AP Chemistry Name ______
Fall Semester Practice Free Response Period _____
Free Response: Answer the questions on a separate sheet of paper.
1. Answer the following questions related to sulfur and one of its compounds.
a. Consider the two chemical species S and S2-.
(1) Write the electron configuration (1s2 2s2 ...) of each.
(2) Explain why the radius of the S2- ion is larger than the radius of the S atom.
(3) Which of the two species would be attracted into a magnetic field? Explain.
b. The S2- ion is isoelectronic with the Ar atom. From which species, S2- or Ar, is it easier to remove an electron? Explain.
c. In the H2S molecule, the H–S–H bond angle is closer to 90° than 109.5o. On the basis of this information, which atomic orbitals of the S atom are involved in bonding with the H atoms?
d. Two types of intermolecular forces present in liquid H2S are London (dispersion) and dipole-dipole forces.
(1) Compare the strength of the London (dispersion) forces in liquid H2S to the strength of the London (dispersion) forces in liquid H2O. Explain.
(2) Compare the strength of the dipole-dipole forces in liquid H2S to the strength of the dipole-dipole forces in liquid H2O. Explain.
2. Using principles of atomic and molecular structure and the information in the table, answer the questions about atomic fluorine, oxygen, and xenon, as well as some compounds.
Atom / F / O / XeFirst Ionization Energy (kJ/mol) / 1,681.0 / 1,313.9 / ?
a. Account for the fact that the first ionization energy of atomic fluorine is greater than that of atomic oxygen. (You must discuss both atoms in your response.)
b. Predict whether the first ionization energy of atomic xenon is greater than, less than, or equal to the first ionization energy of atomic fluorine. Explain.
c. Xenon can react with oxygen and fluorine to form compounds such as XeO3 and XeF4. Draw the complete Lewis electron-dot diagram for each of the molecules.
d. On the basis of the Lewis electron-dot diagrams you drew for part (d), predict the following:
(1) The geometric shape of the XeO3 molecule
(2) The hybridization xenon in XeF4
3. Answer the following questions by using principles of molecular structure and intermolecular forces.
a. Structures of the pyridine molecule and the benzene molecule are shown below. Pyridine is soluble in water, whereas benzene is not soluble in water. Account for the difference in solubility. You must discuss both of the substances in your answer.
Pyridine Benzene
b. Structures of the dimethyl ether molecule and the ethanol molecule are shown below. The normal boiling point of dimethyl ether is 250 K, whereas the normal boiling point of ethanol is 351 K. Account for the difference in boiling points. You must discuss both of the substances in your answer.
Dimethyl Ethanol
Ether
c. SO2 melts at 201 K, whereas SiO2 melts at 1,883 K. Account for the difference in melting points. You must discuss both of the substances in your answer.
d. The normal boiling point of Cl2(l) (238 K) is higher than the normal boiling point of HCl(l) (188 K). Account for the difference in normal boiling points based on the types of intermolecular forces in the substances. You must discuss both of the substances in your answer.
4. A student was assigned the task of determining the molar mass of an unknown gas. The student measured the mass of a sealed 843 mL rigid flask that contained dry air. The student then flushed the flask with the unknown gas, resealed it, and measured the mass again. Both the air and the unknown gas were at 23.0°C and 750. torr. The data for the experiment are shown in the table below.
Volume of sealed flask / 843 mLMass of sealed flask and dry air / 157.70 g
Mass of sealed flask and unknown gas / 158.08 g
a. Calculate the mass, in grams, of the dry air that was in the sealed flask. (density of dry air is 1.18 g•L−1)
b. Calculate the mass, in grams, of the sealed flask itself.
c. Calculate the mass, in grams, of the unknown gas that was added to the sealed flask.
d. Calculate the molar mass of the unknown gas.
After the experiment was completed, the instructor informed the student that the unknown gas was carbon dioxide (44.0 g/mol).
e. Calculate the percent error in the value of the molar mass calculated in part (d).
f. For the two possible occurrences, indicate whether it by itself could have been responsible for the error in the student’s experimental result. Explain
Occurrence 1: The flask was incompletely flushed with CO2, resulting in some dry air remaining in the flask.
Occurrence 2: The temperature of the air was 23.0°C, but the temperature of the CO2 was lower than 23.0°C.
g. Describe the steps of a laboratory method that the student could use to verify that the volume of the rigid flask is 843 mL at 23.0°C.
5. Methane gas reacts with chlorine gas to form dichloromethane and hydrogen chloride, as represented by the equation. CH4(g) + 2 Cl2(g) → CH2Cl2(g) + 2 HCl(g)
a. A 25.0 g sample of methane gas is placed in a reaction vessel containing 2.58 mol of Cl2(g).
(1) Identify the limiting reactant.
(2) Calculate the theoretical yield of CH2Cl2(g) in grams.
The reaction is initiated by shining light on the mixture. It takes 242 kJ to break one mol of Cl–Cl bonds.
b. Calculate the amount of energy, in joules, needed to break a single Cl–Cl bond.
c. Calculate the longest wavelength of light, in meters, that can supply the energy to break the Cl–Cl bond.
6. A student is assigned the task of determining the number of moles of water in one mole of MgCl2•n H2O. The student collects the data shown in the following table.
Mass of empty container / 22.347 gInitial mass of sample and container / 25.825 g
Mass of sample and container after 1st heating / 23.982 g
Mass of sample and container after 2nd heating / 23.976 g
Mass of sample and container after 3rd heating / 23.977 g
a. Explain why the student can correctly conclude that the hydrate was heated a sufficient number of times.
b. Use the data above to
(1) calculate the moles of water lost during heating.
(2) determine the formula of the hydrated compound.
c. A different student heats the hydrate in an uncovered crucible, and some of the solid spatters out of the crucible. This spattering will have what effect on the calculated mass of the water lost by the hydrate? Explain.
7. Answer the following questions about nitrogen, hydrogen, and ammonia.
a. Draw the complete Lewis electron-dot diagrams for N2 and NH3.
b. Calculate the standard free-energy change, ΔGo, that occurs when 12.0 g of H2(g) reacts with excess N2(g) at 298 K according to the reaction:
N2(g) + 3 H2(g) D 2 NH3(g) DGo298 = -34 kJ•mol-1
c. Given that ΔHo298 for the reaction is -92.2 kJ•mol-1, which is larger, the total bond dissociation energy of the reactants or the total bond dissociation energy of the products? Explain.
d. The value of the standard entropy change, ΔSo298, for the reaction is -199 J•mol-1•K-1. Explain why the value of ΔSo298 is negative.
e. Assume that ΔHo and ΔSo for the reaction are independent of temperature. Explain why there is a temperature above 298 K at which the algebraic sign of the value of ΔGo changes.
8. A student performed an experiment to investigate the decomposition of sodium thiosulfate, Na2S2O3, in acidic solution, S2O32- ® SO32- + S(s). In each trial the student mixed a different concentration of sodium thiosulfate with hydrochloric acid at constant temperature and determined the rate of disappearance of S2O32-(aq). Data from five trials are given below in the table on the left and are plotted in the graph on the right.
Trial / Initial Concentration of S2O32-(M) / Initial Rate of Disappearance of S2O32- (M•s-1)
1 / 0.050 / 0.020
2 / 0.075 / 0.030
3 / 0.088 / 0.034
4 / 0.112 / 0.045
5 / 0.125 / 0.051
a. Identify the independent variable in the experiment.
b. Determine the order of the reaction with respect to S2O32-. Justify your answer with the information above.
c. Determine the value of the rate constant, k, for the reaction. Include units in your answer. Show how you arrived at your answer.
d. In another trial the student mixed 0.10 M Na2S2O3 with hydrochloric acid. Calculate the amount of time it would take for the concentration of S2O32- to drop to 0.020 M.
e. On the graph below, sketch the line that shows the results that would be expected if the student repeated the five trials at a temperature lower than that during the first set of trials.
Answers
1 / a1 / S: 1s2 2s22p6 3s23p4, S2-: 1s2 2s22p6 3s23p6a2 / S2- has a greater radius because the two additional electrons increase the electron-electron repulsion but the electron to nucleus attraction is the same \ resulting in a larger radius.
a3 / S is paramagnetic because it has two unpaired electrons in the 3p sublevel, which produce a net magnetic field that is attracted to the external field. S2- has all paired electrons so it is diamagnetic.
b / It is easier to remove an electron from S2- because its valence electrons are less attracted to the nucleus (16) compared to Ar (18).
c / The bonding electrons come from the p sublevel, whose orbitals are 90o apart, rather than sp3 hybrid orbitals, which are 109.5o apart.
d1 / The London forces are stronger in H2S compared to H2O because S has more electrons than O, which are more polarizable.
d2 / The dipole-dipole forces in H2O are stronger compared to H2S because O has a greater electronegativity, which increases the H–O bond polarity and the strength of the attraction between molecules.
2 / a / Pyradine can form H-bonds with water around the :N, but benzene can not because it has no polar region. Therefore pyradine will dissolve in water whereas benzene will not.
b / Ethanol and dimethyl ether form London dispersion and dipole forces, but the dipole forces (H-bonds) that form around the OH in ethanol are much stronger. The stronger forces require more energy to break \ the boiling temperature is greater.
c / In the solid state SO2 forms weak molecular bonds between molecules whereas, SiO2, a network solid, forms covalent bonds throughout. The stronger covalent bonds, which are broken during melting, require a higher temperature to break.
d / Cl2 is held together by London dispersion forces, which although weak increase in strength as the number of electrons increases. HCl is held together by dipole forces and London dispersion forces, but the dipole forces don't make up for the fewer electrons.
3 / a / In both cases the electron removed is from the same energy level (2p), but F has a greater effective nuclear charge due to one more proton in its nucleus \ it takes more energy to remove.
b / Less because the electron removed from F is from a 2p orbital, whereas the electron removed from Xe is from a 5p orbital, which is farther from the nucleus \ it takes less energy to remove.
c / XeO3: single bonds to each O plus a pair of unshared electrons
XeF4: single bonds to each F plus two pairs of unshared electrons
d1 / Trigonal pyramid
d2 / sp3d2
4 / a / 843 mL x 1 L/1000 mL x 1.18 g/1 L = 0.995 g
b / 157.70 g – 0.995 g = 156.71 g
c / 158.08 g – 165.71 g = 1.37 g
d / PV = nRT
(750/760 atm)(843/1000 L) = n(0.0821 atm•L/mol•K)(296 K)
\ n = 0.0342 mol: MM = m/n = 1.37 g/0.0342 mol = 40.1 g/mol
e / percent error = |40.1 g – 44.0 g|/44.0 g x 100 = 8.9 %
f1 / NO, Dry air (mostly N2 and O2) is less dense than CO2 \ the mass would be less and the calculated MM would be less than 40 g.
f2 / Yes, Lower temperature would mean more moles of gas than calculated \ the mass would be greater and the calculated MM would be greater than 40.0 g.
g / Find the mass of the empty flask. Fill the flask with water. Mass the flask with water. Subtract to find the mass of water and use the density at 23oC to determine volume.
5 / a1 / 25.0 g CH4 x 1 mol CH4 x 1 mol CH2Cl2 = 1.56 mol CH2Cl2
16.0 g CH4 1 mol CH4
2.58 mol Cl2 x 1 mol CH2Cl2/2 mol Cl2 = 1.29 mol CH2Cl2
a2 / 1.29 mol CH2Cl2 x 85.0 g CH2Cl2/1 mol CH2Cl2 = 110. g CH2Cl2
b / 242 kJ/1 mol x 1000 J/1 kJ x 1 mol/6.02 x 1023 = 4.02 x 10-19 J/bond
c / E = hc/l
4.02 x 10-19 J = (6.63 x 10-34 J•s)(3.0 x 108 m/s)/l \ l = 4.9 x 10-7 m
6 / a / No additional mass was lost during the third heating, \ all the water of hydration has been driven off.
b1 / 25.825 – 22.977 = 1.848 g H2O x 1 mol/18.02 g = 0.1026 mol H2O
b2 / 23.977 – 22.347 = 1.630 g MgCl2 x 1 mol/95.20 g = .01712 mol MgCl2
0.1026 mol H2O/0.01712 mol MgCl2 = 6 \ MgCl2•6 H2O
c / The calculated mass (moles) of water lost by the hydrate will be too large because the mass of the solid that was lost will be assumed to be water when it actually included some MgCl2 as well.
7 / a / N2: Triple bond between and one pair of unshared electrons per N.
NH3: Single bonds to each H and one pair of unshared electrons.
b / 12.0 g H2 x 1 mol H2/2.0 g H2 x -34 kJ/3 mol H2 = -68 kJ