V2

(0.10 M) (250 mL) = M2 (750 mL)

M2 = 0.03 M

5) If I leave 750 mL of 0.50 M sodium chloride solution uncovered on a windowsill and 150 mL of the solvent evaporates, what will the new concentration of the sodium chloride solution be?

M1 V1 = M2 V2

(0.50 M) (750 mL) = M2 (600 mL)

M2 = 0.63 M

6) To what volume would I need to add water to the evaporated solution in problem 3 to get a solution with a concentration of 0.25 M?

M1 V1 = M2 V2

(0.50 M) (340 mL) = (0.25 M) (x mL)

mL = 680 mL

680 mL – 340 mL = 340 mL

   

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Equilibrium Constant (K) 

Write the expression for the equilibrium constant, K, for the reactions below. 

1. N2 (g) + 3H2 (g) ↔ 2NH3 (g)  K  =  [NH3]2   [N2] [H2]3  

2. 2KClO3 (s) ↔ 2KCl (s) + 3O2 (g)  K  =   [O2]3     

 

3. H2O (l) ↔ H+ (aq) + OH‐ (aq)  Keq = [H+] [OH‐]     

 

4. 2CO (g) + O2 (g) ↔ 2CO2 (g)  K  =    [CO2]2   [CO]2 [O2]   

 

5. Li2CO3 (s)  2Li+ (aq) + CO3‐2 (aq)   Keq = [Li+] [CO3

‐2]    

 

   

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Equilibrium and Kinetics Worksheet

1. What is kinetics? Chemical kinetics, also known as reaction kinetics, is the study of rates of chemical processes. Chemical kinetics includes investigations of how different experimental conditions can influence the speed of a chemical reaction

2. Define activation energy. It is the least amount of energy required to activate

atoms or molecules to a state in which they can undergo a chemical reaction

3. Why do all reactions require activation energy? A reaction won’t occur unless atoms or molecules of reactants come together. This happens only if the particles are moving, and movement takes energy. Often, reactants have to overcome forces that push them apart. This takes energy.

4. Name 4 means of increasing the rate of a reaction.

The classic answers are: 1. Temperature, 2. Pressure, 3. Concentration and 4. Surface Area

5. Why will adding a catalyst speed up a reaction? A catalyst provides an alternative route for the reaction with a lower activation energy

6. Define chemical equilibrium. There is equilibrium when the concentrations of reactants and products are in an unchanging ratio. Another way of saying this is that a system is in equilibrium when the forward and reverse reactions occur at equal rates.

7. What does it mean for a reaction to go to completion? A reaction is at completion when one or more of the reagents in the sample have been completely used up.

8. What does it mean for a reaction to be reversible? A reversible reaction is a

chemical reaction where the reactants form products that, in turn, react

together to give the reactants back. Reversible reactions will reach an

equilibrium point where the concentrations of the reactants and products will

no longer change.

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9. What is the equilibrium expression? The equilibrium constant expression is the ratio of the concentrations of the products over the reactants. Notice how each concentration of product or reactant is raised to the power of its coefficient.

10. What are the only two states that substances can be in to be a part of the equilibrium expression? Gases and aqueous solutions (NOT solids or liquids)

11. What do small K values indicate about a reaction? And a large K value?

Larger K values indicates that the reaction really wants to (and does) occur, proceeds toward completion.

12. What does a small Keq value mean? And a large Keq value? Smaller K values means that the reactants are favored, not many products will form

13. What are the four changes that can be imposed on a system that is at equilibrium?

Temperature ‐ Volume of container ‐ Concentrations of aq/g ‐ Pressure 14. After an equilibrium system has been disturbed, will it come back to equilibrium?

15. For a gaseous reaction that is at equilibrium, if it is disturbed by increasing the

volume of the reaction vessel (or decreasing the pressure), in which direction with the reaction system shift in order to re-establish equilibrium and why? It will shift to the side with the most molecules (moles) because it has more room to stay “larger.”

16. Why does the K value (Equilib constant) change if you disturb a reaction at

equilibrium by increasing or decreasing temperature?

Reaction are normally endothermic or exothermic increasing the temperature will cause the energy formed in the reaction to shift to the side without the energy. + ΔH will be endothermic and exothermic will have a – ΔH.

    

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LeChatelier’s Principle 

LeChatelier’s Principle states that when a system at equilibrium is subjected to a stress, the system will shift its equilibrium point in order to relieve the stress. Complete the following chart by writing left, right, or none for equilibrium shift, and decreases, increases, or remains the same for the concentrations of the reactants and products, and for the value of K.

N2 (g) + 3H2 (g) ↔ 2NH3 (g) + 22.0 kcal

Stress Equilibrium 

Shift [N2]  [H2]  [NH3]  K 

1. Add N2  Right _____

Decreases Increases

Remains the same

2. Add H2 

Right

Decreases

_____

Increases

3. Add NH3 

Left

Left

Increases _____

4. Remove N2 

Left

_____ Decreases Decreases

5. Remove H2 

Left

Left

_____ Decreases

6. Remove NH3 

Right

Right Decreases _____

7. Increase Temperature 

Right

Increases Increases Decreases

8. Decrease Temperature 

Left

Decreases Decreases Increases

9. Increase Pressure 

Right

Decreases Decreases Increases

10. Decrease Pressure 

Left

Increases Increases Decreases