Answers to 1st Sem. AP Chemistry Review 09
General Chemistry
1.(a)6.0000 g C* 1 mol/12.01 g = 0.5000 mol C
1.344 g H * 1 mol /1.008 g = 1.333 mol H
Mole ratio: 0.5000/0.5000 = 1 C : 1.333 mol H/0.5000 =2.666 H
Multiply both by 3 to get whole #’s. Therefore, the Empirical Formula is C3H8.
grams
(b)
i.
PV =
molar mass
RT
L • atm 

(1.96 g/L) 0.08205
298 K

mol • K
1.09 atm
ii.
molar mass =
gramsR T
P*V
s
= 43.99 g/mol
Since the Molar mass = the mass of the empirical formula, the molecular
formula is also C3H8.
Acid/Base Standard Titration
2. (a) too low; molarity NaOH is lower concentration (through dilution with the drops of
distilled water remaining in the buret) than standard leading to a higher volume used
in titration, since MBVB = molA and 0.500 g/molA = molar mass, then a larger
denominator gives a result that is too small.
(b) not affected; extra water changes neither the moles of acid originally measured nor
the volume of base required to reach the end point.
(c) too high; the equivalence point is reached with too little volume of base, since MBVB
= molA and 0.500 g/molA = molar mass, then a smaller denominator gives a result
that is too large.
(N.B., there would be no effect if the NaOH were standardized with the same
indicator)
(d) too low; the volume of NaOH would be higher than expected, since MBVB = molA
and 0.500 g/molA = molar mass, then a larger denominator gives a result that is too
small.
Stoichiometry
3. (a) Not Applicable for final exam – gas laws! 0.145 g CO2 produced.
(b) CaCO3(s)  CaO(s) + CO2(g); MgCO3  MgO(s) + CO2(g)
(c) Find the moles of Ca in the sample: 0.0448 g Ca* 1 mol/40.078 g = 0.00112 mol
Ca. The moles of carbonate are also 0.00112 mol, which has a mass of 0.00112 mol
* 60 g/mol = 0.0671 grams. This means that the Calcium carbonate sample has a
mass of 0.0448 + 0.0671 g = 0.1119 grams of CaCO3 are produced which is
(.1119 g / .2800 g) * 100% = 40.0 % of the original limestone sample.
(d) 0.2800 g sample – 0.112 g CaCO3 = 0.169 g MgCO3
That is 0.00200 mole of MgCO3. Therefore 0.00200 moles of MgO are left over
which = 0.0806 grams of MgO.
Non-numbered Stoichiometry Question
(a) nH3PO4 = 0.600mol Ba(NO3)2(2 mol H3PO4/3 mol Ba(NO3)2) = 0.400 mol H3PO4
required to completely react with 0.600 mol Barium Nitrate. There is 0.300 mol
phosphoric acid available. Therefore, phosphoric acid is the limiting reactant.
MassBa3(PO4)2 = 0.300 mol H3PO4(1 mol Ba(NO3)2/2 mol H2PO4) *(602 g
Ba3(PO4)2/mol Ba3(PO4)2) = 90.3 g Ba3(PO4)2.
(b)moles of HNO3 = 0.300 mol H3PO4(6 mol HNO3/2 mol H3PO4) = 0.900 mol
HNO3. Molarity of HNO3 = 0.900 mol/2.0 L = 0.45 M.
Since all the H3PO4 has reacted, the only acid in the solution is HNO3. Since HNO3
is a strong acid it completely dissociates.
pH = -log[H+] = -log (0.45) = 0.35
(c) The final concentration of NO3- must be the same as the initial concentration.
NNO3- = 0.600 mol Ba(NO3)2 (2mol NO3-/1 mol Ba(NO3)2) = 1.2 mol NO3[NO3-] = 1.2 mol nitrate/2.0 L = 0.60 M nitrate ion
4. (a) CO will decrease. An increase of hydrogen gas molecule will increase the rate of
the reverse reaction which consumes CO. A LeChatelier Principle shift to the left.
(b) CO will increase. Since the forward reaction is endothermic (a H > 0) an
increase in temperature will cause the forward reaction to increase its rate and
produce more CO. A LeChatelier Principle shift to the right.
(c) CO will decrease. A decrease in volume will result in an increase in pressure, the
equilibrium will shift to the side with fewer gas molecules to decrease the pressure,
, a shift to the left.
(d) CO will remain the same. Once at equilibrium, the size of the solid will affect
neither the reaction rates nor the equilibrium nor the concentrations of reactants or
products.
5. (a) Kc =
(b) (i)
(ii)
[H2]2[S2]
[H2S]2
3.7210–2 mol S2
1.25 L
2 mol H2
1 mol S2
–2
M H2

1 mol  
2 mol H S 
3.40 g H S 
3.72 10 mol S 


34.0 g  
1 mol S 
1.25 L
–2
M H2S

5 . 952
5.
(c)
(d)
(e) K’c =
10
Kc =
2 

2 2 3 . 72 10


 1 . 25


0 . 02048 2
PV=nRT = 1.18
1
= 0.251
K c = 2.00
6. -1323kJ per 2 moles of water vapor are released. If those two moles of water vapor
are changed to liquid water, -88 kJ would also be released (2 x -44 kJ/mol). A total
of -1411 kJ would be released.
7. (a) x would be endothermic; Y would be exothermic; Z would be exothermic.
(b) delta H is positive for X and negative for Y and Z
(c) The delta H is the same for both Y and Z, though Z has absorbed more energy to
initiate the reaction.
(d) X has the highest activation energy, therefore it would benefit the most from a
catalyst. A catalyst can change the activation energy, but not the delta H of the
reaction.
(e) TS = transition state which is the molecule produced when all of the reactants are
combined into an unstable structure. Eact is the activation energy. This is the
energy needed to activate the reaction.
8. 3.2 grams of methanol are needed to produce 71.1 KJ of energy.
9. (a) ∆H°f (C7H16) = -191 kJ/mol
(b) qreleased = 0.0108 mol C7H16(-4850 kJ/mol C7H16) = 52.4 kJ of heat released.
10. (a)
(C2H6) - [2(130.7) + 200.9] J/K
(C2H6) = 229.6 J/K
(b)
H
 Hâ
-  Hâ
= -84.7 kJ - [226.7 + 2(0)] kJ = -311.4 kJ
G
H S
= -311.4 kJ - (298K)(-0.2327 kJ/K) = -242.1 kJ
G
G
eous forward reaction.
- G/RT
42
-(-242100/(8.314)(298))
(c)
Keq = e
=e
= 2.74 10
(d)
H
f products - bond energy of reactants
-311.4 kJ = [(2)(436) + Ecc + (2)(414)] - [347 + (6)(414)] kJ
Ec≡c = 820 kJ
(products)
(reactants)
Atomic Structure/Bonding
11. (a) The valence electrons in a calcium atom are the 4s2. In a calcium ion these
electrons are absent and the highest energy electrons are 3p, which has a much
smaller size because the (-)/(+) charge ratio is less than 1 causing a contraction of the
electron shell.
Q 1Q 2
r ,
(b) Lattice energy can be represented by Coulomb’s law: lattice energy = k
where Q1 and Q2 are the charges on the ions, in CaO these are +2 and -2 respectively,
while in K2O they are +1 and -2. The r (the distance between ions) is slightly smaller
in CaO, combined with the larger charges, thus accounts for the larger lattice energy.
(c) Electron arrangements: K = [Ar] 4s1, Ca = [Ar] 4s2
(i) Potassium has a single 4s electron that is easily removed to produce an [Ar] core,
whereas, calcium has paired 4s electrons which require greater energy to remove
one.
(ii) a K+ ion has a stable [Ar] electron core and requires a large amount of energy to
destabilize it and create a K2+ ion. Ca+ has a remaining 4s1 electron that is more easily
removed than a core electron, but not as easily as its first 4s electron.
(d) Electron arrangements,
Mg = [Ne] 3s2, Al = [Ne] 3s2, 3p1
It is easier to remove a shielded, single, unpaired 3p electron from the aluminum
than to remove one electron from a paired 3s orbital in magnesium.
12. H2 (g) + Cl2(g)  2HCl(g) Calculate Hrxn = bonds broken - bonds formed
(1 mol H2)(440kJ/mol) + (1 mol Cl2)(240 kJ/mol)]-(2mol HCl(430 kJ/mol)) = -180 kJ
Periodic Table
13. (a) Both Ca2+ and Cl-1 ions have 18 electrons. Their electron configuration is
1s22s22p63s23p6. However, they differ by the # of protons in the nucleus.
Calcium has 20 protons and chlorine has 17 protons.
The valence electrons are shielded by the same # of electrons in each ion, so the
effective nuclear charge experienced byt eh valence electrons in the calcium ion is
+10 and for the chloride ion it is +7. The valence electrons in chloride ion
experience a smaller attraction to the nucleus due to the smaller nuclear charge, so
Cl- has the larger ionic radius.
(b)Binary compounds of carbon exhibit covalent character (property of a nonmetallic
element), whereas binary compounds of lead exhibit ionic character (property of
a metallic element).
(c) Helium has a filled shell (the first shell), so does not tend to lose or gain
electrons. Therefore, helium does not react. Krypton, while having filled 4s and
4p sublevels, has empty 4d and 4f sublevels. These empty orbitals affect the
reactivity of Kr. Also acceptable is a comparison of the ionization energies of
helium and krypton and then the justification for krypton being more reactive.
(d) The electron configuration for Be is 1s22s2, whereas the electron configuration
for B is 1s22s22p1. The first electron removed in boron is in a sp subshell, which
is higher in energy than the 2s subshell, from which the first electron is removed
in beryllium. The higher in energy the subshell containing the electron to be
removed (ionized), the lower the ionization energy.
Equations
1) Solid ammonium carbonate is heated.
(NH4)2CO3  2NH3 + CO2 + 2H2O
2) a piece of nickel metal is immersed in a solution of copper(II) sulfate.
Ni + Cu2+  Ni2+ + Cu
3) A small piece of sodium metal is added to distilled water.
Na + 2H2O 2 Na+ + 2OH- + H2
4) Ethanol is burned in oxygen.
C2H5OH +3O2  2CO2 +3 H2O
5) Solid calcium sulfite is heated in a vacuum.
CaSO3  CaO + SO2
6) Phosphorus(V) oxide powder is sprinkled over distilled water.
P2O5 + 3H2O  2H3PO4
7) A 0.1 M nitrous acid solution is added to the same volume of a 0.1 M sodium
hydroxide solution.
HNO2 + OH–  H2O + NO28) A solution of sodium phosphate is added to a solution of aluminum nitrate.
3–
PO4 + Al3+  AlPO4
9) Magnesium pellets are added to HCl
Mg + 2H+  Mg2+ + H2