Chemistry Free Response Review
PSI Chemistry
Name___________________________
1. Answer the following questions related to sulfur and one of its compounds.
(a) Consider the two chemical species S and S2-.
i. Write the electron configuration (e.g., 1s2 2s2 . . .) of each species.
ii. Explain why the radius of the S2− ion is larger than the radius of the S atom.
iii. 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 close to 90°. 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) forces
and dipole-dipole forces.
i. Compare the strength of the London (dispersion) forces in liquid H2S to the
strength of the London (dispersion) forces in liquid H2O. Explain.
ii. Compare the strength of the dipole-dipole forces in liquid H2S to the strength of
the dipole-dipole forces in liquid H2O. Explain.
2. 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.
(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.
(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.
3. Use your knowledge of the periodic table of the elements to answer the following questions.
(a) Explain the trend in electronegativity from P to S to Cl.
(b) Explain the trend in electronegativity from Cl to Br to I.
(c) Explain the trend in atomic radius from Li to Na to K.
(d) Explain the trend in atomic radius from Al to Mg to Na.
4. 2 KClO3(s) → 2 KCl(s) + 3 O2(g)
The above reaction took place and 1.45 liters of oxygen gas were collected over water at a
temperature of 29o C and a pressure of 755 millimeters of mercury. The vapor pressure of
water at 29o C is 30.0 millimeters of mercury.
(a) What was the partial pressure of oxygen gas collected?
(b) How many moles of oxygen gas were collected?
(c) What would be the volume of the oxygen gas at a pressure of 760 millimeters of mercury
and a temperature of 273 K?
(d) What was the mass of the KClO3 consumed in the reaction?
5. Equal molar quantities of two gases, O2 and H2O are confined to a closed vessel at constant
temperature.
(a) Which gas, if any, has the greater partial pressure?
(b) Which gas, if any, has the greater density?
(c) Which gas, if any, has the greater concentration?
(d) Which gas, if any, has the greater average kinetic energy?
(e) Which gas, if any, will show the greater deviation from ideal behavior?
(f) Which gas, if any, has the greater average molecular speed?
6. The phase diagram for a substance is shown above. Use the diagram and your knowledge of
phase changes to answer the following questions.
(a) Describe the phase change that the substance undergoes as the temperature is
increased from points B to C to D at constant pressure.
(b) Which of the phases are in equilibrium at point F? Give a name for one of the phase
changes between these two phases.
(c) Could this phase diagram represent water? Explain why or why not.
(d) What is the name given to point A, and what is the situation particular to this point?
7. N2(g) + 3 H2(g) → 2 NH3(g)
The heat of formation, ∆Hfo, of NH3(g) is -46.2 kJ/mol. The free energy of formation, ∆Gfo, of
NH3(g) is -16.7 kJ/mol.
(a) What are the values of ∆Hf and ∆Gf for the reaction?
(b) What is the value of the entropy change ∆So for the above reaction at 298 K?
(c) As the temperature is increased, what is the effect on ∆G for the reaction? How does
this affect the spontaneity of the reaction?
(d) At what temperature can N2, H2, and NH3 gases be maintained together in equilibrium,
each with a partial pressure of 1 atm?
8. A(g) + B(g) = C(g)
The reaction above is second order with respect to A and zero order with respect to B.
reactants A and B are present in a closed container. Predict how each of the following
changes to the reaction system will affect (1) the rate and (2) the rate constant and explain
why.
(a) More gas A is added to the container.
(b) More gas B is added to the container.
(c) Temperature is increased.
(d) An inert gas D is added to the container.
(e) The volume of the container is decreased.
9. Use your knowledge of kinetics to explain each of the following statements.
(a) An increase in the temperature at which a reaction takes place increases the reaction
rate.
(b) The addition of a catalyst increases the rate at which a reaction will take place.
(c) A catalyst that has been ground into powder will be more effective than a solid block of
the same catalyst.
(d) Increasing the concentration of reactants increases the rate of the reaction.
10) 85g of Iron (III) oxide reacts with 90g of carbon monoxide.
Fe2O3 (s) + 3CO(g)  2Fe(s) +3CO2(g)
a) Which one is the limiting reagent.
b) How many grams of iron will be produced?
c) How many moles of iron is produced?
d) If 48g of Iron is produced, what is the percent yield of this reaction?
Answers
Question 1
1. i) S: 1s2 2s2 2p6 3s2 3p4
S2-: 1s2 2s2 2p6 3s2 3p6
ii) Two electrons have been added to the S atom to form the S2- anion. The added electrons in
anions increase the repulsions between electrons, making the anion larger than its parent atom.
2. It is easier to remove an electron from the S2- ion than the Ar atom. The Ar atom has a greater
effective nuclear charge and smaller radius, allowing it to hold the electrons in with greater
force.
3. The 3p electrons from the S atom bond with the H atom.
4. i) H2S and H2O have the same molecular geometry. Because the H2S molecule is larger, it will
experience greater London (dispersion) forces.
ii) H2O has hydrogen bonding. This is a specialized, strong case of dipole-dipole forces. H2S
forms a dipole as well, but it does not have the hydrogen bonding, giving it weaker dipole-dipole
forces.
Question 2
(a) Pyridine is polar (and capable of forming hydrogen bonds with water), while the nonpolar benzene is not
capable of forming hydrogen bonds. Pyridine will dissolve in water because of the strong hydrogen bonds
(or dipole-dipole intermolecular interactions) that exist between the lone pair of electrons on pyridine’s
nitrogen atom and the solvent water molecules. No such strong intermolecular interaction can exist
between benzene and water, so benzene is insoluble in water.
One point is earned for identifying a relevant structural difference between pyridine and benzene.
One point is earned for indicating that pyridine is soluble in water because pyridine can form
strong dipole-dipole interactions (or hydrogen bonds) with water, while benzene cannot.
(b) The intermolecular forces of attraction among molecules of dimethyl ether consist of London
(dispersion) forces and weak dipole-dipole interactions. In addition to London forces and dipole-dipole
interactions that are comparable in strength to those in dimethyl ether, ethanol can form hydrogen bonds
between the H of one molecule and the O of a nearby ethanol molecule. Hydrogen bonds are particularly
strong intermolecular forces, so they require more energy to overcome during the boiling process. As a
result, a higher temperature is needed to boil ethanol than is needed to boil dimethyl ether.
One point is earned for recognizing that ethanol molecules can form intermolecular hydrogen
bonds, whereas dimethyl ether molecules do not form intermolecular hydrogen bonds.
One point is earned for recognizing that, compared to the energy required to overcome the
weaker intermolecular forces in liquid dimethyl ether, more energy is required to overcome the
stronger hydrogen bonds in liquid ethanol, leading to a higher boiling point.
(c) In the solid phase, SO2 consists of discrete molecules with dipole-dipole and London (dispersion) forces
among the molecules. These forces are relatively weak and are easily overcome at a relatively low
temperature, consistent with the low melting point of SO 2 . In solid SiO2 , a network of Si and O atoms,
linked by strong covalent bonds, exists. These covalent bonds are much stronger than typical
intermolecular interactions, so very high temperatures are needed to overcome the covalent bonds in
SiO2. This is consistent with the very high melting point for SiO 2 .
One point is earned for recognizing that SO2 is a molecular solid with only weak dipole-dipole and
London forces among SO2 molecules.
One point is earned for recognizing that SiO2 is a covalent network solid, and that strong covalent
bonds must be broken for SiO2 to melt.
(d) The intermolecular forces in liquid Cl2 are London (dispersion) forces, whereas the intermolecular forces
in liquid HCl consist of London forces and dipole-dipole interactions. Since the boiling point of Cl2 is
higher than the boiling point of HCl, the London forces among Cl2 molecules must be greater than the
London and dipole-dipole forces among HCl molecules. The greater strength of the London forces
between Cl2 molecules occurs because Cl2 has more electrons than HCl, and the strength of the London
interaction is proportional to the total number of electrons.
One point is earned for recognizing that the London forces among Cl2 molecules must be larger
than the intermolecular forces (London and dipole-dipole) among HCl molecules.
One point is earned for recognizing that the strength of the London forces among molecules is
proportional to the total number of electrons in each molecule.
Question 3
(a) Electronegativity is the pull of the nucleus of one atom on the electrons of other atoms; it increases from
P to S to Cl because nuclear charge increases. This is because as you move from left to right across the
periodic table, atomic radii decrease in size. Increasing nuclear charge means that Cl has the most
positively charged nucleus of the three and will exert the greatest pull on the electrons of other atoms.
(b) Electronegativity is the pull of the nucleus of one atom on the electrons of other atoms; it decreases from
Cl to Br to I because electron shells are added. The added electron shells shield the nucleus, causing it to
have less of an effect on the electrons of other atoms. Therefore, iodine will exert the least pull on the
other atoms.
(c) Atomic radius increases from Li to Na to K because electrons are being added in higher energy levels,
which are father away from the nucleus; therefore the K atom is the largest of the three.
(d) Atomic radius increases from Al to Mg to Na because protons are being removed from the nucleus while
the energy levels of the valence electrons remain unchanged. If there are fewer positive charges in the
nucleus, the electrons of Na will be less attracted to the nucleus and will remain farther away.
Question 4
(a) Use Dalton’s Law:
Ptotal = Poxygen + Pwater
755 mmHg = Poxygen + 30 mmHg
Poxygen = 725 mmHg
(b) Use the ideal gas law:
PV = nRT
(c) At STP, moles of a gas and volume are directly related:
(d) We know that 0.056 moles of O2 were produced in the reaction. From the balanced equation, we know
that for every 3 moles of O2 produced, 2 moles of KClO3 are consumed. So there are 2/3 as many moles
of KClO3 as O2.
Grams = moles (MW)
Grams = 0.037 mol(122 g/mol) = 4.51g
Question 5
(a) The partial pressures depend on the number of moles of gas present. Because the number of the moles
of the two gases are the same, the partial pressures are the same.
(b) O2 has the greater density. Density is mass per unit volume. Both gases have the same number of moles
in the same volume, but oxygen has heavier molecules, so it has greater density.
(c) Concentration is moles per unit volume. Both gases have the same number of moles in the same volume,
so their concentrations are the same.
(d) According to kinetic-molecular theory, the average kinetic energy of a gas depends only on the
temperature. Both gases are at the same temperature, so they have the same average kinetic energy.
(e) Either answer would be accepted:
 H2O will deviate most from ideal behavior. Ideal behavior assumes that there will be no intermolecular
interactions. H2O undergoes hydrogen bonding and O2 does not. So H2O has strong intermolecular
interactions causing it to deviate more.
 O2 will deviate most from ideal behavior. In general, the smaller the molecule, the less it deviates from
ideal behavior. O2 is a larger molecule, so it would be expected to deviate more.
(f) H2O has the greater average molecular speed. If two gases are at the same temperature, the one with the
smaller molecular weight will have the greater average molecular speed (KE = ½ m v 2).
Question 6
(a) At point B, the substance is a solid.
At point C, the substance is in equilibrium between solid and liquid phase; it’s melting.
At point D, the substance is entirely in liquid phase.
(b) At point F, The substance is in equilibrium between gas and solid phases. When a substance changes
from solid to gas phase, it’s called sublimation. When a substance changes from gas to solid phase it’s
called deposition.
(c) The diagram cannot represent water.
For water the slope of the solid-liquid equilibrium line must be negative, indicating that the liquid phase of
water is denser then the solid phase.
(d) Point A is called the triple point.
At the triple point, all three phases can exist simultaneously in equilibrium.
Question 7
(a) By definition, ∆Hfo and ∆Gfo for N2(g) and H2(g) are equal to zero.
(b) Use ∆G = ∆H + T ∆S
∆S = -198 J/K
(c) Use ∆G = ∆H + T ∆S
From (b), ∆S is negative, so increasing the temperature increases the value of ∆G, making the reaction
less spontaneous.
(d) Use ∆G = ∆H + T ∆S
At Equilibrium ∆G = 0.
Question 8
(a)
1. The rate of the reaction will increase because the rate depends on the
concentration of as given in the rate law: Rate = k[A]2.
2. The rate constant, k, is independent of the concentration of the reactants and will not change.
(b)
(c)
1. The rate of the reaction will not change. If the reaction is zero order
with respect to B, then the rate is independent of the concentration of B.
2. The rate constant, k, is independent of the concentration of the reactants and will not change.
1. The rate of the reaction will increase with increasing temperature
because the rate constant increases with increasing temperature.
2. The rate constant increases with increasing temperature because at higher temperature more gas
molecules collide with enough energy to overcome the activation energy got the reaction.
(d)
Neither the rate nor rate constant will be affected by the addition of an
inert gas.
(e)
1. The rate of reaction will increase because decreasing the volume of
the container will increase the concentration of A and Rate = k[A]2.
2. The rate constant, k, is independent of the concentration of the reactants and will not change.
Question 9
(a)
An increase in the temperature means an increase in the energy of the
molecules present. If the molecules have more energy, then more of them will collide more often with
enough energy to overcome the activation energy required for the reaction to take place, causing the
reaction to proceed more quickly.
(b)
A catalyst offers a reaction an alternative pathway with a lower activation
energy. If the activation energy is lowered, then more molecular collisions will occur with enough energy
to overcome the activation energy required for the reaction to take place, causing the reaction to proceed
more quickly.
(c)
The effectiveness of a solid catalyst depends on the surface area of the
catalyst that is exposed to the reactants. Grinding a solid into powder greatly increases the surface area.
(d)
Increasing the concentration of the reactants crowds the reactants more
closely together, making it more likely that they will collide with one another. The more collisions that
occur, the more likely that collisions that will result in reactions will occur.
Question 10
a) 85g/149.7g = 0.57mol
90g/55.85= 1.61mol
Fe2O3 LR
b) 1mol Fe2O3 /2mol Fe = 0.57/ xmol
x = 1.14mol of Fe
c) 1.14mol x 55.85g = 63. 67g of Fe
d) percent yield = 48g x100/63.67gx 75.38%