HONORS Chemistry Final 2013
Study Guide
Academic Chemistry Final:
75 Multiple Choice Questions
7 Stoichiometry
9 Gases
12 Solutions
19 Acids & Bases
8 Equilibrium
7 Kinetics, Rxn Rates
7 Electrochemistry
6 Thermodynamics
Calculations
1 Stoichiometry, 2 Gas Laws, 1 Solutions, and 3 Acid-Base problems
Short Answer (Essay)
1 Gases, 1 Solutions, 1 Acids/Bases, 1 Reaction Rates, 1 Heat/Energy
Resources for the final exam
 You will be given a periodic table/ion chart and a sheet of equations (below)
Equations, Useful Info:
R = .0821 L  atm
mol  K
22.4L in 1 mole at STP
PV = nRT
STP = 273K and 1 atm
P1V1 = P2V2
T1
T2
Molarity = mol
L
1 L = 1000 mL
pH = -log [H+]
1 atm = 760 mmHg
pOH = -log [OH-]
1 atm = 101.3 kPa
pH + pOH = 14
MaVa = MbVb
Chapter 10: Stoichiometry
Vocabulary
 stoichiometry

molar ratios
Be able to….
 Determine molar ratios
 Perform mole to mole conversions
 Perform multi-step conversions
o (mass-mass, mass-volume, volume-volume)
Practice Problems:
1. Use the following balanced equation to answer questions (2 points each)
2Al2(CO3)3 + 3H2SO4  2Al2(SO4)3 + 3 H2O + 3CO2
a. 2:3
What is the molar ratio for Al2(CO3)3 and H2O?
b. 3 mol
How many moles of CO2 are produced when 3 moles of H2SO4 react?
c. 2 mol
How many moles of H2O are produced when 2 moles of H2SO4 reacts?
2.
How many moles of oxygen are needed to react with 87g of lithium?
4Li + O2  2Li2O
87g Li x 1 mol x 1 mol O2 x 32.00g O2 = 100.2 g O2
6.94g Li 4 mol Li
1 mol
3.
Use the equation to determine what mass of FeS must react to form 326g of FeCl 2.
FeS + 2HCl  H2S + FeCl2
326 g FeCl2 x
1 mol
x 1 mol FeS x 87.91g FeS = 226 g FeS
126.75g FeCl2 1 mol FeCl2 1 mol
4.
If a piece of magnesium with a mass of 2.76g is added to a solution of hydrochloric acid (HCl),
what volume of hydrogen gas would be produced at STP?
2.76g Mg x
1 mol x 1 mol H2 x 22.4 L H2 = 2.54 L H2
24.31g Mg 1 mol Mg
1 mol
5. How many grams Au can be produced when 500 g of Rb is used to reduce it:
Au 2 O3  6Rb  3Rb2 O  2 Au
500 g Rb x
1 mol
x 2 mol Au x 196.97g Au = 384 g Au = 400 g (Sig Figs)
85.47 g Rb
6 mol Rb
1 mol
6. How many grams of CO2 are liberated when 400 g of Propane is burned?
C3 H 8  5O2  4H 2 O  3CO2
400 g C3H8 x
1 mol
x 3 mol CO2 x 44.01 g CO2 = 1197 g CO2 = 1000 g (Sig Figs)
44.11 g C3H8 1 mol Mg
1 mol
7. How many grams in 4.2 moles of KNO2?
4.2 mol x 85.11g = 187 g = 190 g (Sig Figs)
1 mol
8. How many moles in 2.7 L of CO at STP?
2.7 L x 1 mol = .12L
22.4 L
Chapter 12: Gases
Vocab:
 Kinetic Molecular Theory
 Atmospheric pressure
 Temperature



Kinetic Energy
Pressure
STP
Be able to….
 Compare/ contrast the characteristics of solids/ liquids/ gases
 Describe the Kinetic-Molecular theory and explain how it explains gas behavior.
 Explain what gas pressure means and describe how it is measured.
 Solve problems using the gas laws.
Practice Problems:
1. Compare the characteristics of solids/ liquids/ gases:
Solids
definite shape, definite volume; particles vibrate in position
Liquids
indefinite shape, definite volume
Gases
indefinite shape, indefinite volume; particles are far apart (low density, compressible);
particles move in straight lines and collide with walls of the container
2. 91 K A 3.00 liter (V1) sample of neon gas at 0C (T1 = 273 K) and 1.25 atm (P1) is compressed
into a 1.00 liter (V2) container. If the pressure remains constant, what temperature will the
container be?
T2 = P2V2T1 
T2 = V2T1 = 1.00L x 273 K = 91 K
P1V1
3. 96.22atm What is the pressure (P = ?) exerted by 64 grams (convert to moles - n) of oxygen
confined to a volume of 500 mL (V = .500L) at 20 C (T = 293 K)?
64 g O2 x 1 mol = 2.0 mol (n)
32.00 g
P = nRT = 2.0mol x .0821 x 293 K = 96.22atm (this Pressure seems really high?)
V
.500L
4. 1.3 mol How many moles of gas are in a 52 L (V) sample collected at 220 K (T) and .444atm?
n = PV = .444atm x 52L = 1.3 mol
TR 220K x .0821
5. 4.9 L Find the new volume (V2 = ?) when a 2.1 L (V1) sample of a gas collected at 245 Kelvin (T1)
and 2.1atm (P1) is changed to standard conditions (STP: T2 = 273 K, P2 = 1 atm).
V2 = P1V1T2 = 2.1 atm x 2.1 L x 273 K = 4.9 L
T1P2
245 K x 1 atm
6. 23 mL Find the new volume (V2= ?) of a gas that changes 65 ml (V1) at 150 mmHg (P1) to 425
mmHg (P2).
V2 = P1V1T2 = V2 = P1V1 = 150 mmHg x 65 mL = 23 mL
T1P2
P2
425 mmHg
7. Explain the relationship between each of the variables for the following gas laws:



Boyle’s Law: as P increases, V must increase if T and n are constant
Charles’ Law: as T increases, P increases (greater T = more movement = more collisions)
Avogadro’s Law: as n (moles) increases, V increases (think about blowing up a balloon)
Chapter 13: Solutions
Vocab








solute
solvent
saturated
unsaturated
supersaturated
polar/non-polar
soluble
insoluble







miscible
immiscible
solubility curve
solution
colloid
suspension
molarity
Be able to…
 Identify mixtures as solutions, colloids, or suspensions
 Explain colligative properties and relate to practice applications
 salting roads, ice cream
 Interpret a solubility curve
 Identify substances as saturated, unsaturated, or supersaturated
 Calculate molarity
Practice Problems:
1. Intrepret solubility curves:
a. What substance is most soluble at 20oC?
KClO3
b. What substance is least soluble at 90 oC?
Ce2(SO4)3
c. What is the solubility of KNO3 at 50 oC?
80 g
d. How many grams of NaNO3 can dissolve in 100
grams of water at 60° C?
122 - 123 g
e. If 70g of KCl is dissolved at 70 oC, is the solution
saturated, unsaturated, or supersatured?
supersaturated
2. .33M Calculate the molarity when 2 mol of CuSO4 dissolves in 6L of water.
2 mol = .14M
6L
3. .14M Find the molarity of NaCl when 20 grams are mixed with 2500 ml of water.
20 g x 1 mol = .3422mol
58.44g
.3422mol = .14M
2.500 L
4. .18g What mass of HCl is needed to prepare 1.5 L of a 0.010 M solution.
1.5L x .010 mol = .005mol
1L
.005mol x 36.46g = .18g
1 mol
5. .5m What is the molality if 2.0 mol of solute dissolves in 4.0kg of solvent?
2 mol = .5m
4.0kg
Chapter 14: Equilibrium
Vocabulary
 equilibrium
 Le Chatelier’s principle
Be able to….
 Interpret a graph to identify when equilibrium occurs
 Predict shifts in equilibrium based on Le Chatelier’s principle
 Write equilibrium expressions and calculate equilibrium constants
Practice Problems:
3. In the graph below, at what time does equilibrium occur?
4. Write the equilibrium expression for the following reactions:
2 NH3(g)  N2(g) + 3 H2(g)
K = [H2]3 [N2]
[NH3]2
CaH2 (s) + 2 H2O (g)  Ca(OH)2 (s) + 2 H2 (g)
Recall: ignore solids and liquids when writing equilibrium expressions
K = [H2]2
[H2O]2
5. Calculate the equilibrium constant if [N2O4] = .0337 and [NO2] = .0125.
N2O4(g)  2 NO2(g)
K = [NO2]2
[N2O4]
K = [.0125]2
[.0337]
K = .0046
6. Predict the shift in the reaction using Le Chatelier’s principle:
Fe2O3(s) + 3 CO(g)  2 Fe(l) + 3 CO2(g)
a. adding Fe2O3

b. removing CO

c. adding CO2

d. increasing pressure no change (equal # of gas molecules)
Chapter 15: Acids and Bases:
Vocabulary
 Arrhenius Acid, Base
 Bronsted Lowry Acid, Base
 dissociation
 titration
Be able to…
 Identify substances in chemical equations as acids or bases
 Identify the common physical and chemical properties of acids and bases
 Calculate pH of solutions
 Determine an unknown concentration of an acid/base using titrations
Practice Problems:
1. Label the properties of acids and bases:
Dissociates into ____ ions
pH range?
Taste?
Feels?
Conducts Electricity?
Bronsted-Lowry definition
Turns litmus paper ______
Turns Phenolphthalein ______
Acids
H+
0-7
Sour
Yes
Proton Donor
Pink
Bases
OH7-14
Bitter
Slipper
Yes
Proton Acceptor
Blue
Clear
Pink
2. 3 Find the pH of a 1.0 x 10-3 M solution of HCl
pH = -log [1.0 x 10-3] = 3
3. 11.70 Find the pH of a .005 M NaOH solution.
pOH = -log [.005] = 2.30
pH + pOH = 14
pH = 14 – 2.30
pH = 11.7
4. 2 M What is the unknown concentration of base if 40mL of NaOH is titrated with
80mL of 1M solution of standardized HCl?
MaVa = MbVb
1 M (80mL) = Mb (40mL)
5. Identify the Bronsted-Lowry acid/ base, and the conjugate acid/ base.
NH3 + H2O  NH4+ + OH¯
B
A
CA
CB
HC2H3O2 + H2O  H3O+ + C2H3O2¯
A
B
CA
CB
Chapter 16: Reaction Rates
Vocabulary
 Collision Theory
 Catalyst
Be able to…
 Describe the two conditions for a successful chemical reaction
 Understand chemical reactions in terms of collision theory
 Be able to list and describe the factors that affect reaction rates
 Interpret potential energy diagrams
Practice Problems:
1. The following data is for the reaction:
A(g) + 2B(g)  C(g) + D(g)
Experiment
1
2
3
[A] (mol/L)
.15
.30
.15
[B] (mol/L)
.10
.10
.20
Rate (M/s)
.45
1.8
.9
Write the law expression and calculate k for the reaction:
A: double, quadruple = 2nd order
B: double, double = 1st order
rate = k [A]2 [B]1
calculate k:
.45 = k [.15]2 [.10]
k = 200
2. What are the two conditions for a successful reaction?
Orientation
Enough energy to overcome the activation energy for rxn to occur
3. Explain how the following factors change reaction rates:
•
surface area of a solid reactant
increase surface area (powder) because a greater surface area means there
are more possible sites for collisions = faster reaction
•
concentration of a reactant
increase concentration so that there are more reactant particles which means
more collisions = faster reaction
•
temperature
increase temperature because temperature is a measure of the average kinetic
energy, so particles move faster and collide more often = faster reaction
•
presence of a catalyst
catalysts speed up reactions by lowering the activation energy
Chapter 17: Electrochemistry
Vocabulary
 oxidation
 reduction
 anode
 cathode




voltaic/galvanic cell
electrolytic
spontaneous/ non-spontaneous
battery
Be able to…
 Assign oxidation numbers
 Identify half-reactions as oxidation or reduction
 Calculate the cell potential
 Compare/ contrast voltaic/galvanic and electrolytic electrochemical cells
Practice Problems:
1. Assign oxidation numbers:
a. Cr in Cr2O7-2
+6
b. S in SO4-2
+6
c. S in S8
0
d. Fe in Fe2O3
+3
2. Write the oxidation and reduction half-reactions:
a. Fe(s) + 2H+(aq)  Fe+2(aq) + H2(g)
0
+1
+2
0
ox:
Fe  Fe2+ + 2ered: 2 H+ + 2e-  H2
b. Mg + Ca(OH)2 → Ca + Mg(OH)2
0
ox:
red:
+2
-1
0
Mg  Mg2+ + 2eCa2+ + 2e-  Ca
+2 -1
3. Calculate the cell potential and determine if the reaction is spontaneous or not:
a. 2 Ag+ (aq) + Cu (s)  Cu2+ (aq) + 2 Ag (s)
ox:
Cu  Cu2+ + 2e-.34 V (oxidation = flip the sign)
+
red: Ag + e  Ag
.80 V
-.34 V + .80 V = .46V spontaneous (positive)
b. 2 Ag+ (aq) + Zn (s)  Zn2+ (aq) + 2 Ag (s)
ox:
Zn  Zn2+ + 2e.76 V (oxidation = flip the sign)
red: Ag+ + e-  Ag
.80 V
.76 V + .80 V = 1.56V spontaneous (positive)
Chapter 11: Heat & Energy
Vocabulary
 heat

temperature
Be able to….
 Identify exothermic vs endothermic reactions.
 Calculate heat absorbed or released by a reaction using q = m ΔCT
 Use Hess’ Law to calculate the heat of a net reaction
Practice Problems:
1. Label each example as exothermic or endothermic
a. 2H2(g) + O2(g) → 2H2O(g)
ΔH°= -243 kJ
b. H2B4O7(s) → B2O3(s) + H2O(l) feels warm
c. H2B4O7(s) + H2O(l) → 4HBO2(aq) + 11.3 kJ
d.
exothermic (ΔH is negative)
exothermic (exo = warm)
exothermic (energy is a product)
exothermic
2. How much heat is required to heat 40.0g of water from 25.0oC to 75oC? (C=4.18 J/g0C)
q = mCΔT
q = 40.0g (4.18 J/goC) (75-25oC)
q = 8360 J
3. Calculate the heat for the net overall reaction: CH4(g) + H2O(l) → CO(g) + 3H2(g)
1) CO(g) + H2(g) → C + H2O(g)
- ΔH° = -131.3 kJ
2) C+ 2H2(g) → CH4(g)
- ΔH° = -74.8 kJ
=206.1 kJ
4. Be able to interpret a phase change diagram:
Heat is still being applied as phase changes (though there is no change in temp)
Recognize fusion, vaporization