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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 NONpolar 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
Intermolecular Forces
Intramolecular
Forces
Covalent Bonds
400 kcal
H-Bonds
Dipole-dipole
London Dispersion
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
1 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
2. freezing = liquid - solid
3. evaporation = liquid - gas
4. condensation = gas - liquid
5. deposition = gas - solid
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
2 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.
3 Part 2: Solutions
Vocabulary
Solution - a homogeneous mixture of two or more substances in a single physical state
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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 4 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
98.08 g H2SO4
= 0.102 mol H2SO4
2. Calculate liters of solution.
100.0 mL
3. Calculate molarity.
M = mol
L

= 0.1000L
0.102 mol H2SO4= 1.02 M
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


1L
1000 mL
85.00 g
1 mol
= 21.25 g NaNO3
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 kg 11.70 mol NaCl = 585 m 0.0200kg 5 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?
M2 = 0.10 M
M1 = 12.0 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.
6 Solubility - We can predict miscibility using the rule “LIKE DISSOLVES LIKE”
 Polar + ionic = miscible
 Non-polar + ionic = immiscible
 Polar + Polar = miscible
 Non-polar + Non-polar = miscible
 Polar + non-polar = 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
7 Oil 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
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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
100g 200 g
so 140 g of KNO3
4. What is the solubility of NaCl at 99°C in 50g of water? 40 g = x g
100g
50 g
so 20 g of NaCl
8 Dilution Problems 1) How much concentrated 18 M sulfuric acid is needed to prepare 250 mL of a 6.0 M solution? M1V1 = M2V2
M2 = 6.0 M
M1 = 18 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
M2 = 2.0 M
M1 = 12 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
1 mol NaCl
58 g 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 9 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
1 mol AgNO3
10.0 g AgNO3
169.88 g AgNO3
= 0.0589 mol AgNO3
Step 2 convert to Liters
500.0 mL
1L
1000 mL
= 0.5000 L
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 L 1.00 mol KNO3
0.500 M = . X
= 1.00 mol KNO3 2.00 L 101.11 g
1 mol
= 101.11 g KNO3
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
74.55 g
= 0.067 mol KCl
M = mol L 0.25 M = 0.067 mol = 0.268 L KCL X 10 Part 3: Colligative Properties Notes
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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) i = The Number of Ions
Covalent molecules do not
Boiling Point ΔTb = i Kb m disassociate in water so
C6H12O6 is covalent i = 1, kb = 0.515 oC/m, m = moles
i=1
Kg
Ionic compounds
Convert 45.0 g of glucose to moles and 255g of water to kg disassociate in water so
ΔTb = i Kb m = (1) (0.515 oC/m) 0.25 moles
i = the number of ions
0.255 kg
(elements) in the compound.
ΔTb = 0.505 oC so new boiling point is
NaCl i = 2
100 oC + 0.505 oC = 100.5 oC
11 MgCl2 i = 3
AlCl3 i = 4
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
o
ΔTf = 1.82 C
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
1 mol
12 = 400 g NH4Cl
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
1 mol
= 925 g Ca(OH)2
After we have converted to grams then we can follow our regular stoichiometry
steps to solve any stoichiometry problem.
13 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
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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
14 
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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.
15 The Equilibrium Constant, K or Keq
The constant, K, was derived from experimentation.
For the reaction aA + bB ↔ cC + dD
uppercase letters are chemical substances:
where lower case letters are coefficients and
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
16 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]
17 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)
a. adding N2
Shift to Products
Ex:
at equilibrium, disturb the system by:
b. adding NH3
Shift to Reactants
CaCO3 (s)  CaO (s) + CO2 (g)
a. removing CO2
Shift to Products
b. adding CaO
Shift to Reactants
18 c. removing NH3
Shift to Products
at equilibrium, disturb the sys by:
c. adding CaCO3
Shift to Products
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
Ex:
- shift to L, 4 molecules
CaCO3 (s)  CaO (s) + CO2 (g)
0 molecules
at equilibrium, disturb the system by:
1 molecule
a. decreasing volume of container
b. increasing volume of container
- shift L, 0 gas molecules
- shift R, 1 gas molecules
19 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)
20 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:
1.
2.
3.
4.
5.
6.
7.
8.
9.
Change
Add N2O4
Remove N2O4
Add NO2
Remove NO2
Add Ne
Inc container vol
Dec container vol
Inc pressure
Dec pressure
Shift
Left
Right
Right
Left
Right
Left
Right
Right
Left
10. Inc temp
Left
11. Dec temp
Right
(inert gas)
(2 gas molecules vs. 1 gas mlc)
(2NO2 (g) ↔ N2O4 + 58 kJ exo
reaction)
21 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 mol
= 1.60 g CuSO4
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
M = mol L 1 mol
= 0.3499 mol NaCl
58.45 g
x M = 0.3499 mol = 0.4998 M 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
M = mol L 1 mol
= 0.4573 mol NaCl
342.22 g
x M = 0.4573 mol = 2.23 M 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
1 mol
= 249 g KI
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
1 mol
22 = 5.46 g NH4Cl
6. Calculate the molarity of a solution containing 106 g of naphthalene (C10H8)
dissolved in 3.15 L of water.
128
106 g C10H8
M = mol/L
1 mole C10H8
128 g C10H8
0.828 mol C10H8 / 3.15 L
23 = 0.828 mol C10H8
M = 0.263 M
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°C100°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__
24 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)
37.5 mol
+
Ca(OH)2 (s) 
Excess
Ca(H2PO4)2 (aq) +
2 H2O (l)
?g
mol
37.5 mol H3PO4
2 mol H2O
2 mol H3PO4
18.02 g H2O
1 mole H2O
= 675 g 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
180 g
= 0.0167 mol C6H12O6
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
25 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
26 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) 27 Keq = [Li+] [CO3‐2] 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.
28 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.
29 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 Decreases
Increases
Remains the
same
1. Add N2 Right
2. Add H2 Right
Decreases
3. Add NH3 Left
Left
Increases
_____
4. Remove N2 Left
_____
Decreases
Decreases
5. Remove H2 Left
Left
_____
Decreases
Right
Decreases
_____
_____
Increases
Right
6. Remove NH3 7. Increase Temperature Right
Increases
Increases
Decreases
8. Decrease Temperature Left
Decreases
Decreases
Increases
Decreases
Decreases
Increases
Increases
Increases
Decreases
9. Increase Pressure 10. Decrease Pressure Right
Left
30