CP Chemistry
Final Exam Review Sheet - 2011
I.
Vocabulary
Dimensional analysis
mole ratio
stoichiometry
limiting reagent
solid
liquid
vapor (gas)
melting / freezing
boiling /condensing
heat of fusion
heat of vaporization
specific heat capacity
heat
enthalpy
exothermic
endothermic
activation energy
reaction coordinate
catalyst
ideal gas
pressure
barometer
volume
temperature
absolute zero
STP
dissolve
dissociation
solvation
“like dissolves like”
concentration
molarity
molality
solubility
saturated
unsaturated
supersaturated
boiling point elevation
freezing point depression
bronsted-lowry acid
bronsted-lowry base
conjugate acid/base
hydronium/hydrogen ion
weak / strong
neutralization
salt
amphoteric
equilibrium
equilibrium constant (Keq)
Le Chatelier’s Principle
pH, pOH, Kw
titration
chemical kinetics
reaction rate
II. Problems
A. Stoichiometry
Write the equation.
Balance the equation.
Start with what you know.
Always go through moles!
Don’t stop until you find what you were looking for.
Sample Problems
Copper (II) sulfate reacts with sodium sulfide to produce copper (II) sulfide and
sodium sulfate.
1
P1a. How many moles of sodium sulfate are formed from 0.3 moles of sodium
sulfide?
P1b. How many grams of copper (II) sulfate are needed to react with 20.3
grams of sodium sulfide?
B. Thermochemistry
1. Heat of Fusion / Heat of Vaporization / Heating Curve of Water
q = n DHfusp or vap
e.g. How much heat is needed to melt 90.0 grams of aluminum?
q = (90.0 g) (1 mol / 27g)(10,665 J/mol)
q = 35,600 J
Use q = m C DT when determining the heat during a specific phase and
q = nΔH during a phase change
e.g. How much energy is needed to heat 300g of water from -10°C to
30°C (ΔHfus = 6 kJ/mol)
3 steps are required: 1) Heat solid water from -10°C to 0°C
2) Melt water
3) Heat liquid water from 0°C to 30°C
Step 1: q = m Cp ΔT = (300g)(2.06 J/g°C)(0 - -10°C) = 6160J = 6.16 KJ
Step 2: q = n ΔH = (300g/1)(1mol/18g)(6.0 KJ/mol) = 100 KJ
Step 3: q = m Cp ΔT = (300g)(4.184 J/g°C)(30-0°C) = 37656J = 37.66 KJ
Total = 6.16 + 100 + 37.66 = 143.82 KJ
Sample Problems
P2. How much energy is needed to fully vaporize 36g of water (ΔHvap
= 40.7 KJ/mol)
2
P3. How much energy is needed to fully condense 36g of water at
100°C (remember sign for ΔH)
P4. Determine the amount of energy released when 72g of water is
cooled from 45°C to -20°C. Draw the heating curve associated with
this question. Show the state of the water and the boiling and/or
freezing point in the diagram.
2. Specific Heat Capacity – Cp
q = m Cp DT
Eg. How much heat is needed to bring 90.0 grams of aluminum
from 25°C to 75°C?
q = (90.0 g) (0.902 J/g.K) (75°C - 25°C)
q = 4060 J
Using m Cp ΔT = - m Cp ΔT
e.g. A 25g sample of copper at 128°C is dropped into 82g of water at
34°C. The final temperature reaches 40°C. Determine the specific heat of
the copper.
m C ΔT = - m C ΔT (Heat in = heat out)
(82g)(4.184 J/g°C)(40-34°C) = -(25g)(Cp)(49-128°C)
Cp = 1.04 J/g°C
3
Sample Problems
P5. How much heat is required to change 50.0 grams of liquid water at 60°C
to 90°C?
P6. A 34g sample of aluminum (Cp = 0.902 J/g°C) is heated to 100°C and
placed into 100g of benzene at 20°C. The final temperature of the mixture is
46°C. What is the specific heat of the benzene?
3. Thermodynamics
Enthalpy - ΔH°rxn
eg. What is the heat of reaction based on this enthalpy diagram? -100 KJ
Sample Problem
P7. What is the activation energy for the reaction above?
400-100 = 300 kJ
4
4. ΔH°rxn from ΔHf
a. ΔH°rxn = ΣΔHproducts - ΣΔHreactants (Remember that ΔHf for an element
is zero)
e.g. Using heats of formation (with your homework on this), determine
ΔH°rxn for the following reaction:
2 CH3OH(l) + 3 O2 → 2CO2(g) + 4 H2O(l)
ΔH°rxn = [(2 x -393.5)+(4 x -285.8)] – [(2 x -238.4) + (3 x 0)] = 1453.4 KJ/mol
Sample Problem
P8. Determine the ΔH° rxn for the following reaction:
C3H8 (g) + 5 O2 (g) → 3 CO2 (g) + 4 H2O (g)
5. Heat Stoichiometry
Using q = n ΔH in balanced chemical equations. Remember that the ΔHrxn
gets divided by the number of moles from the balanced equation of the compound
you are solving for.
e.g. Using information from the example in section 4a above, determine
the amount of heat released when 132g of CO2 is produced.
ΔH°rxn from 4a above is -1453.4 KJ/mol
q = (132g/1)(1 mol / 44 g)(-1453.4 KJ / 3 mol CO2) = -4360.2 KJ
Sample Problem
P9. Using information from the sample problem in section 4a above,
determine the amount of heat released when 132g of H2O is produced.
C. Gases
1. Conversions
a. Pressure conversions
1 atm = 760 mm Hg = 101.3 kPa
eg. 2.5 atm = 1900 mm Hg = 253.25 kPa
5
2.
b.
Volume conversions
1 L = 1 dm3 = 1000 cm3
eg. 4.8 L = 4.8 dm3 = 4800 cm3
c.
Temperature conversion
K = °C + 273
eg. -50°C = 223 K
Gas Laws
a. Dalton’s Law of Partial Pressure
PT = P1 + P2 + P3 +…
eg. A balloon is filled with neon and helium. If the partial pressure
of neon is 300 mm Hg and the partial pressure of helium is 900
mm Hg, what is the total pressure of the gases inside the balloon?
PT = PNe + PHe
PT = 300 mm Hg + 900 mm Hg = 1200 mm Hg
b.
Boyle’s Law
P1V1 = P2V2
eg. A balloon has a volume of 0.9 L and a pressure of 3.1 atm. If
the pressure is increased to 7.1 atm, what will the new volume be?
P1V1 = P2V2
(3.1 atm)(0.90 L) = (7.1 atm) V2
V2 = 0.39 L
c.
Charles’ Law
V1 / T1 = V2 / T2
eg. A balloon has a volume of 0.90 L at a temperature of 5°C.
What will the volume be if the temperature is increased to 25°C?
V1 / T1 = V2 / T2
0.90 L / 278 K = V2 / 298 K
V2 = 0.96 L
Gay-Lussac’s Law
P1 / T1 = P2 / T2
eg. A tank with a fixed volume has a pressure of 1.00 atm at a
temperature of 27°C. What will the pressure be if the temperature is
increased to 227°C?
P1 / T1 = P2 / T2
1.00 atm / 300 K = P2 / 500 K
P2 = 1.67 atm
d.
e.
Combined Gas Law
P1V1 = P2V2
T1
T2
eg. A balloon has a volume of 0.90 L at STP. At what pressure will
the balloon have a volume of 0.40 L and a temperature of 30°C?
P1V1 = P2V2
6
T1
T2
(1atm)(0.90 L) = P2(0.40 L)
(273 K)
(303 K)
P2 = 2.5 atm
f.
Ideal Gas Law
PV = nRT PVM = mRT PM = DRT
eg. A balloon contains 2.7 moles of gas at a pressure of 450 kPa and
a temperature of -20°C. What is the volume of the balloon?
PV = nRT
(450 kPa) V = (2.7 moles)(8.314 kPa L)/(mol K) )(253 K)
V = 13 L
f. Avogadro’s Principle
The volume of one mole of ANY gas at STP is 22.4L – KNOW THIS
AND USE THIS!
Sample Problems
P10. Perform the following conversions:
a. 308 kPa = _____ mm Hg
c. 59 kPa = _____ atm
b. 1040 mm Hg = ___ atm
d. 56°C = _____ K
P11. A tank is filled with hydrogen, nitrogen, and oxygen. If the partial
pressure of hydrogen is 208.1 kPa, the pressure due to oxygen is 502.9 kPa,
and the total pressure of the gases in the tank is 780 kPa, what is the partial
pressure of nitrogen?
7
P12. Gas in a 78 cm3 container has a pressure of 8.7 atm. What will the
pressure be if the gas is transferred to a 150 cm3 container?
P13. A gas occupies 5.1 dm3 at standard temperature. At what temperature
will the gas occupy 15.3 dm3?
P14. An air-tight, 1500-mL box contains gas at 370 kPa and 76°C. What will
the volume be at a pressure of 660 kPa and -6°C?
P15. How many moles of gas are in a 30.-L container held at 4.8 atm and
40°C?
P16. 56.0 grams of nitrogen dioxide occupies 2.8 L. What will be the volume
of 3.1 moles of nitrogen dioxide?
3. Gas Stoichiometry – Using stoichiometry and the Ideal Gas Law
g A → mol A → mol B → volume B
volume A → mol A → mol B → g B
volume A → mol A → mol B → volume B
e.g. Determine the mass of aluminum chloride produced when 44.8 L of
hydrogen chloride is reacted with aluminum at 2 atm and 45°C.
8
2Al + 6HCl → 2AlCl3 + 3H2
PV = nRT
! 0.082Latm $
(2atm)(44.8L) = n #
(318K )
" molK &%
n = 3.44mol
! 3.44molHCl $ ! 2molAlCl3 $ ! 133.5g $
#"
&% #"
&#
& = 153.08gAlCl3
1
6molHCl % " 1mol %
Alternately, if STP were specified, you could find the number of moles or
liters simply by remembering that 1 mole of any gas at STP has a volume
of 22.4L, thus using ! 22.4L $ or ! 1mol $ as needed.
#"
&
1mol %
#"
&
22.4L %
Sample Problems
P17. What volume of oxygen at 25 degrees Celsius and 3 atm is needed to
completely burn 13g of methanol (CH3OH) in an engine?
2CH3OH + 3 O2 →
2CO2 +
4H2O
P18. What mass of magnesium oxide will be formed if 125L of oxygen is used
to burn magnesium at STP?
2Mg + O2 → 2MgO
D. Solutions
1. Concentration
a. Molarity - M - moles solute per liter solution
eg. What is the molar concentration of a solution made by dissolving
6.0 grams of hydrogen gas in 250 mL of water?
6.0 grams H2 | 1 mol H2 = 0.75 M H2
0.25 L | 2 g H2
b.
Molality - m - moles solute per kilogram solvent
eg. What is the molal concentration of a solution made by dissolving
6.0 grams of hydrogen gas in 250 grams of water?
6.0 grams H2 | 1 mol H2 = 0.75 m H2
9
0.25
c.
kg
| 2 g H2
Level of saturation
eg. What mass of KNO3 is needed to make a saturated solution with
320 grams of water at 50°C? Refer to solubility curve
320 grams H2O | 85 g KNO3 = 272 g KNO3
| 100 g H2
Sample Problems
P19. Calculate the molarity of a solution with 31.1 grams of CaCl2
dissolved in 100. mL.
P20. Calculate the molality of a solution made by dissolving 20.
grams of NH4Cl in 400. grams of water.
10
P21. 360 grams of KCl are used to make a saturated solution at 70°C.
How much water was needed?
P22. Determine whether the following solutions are saturated,
supersaturated, or unsaturated:
a. 123g of KI in 100mL water at 30°C
unsaturated
b. 23g of KNO3 in 25mL water at 50°C
supersaturated
c. 90g of NaCl in 50mL water at 10°C
supersaturated
2.
Freezing point depression / Boiling point elevation
ΔTb or f = Kb or f mi
eg. What is the boiling point of a solution in which 20. grams of CaBr2 is
dissolved in 500 grams of water?
ΔT = (0.10 moles / 0.500 kg H2O) x 0.52 °C kg/mol x 3
ΔT = 0.31°C
boiling point = 100°C + 0.31 °C = 100.31°C
Sample Problem
P23. What is the boiling point of a solution with 64.2 grams CuCl2
dissolved in 250 g H2O?
P25. Ice cream is best made at -6°C. If Ca(NO3)2 were used in water
as the salt to cool down the mixture to make ice cream, what molality
solution would be needed?
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E. Phase Diagrams
C
D
B
A
Be able to interpret a phase diagram
Sample problems
P26. Determine on which line/curve segment the following phase changes occur:
a. Melting (Fusion) ___AB____________
b. Evaportation (Vaporization) ____________BC_________
c. Condensation _______BD____________
P27. Phase change(s) for CO2 as ΔT goes from -60ºC to -40ºC at 6atm
______solid to liquid to gas________________
P28. What is the effect of increasing pressure on the boiling points of CO2?
_____BP’s go up_____________
P29. What state of matter is CO2 at 20 atm and -70ºC?
___________solid_____________
P30. Name of the place on the graph where CO2 is a solid, liquid, and gas?
_____B (triple point)_________
F. Acids and Bases
1.
pH = -log [H3O+]
pOH = -log [OH-]
pH + pOH = 14
a. What is the pH of a solution where the hydronium ion concentration
is 3.3 x 10-6M?
12
pH = -log(3.3 x 10-6M) = 5.5
b. What is the pOH of the above solution?
pOH = 14 – pH
pOH = 14 – 5.5 = 8.5
c. What is the hydroxide ion concentration of the above solution?
pOH = -log[OH-]
8.5 = - log [OH-]
[OH-] = 10-8.5 = 3.16 x 10-9 M
2.
Titrations
What volume of .5 M KOH is needed to neutralize 100 ml of .4 M HBr?
MaVa = MbVb
(.4 M)(100 ml) = (.5 M)Vb
Vb = 80 ml
Sample problems
P31. What is the pH of a solution that has a hydroxide ion concentration of
8.9x10-10M? Is this solution acidic or basic?
P32. What is the hydronium ion concentration of a 3.6x10-3 M solution of
HI?
13
P33. What is the hydroxide ion concentration of a 3.6x10-3 M solution of
HI?
P34. A solution was titrated with 34ml of 0.75 M NaOH. The sample
contained 31ml HCl. What is the molarity of the acid solution?
P35. A solution was titrated with 25 ml of 2 M NaOH. The sample
contained 44ml H2SO4. What is the molarity of the acid solution.
G. Chemical Equilibrium
1. Determining Keq
- write a balanced chemical equation
- write an equilibrium expression: Keq = [products]
[reactants]
Sample Problem
An aqueous solution of carbonic acid reacts to reach equilibrium as described
below:
H2CO3(aq) + H2O(l) ⇄ HCO3- (aq) + H3O+(aq)
The solution contains the following solute concentrations: carbonic acid 3.3 x
10-2 M; bicarbonate ion, 1.19 x 10-4 M. Determine the Keq.
Keq = [HCO3-][H3O+]
[H2CO3]
= (1.19 x 10-4) x (1.19 x 10-4) = 4.3 x 10-7
14
(3.3 x 10-2)
P36. If Keq is 1.65 x 10-3 at 2027 °C for the reaction below, what is the
equilibrium concentration of NO when [N2] = 1.8 x 10-3 and [O2] = 4.2 x 10-4?
H. Reaction Rates
Describes the change in the concentrations of reactants/products over time
Sample Problem
The data below were collected during a study of the following reaction.
2Br –(aq) + H2O2 (aq) + 2H3O+(aq) ---> Br2(aq) + 4H2O(l)
Time (s)
0
85
95
105
[H3O+] (M)
0.0500
0.0298
0.0280
0.0263
[Br2] (M)
0
0.0101
0.0110
0.0118
Rate = -Δ[H3O+] = Δ[Br2] = -(-0.0017 M) = 8.5 x 10-5 M/s
2Δt
Δt
2(10s)
P37. The folllowing reaction was studied by measuring the concentration of
H3O+.
CH3OH(aq) + HCl(aq) ----> CH3Cl(aq) + H2O(l)
[H3O+] (M)
1.85
1.67
1.52
1.30
1.00
Time (min)
0
79
158
316
632
15
Use the data in the table to determine the initial rate of the reaction in the
first 79 minutes.
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