Answers (and some solutions) to questions posed by the demonstrations

The fruit of physics

Answer: Mathematically, we can state the problem as

    dobject / dmolecule = dEarth / dobject
and solve for the size of the object

 
    dobject = ( dEarth x dmolecule )1/2 = 6 cm
Richard Feynman, a physicist of remarkable intuition and accomplishments, chose an apple.

The atomic necklace

Answer: In three dimensions, the volume of the mole of close packed Argon atoms is only 18 cm3. The mole would fit comfortably in a juice glass.

In two dimensions, the area covered by the Argon sheet is 58,000 m2 which is the area of over 10 football fields!

In one dimension, the necklace would be 1.9 x 1014 m. The distance from the Earth to the Sun, 1.5 x 1011 m, is much shorter than the atomic necklace. The atomic necklace can make more than 500 round trips to the Sun!

Low energy physics

Answer: The volume of the oil drop must equal the volume of the oil disk, Vdisk =Vdrop. Solving for h provides an estimate of the height, or the diameter, of the molecule. 13 Angstroms.

Mole hills

Answer: One mole. Remarkable! Or is it just making a mole hill out of mountain bikes?

Around the world in 46 seconds

Answer: The ratio 
 
     rnucleus / ratom = 10-5
and since most of the mass of the atom is the nucleus, and the density (= mass/volume) is inversely proportional to the volume,

    dnucleus / datom = 1015.
Most of the volume of an atom is ``empty'' -- more than 99.9% of the atom's mass is concentrated in the nucleus!

Working from the circumference of the Earth, its radius is approximately 6.4 x 106 m and its volume is approximately 1.0 x 1021 m3. That means its mass density is 6000 kg/m3. The density of the carbon atom is similar at 4700 kg/m3. The density of the carbon atom nucleus is 15 orders of magnitude larger at 4.7 x1018 kg/m3.

    MEarth /  VEarth = density of nucleus = 4.7 x 1018 kg/m3
At nuclear density, the Earth's mass would occupy a volume of 1.3 x 106 m3 so that the radius would be only 67 m! The compressed Earth would fit comfortably into Foxboro stadium. Michael Johnson in his prime could run around this compressed Earth at its equator in 46 seconds!

Missing mass

Answer: The answer to this problem is found in a remarkable concept that was first understood by Albert Einstein. His understanding and the subsequent applications of the idea have changed the face of war and life on Earth. It is that

2 p + 2 n + 2 e → 4He + energy

How much energy? Einstein proposed the equivalence

E = m c2

where m is the mass of a resting particle and c = 2.99 x 108 m/s is the speed of light.

Let's take the difference in mass between the atom of 4He and the six particles that it is composed of. That difference is m = 0.05044 x 10-27 kg and the corresponding energy is E = m c2 = 4.5 x 10-12 J = 2.7 x 1012 J/mol.

Yow! Chemical reactions that yield a significant amount of energy, such as combustion reactions, produce on the order of 500,000 Joules of energy per mole. But that is a mass equivalent of m = 500,000 J/mol / c2 = 5.5 x 10-9 g/mol which is very, very small.

This is what Einstein realized. Small changes in the arrangements of nucleons can lead to the release of vast quantities of energy that far exceed what can be produced by chemical means. Nuclear bombs can far exceed the power of conventional chemical explosives. That knowledge led to the production of the atomic bombs that were exploded over the cities of Hiroshima, Japan on August 6, 1945 and over Nagasaki, Japan on August 9, 1945.

Identification of the reductant

Answer: In the final reaction

BrCH(CO2H)2(aq) + 4 Mn3+(aq) + 2 H2O(l) → Br-(aq) + HCO2H(aq) + 2 CO2(g) + 4 Mn2+(aq) + 5 H+(aq)

the bromine atoms remain in an oxidation state of -1. All hydrogen atoms are in the +1 oxidation state, and all oxygen atoms are in the -2 oxidation state. As the manganese ions are reduced the carbon atoms are oxidized.

The molecular structure of smog

Answer: Nitric oxide, NO, is a radical with 11 valence electrons. As the oxygen is more electronegative, we place the lone electron on the nitrogen atom. The molecule is polar. It is also a radical and highly reactive.

Nitrous oxide, N2O, has 16 valence electrons with the oxygen at the center, double bonded to each nitrogen atom. The molecule is linear and non-polar.

Nitrogen dioxide, NO2, has 17 valence electrons with the nitrogen at the center. As in the case of nitric oxide, the lone electron is placed on the nitrogen atom. There are two resonance structures, for the two possible placements of the double bond. The molecule is bent and polar. It is also a radical and highly reactive.

Dinitrogen tetroxide, N2O4, has 34 valence electrons. The molecule is the dimer of two nitrogen dioxide molecules, where the nitrogen atoms have paired their unpaired electrons to form a single nitrogen-nitrogen covalent bond.

Dinitrogen pentoxide, N2O5, has 40 valence electrons with a central oxygen atom bonded to two nitrogen dioxide moieties.

Nitrous acid, HONO, has 18 valence electrons. There is a lone pair of electrons on the central nitrogen atom, and the O-N-O bond angle is bent.

Nitric acid, HONO2, has 24 valence electrons. The central nitrogen atom is bonded to three oxygen atoms. Fixing the position of the hydrogen atom, there are two resonance structures corresponding to the two possible placements of the nitrogen-oxygen double bond.

Peroxynitric acid, O2NOOH, has 30 valence electrons. The structure is that of nitric acid where the proton is replaced with a hydroxy, OH, moiety.

Clear skies

Answer: Not necessarily! If the temperature is cold enough, the equilibrium

2 NO2(g) (Reddish-Brown) → N2O4(g) (Clear)

will shift to the right, favoring the clear dinitrogen tetroxide. While we may not be able to see it, the concentration of the nitrogen oxide and the potential for acid raid may still be high.

A quantitative measure of reactivity

Answer: We find for the dehydration reaction

CuSO4·5H2O(s) → CuSO4(s) + 5H2O(g)         K = aH2O5

where the activity of the gaseous water is aH2O = PH2O / Pref. Pure solids do not enter the expression. For the complexation reaction

CuSO4(aq) + 4 NH4OH(aq) → Cu(NH3)4SO4H2O(s) + 3 H2O(l)         K = 1/[aCu2+ aSO42- (aNH4+ aOH- )4]

where pure solids and liquids do not enter the expression. The activity of a given ion in solution is the molarity of the ion divided by the reference concentration. For the nothingness reaction

Cu2+(aq) + SO42-(aq) + K+(aq) + H2PO4-(aq) → Cu2+(aq) + SO42-(aq) + K+(aq) + H2PO4-(aq)         K = 1

as there is no net reaction.

Stable paste

Answer: Begin by drawing the Lewis electron dot structure of the nitrogen triiodide molecule. You will note that the structure is isomorphic with the structure of the ammonia molecule. Both molecules are trigonal pyramids, with a lone pair of electrons at the top of the pyramid. As is the case with ammonia, the nitrogen triiodide molecule has a substantial dipole moment, making the molecule polar.

When the nitrogen triiodide molecule is solvated in water, the polar water molecules make energetically favorable arrangements with the polar nitrogen triiodide molecule. Those intermolecular interactions stabilize the nitrogen triiodide molecules in aqueous solution. However, when the water evaporates, the nitrogen triiodide is left "high and dry" - dry in the absence of water and high in energy! The molecule becomes unstable, readily reacting to create much molecular disorder through the formation of nitrogen and iodine gases.

Lewis acids and bases

Answer: In the first reaction, the cobalt ion acts as a Lewis acid, accepting the electron pair of the water, which acts as a Lewis base.

In the second reaction, the cobalt once again acts as the Lewis acid, accepting the electron pair donated by the chloride ion, which acts as a Lewis base.

What is oxidized and what is reduced?

Answer: In the first reaction, the Cu(II) is reduced to Cu(I), while oxygen is oxidized from -1, in hydrogen peroxide, to 0, in molecular oxygen. We will ignore any change in oxidation state of the carbon atoms.

In the second reaction, the Cu(I) is oxidized to Cu(II), while oxygen in the peroxide is reduced from -1, in hydrogen peroxide, to -2, in water. If molecular oxygen is also a product, then some of the hydrogen peroxide reacts through the oxidation of oxygen, from -1 in the peroxide to 0 in dioxygen.

Pushing and pulling the equilibrium

Answer: One possibility is water. Increasing the volume of solution will dilute the iodide ion. The lower concentration of iodide ion should push the equilibrium

HgI2(s) + 2 I-(aq) → HgI42-(aq)

back to the left and the recreation of the orange mercury(II) iodide precipitate.

QUALitative and QUANTitative understanding of titrations

Answer: For the initial solution, the dominant species are H+ and H2O. The chemical equilibrium that defines the pH is the presence of excess hydrogen ion contributed by the essentially complete dissociation of hydrogen chloride in solution

HCl(aq) + H2O(l) → H3O+(aq) + Cl-(aq)

where the equilibrium constant is VERY large, and it is acceptable to assume full dissociation of the acid.

In region IV, the titration has reached the equivalence point and there is no strong acid remaining in solution. There is also no excess strong base. The dominant species in solution will be H2O. The equilibrium that will determine the pH at the equivalence point is the autoionization of water

H2O(l) → H+(aq) + OH-(aq)         Kw = 1.00 x 10-14

In region V, the titration is beyond the equivalence point. No strong acid remains. The dominant species in the solution will be the excess OH- (and water). The equilibrium that determines the pH will be the presence of excess hydroxide ion in solution contributed by the essentially complete dissociation of sodium hydroxide in solution

NaOH(aq) → Na+(aq) + OH-(aq)

The pH of the solution at the following volumes of strong base added in the titration will be 0.00 (0.6990), 10.00 (0.8752), 25.00 (1.1761), 40.00 (1.6532), 49.00 (2.695), 49.90 (3.70), 49.99 (4.7), 50.00 (7.00), 50.01 (9.3), and 55.00mL (11.979).

The significant figures pose a challenge in this problem. Recall the rule for converting the activity of hydrogen ion, aH+, to pH=-log10(aH+). The activity is expressed in scientific notation as

0.mantissa x 10-characteristic

In the pH, the numbers before the decimal point represent the characteristic. The numbers after the decimal point represent the logarithm of the mantissa.

pH = characteristic - log10(0.mantissa)

The number of significant figures in the log10(0.mantissa) is equal to the number of significant figures in the mantissa. Since the characteristic is an exact counting number, the number of significant figures that follow the decimal point in the pH is the number of significant figures in the mantissa.

In this problem, the pH at 0.00, 10.00, 25.00, and 40.00mL will have the same number of significant figures. In each case, the activity has 4 significant figures, and the pH reports four figures following the decimal point.

At 49.00mL there are only three significant figures in the activity, so there are only three figures reported beyond the decimal point in the pH. At 49.90mL there are only two significant figures in the activity, and at 49.99 and 50.01mL there is only one significant figure in the activity.

At the equivalence point, when 50.00mL has been added, the number of significant figures is determined by the Kw = 1.00 x 10-14 so that pH=7.00.

Exploring the chemistry of solutions of acids and bases

Answer: In region I, the solution is a pure weak acid. The dominant species in the solution are the acetic acid and water. The equilibrium that determines the pH at that point is the ionization of the weak acid

HAc(aq) + H2O(l) → H3O+(aq) + Ac-(aq)         Ka = 1.8 x 10-5

Acetic acid is a weak acid, and we can assume that little of the initial acetic acid dissociates in solution. To a good approximation the concentration of acetic acid at equilibrium will be its initial concentration. The activity of the hydrogen ion will be approximately aH+ = (Ka [weak acid]0)½. The pH will be well approximated by

pH = ½ pKa - ½ log10 ([weak acid]0)

In region II, the solution is near the half-equivalence point where the dominant species in solution are the acetic acid and its conjugate weak base, the acetate ion. The equilibrium that determines the pH is the same as for the pure weak acid solution

HAc(aq) + H2O(l) → H3O+(aq) + Ac-(aq)         Ka = 1.8 x 10-5

As the acetate ion is a weak base and acetic acid is a weak acid, we assume that the concentrations of those species at equilibrium will simply be their initial concentrations. The activity of the hydrogen ion will be well approximated by aH+ = (Ka [weak acid]0/[conjugate weak base]0), and the pH will be given by the Henderson-Hasselbalch equation

pH = pKa + log10 ([conjugate weak base]0/[weak acid]0)

As long as the concentration of weak acid and its conjugate weak base are nearly equal, the pH of the buffer solution will never stray too far from the pKa = -log10(Ka).

In region III, the original equivalents of weak acid have been neutralized by equivalents of strong base added. At this equivalence point, all of the original weak acid has been converted to conjugate weak base. The dominant species in the solution are the acetate ion and water. The equilibrium that determines the pH will be the hydrolysis of the weak base

Ac-(aq) + H2O(l) → OH-(aq) + HAc(aq)         Kb = Kw/Ka = 5.6 x 10-10

In region V, the titration is beyond the equivalence point. No weak acid remains. The dominant species in the solution will be the excess OH- (and water and acetate ion). The equilibrium that determines the pH will be the presence of excess hydroxide ion in solution contributed by the essentially complete dissociation of sodium hydroxide in solution

NaOH(aq) → Na+(aq) + OH-(aq)

The pH of the solution at the following volumes of strong base added in the titration will be 0.00 (2.72), 10.00 (4.14), 25.00 (4.74), 40.00 (5.35), 50.00 (8.87), 50.01 (9.3) and 55.00mL (11.979).

What about those &#%$ significant figures? In this problem, the pH at 0.00, 10.00, 25.00, 40.00, and 50.00mL will all have the same number of significant figures, limited by the two significant figures in the Ka. In each case, the activity of hydrogen ion has 2 significant figures, and the pH reports two figures following the decimal point.

At 50.01mL there will be one significant figure in the excess base concentration and there is one figure reported beyond the decimal point in the pH.

At 55.00mL there are three significant figures in the excess base concentration and there are three figures reported beyond the decimal point in the pH.

Medicinal use of silver nitrate

Answer: Silver nitrate acts as an antibacterial agent. For many years in American hospitals, it was applied to the eyes of babies within a few hours of birth to prevent eye infections. A common source of eye infections was gonococcal infection resulting from gonorrhea. At the turn of the 20th century, gonococcal ophthalmia was the cause of one out of four cases of blindness in the United States.

Dissolution of silver chloride

Answer: The Lewis structures of ammonia, NH3, thiosulfate ion, S2O32- (which has a sulfur at the center of the three oxygen and terminal sulfur atoms), and cyanide ion, CN- , will leave all atoms satisfying the octet rule. These species are quite stable. In each case, the molecule or molecular ion presents lone pairs of electrons and acts as a Lewis base.

The ammonia donates an electron pair on its nitrogen

AgCl(s) + 2 NH3(aq) → [Ag(NH3)2]+(aq) + Cl-(aq)

creating the stable diamminesilver(I) complex ion. Increasing the concentration of ammonia pushes the equilibrium to the right, increasing the concentration of the complex ion.

In the case of the cyanide ion, either the carbon or nitrogen can donate a lone pair of electrons, acting as the Lewis base

AgCl(s) + CN-(aq) → AgCN(aq) + Cl-(aq)

Solutions of silver cyanide are used in electroplating to create a soluble form of silver ion.

The terminal sulfur atom of the thiosulfate ion acts as a Lewis base, donating an electron pair to the silver ion

AgCl(s) + S2O32-(aq) → AgS2O3-(aq) + Cl-(aq)

AgS2O3-(aq) + S2O32-(aq) → Ag(S2O3)23-(aq)

Photographers use solutions of sodium thiosulfate, which they refer to as hypo, to solubilize and remove light sensitive silver ion during the photographic process.

Bucky Badger, Chemistry, and Boston University

Answer: Bassam Shakhashiri, a native of Lebanon, is now a distinguished professor of a university whose mascot takes its name from the lead miners of Wisconsin. He first learned of the remarkable chemistry of lead while an undergraduate chemistry major at Boston University.

"I was astonished to discover that lead is the end product of three series of naturally occurring radioactive elements and that lead has more than two dozen radioactive isotopes. In my sophomore year at Boston University, I learned that lead was in a gasoline additive--an organometallic compound called tetraethyl lead. I learned that lead is a major component of automobile batteries, that it was widely used in paint, and that it is used in soldering. Later, I learned that even in ancient times, some physicians believed that lead was poisonous, but it continued to be used in medicines and cosmetics until the 20th century." [Chemical and Engineering News]

The most hydrophobic object around

Answer: Introducing a hydrophobic object into solution disrupts the energetically favorable network of hydrogen bonds between water molecules. If the bubble is introduced, there is essentially no compensation through favorable attractive interaction between the bubble and the water molecules. However, if a molecule of the same size is introduced, even if it is nonpolar, the water will still be attracted through dispersion interactions. There is some compensation. This makes the bubble the most hydrophobic object.

Hot and cold water

Answer: Think of the microscopic motion of molecules in the liquid. Compared with the hot water, on average the water molecules in the cold water are moving relatively slowly. When the potassium permanganate crystals are introduced to the water, the water molecules begin colliding with the surface of the crystals. The polar water molecules will be successful in solvating individual charged ions of potassium and permanganate, surrounding them and dissolving them in the solution. The same molecular dynamics occurs in the hot water. It simply happens much faster.

The crystals of potassium permanganate are more dense than the liquid water. When the crystals are dropped into the cylinder, they fall by the force of gravity toward the bottom of the cylinder. They are falling at a particular rate, and dissolving at a particular rate.

In the cold water, the rate of falling is relatively fast compared to the rate of dissolving, and the crystals leave little trace at the top of the cylinder. In the hot water, the dissolution is quite rapid and much of the crystals are dissolved at the top of the cylinder.

Interpreting the molecular simulation at the atomic scale

Answer: The plastic balls represent argon atoms. The actual argon atom has a diameter of 3.1 x 10-10 m, while the plastic argon "atom" has a diameter of 0.50 cm. That ratio of lengths gives us the scale of our simulation. Just as in the scale of a road map, that might be 1 inch: 10 miles, we have a scale of our simulation that is 0.50 cm: 3.1 x 10-10 m. The road map must be scaled up, while our simulator must be scaled down.

That scale allows us to say that the side of the box of atoms, which is 10. cm, is the equivalent of 10. cm x (3.1 x 10-10 m/0.50 cm) = 6.2 x 10-9m. The volume of the box is then 2.4 x 10-25 m3.

As there are 10 argon atoms in the box, the number density is 4.2 x 1022 atoms/L. Dividing by Avogadro's number and multiplying by the molar mass of argon, 39.95 g/mol, we find a mass density of our simulated argon gas to be 2.8 g/L.

At STP the mass density of argon gas is 39.95 g/mol/22.4 L/mol = 1.78 g/L. So our simulated argon gas is about 50% more dense than actual argon gas at STP. Not bad!

Using our length scale, we can convert the atomic speeds of the argon "atoms." The kinetic theory of gases tells us that the average speed of an argon atom at STP is 380 m/s. We find that the speed of the "cold" atoms are on average 3.7 x 10-9 m/s. Awfully slow! Let's check the "hot" atoms ... 6.2 x 10-8 m/s. Still awfully slow! What gives?

To have an accurate interpretation of the gas dynamics, we need to add to our length scale a time scale. Let's say that the average speed of the "cold" gas "atoms" is the equivalent of 214 m/s, which is the average speed of an argon atom a few degrees above the triple point (84 K). We find a time scale of 1 second:17 picoseconds (which is 17 x 10-12 seconds). So the passing of one second in our simulation is the equivalent of 17 picoseconds in the actual argon gas.

The hot gas moves on average (100cm/s)/(6cm/s)=16.7 times faster. And since uave ∝ T1/2, the temperature of the simulated hot gas is approximately 24,000 K!

Prove that result

Answer: The rate of diffusion is inversely proportional to the square root of the mass of the atom or molecule. The heavier the atom or molecule, the slower its average speed and the slower its rate of diffusion.

The distance traveled by a molecule is proportional to the rate of diffusion. That means that

distance traveled by NH3/ distance traveled by HCl = (MHCl/MNH3)1/2

Since the mass ratio is 36.5/17.0, the ratio of the distances traveled is (36.5/17.0)1/2=1.46. That means that the ammonia travels 1.46 cm for ever 1.00 cm traveled by the heavier hydrochoric acid.

If the tube is 122.5 cm long, the reaction will occur rougly 50 cm from the hydrochloric acid, and 72.5 cm from the ammonia end of the tube.

Variations on the theme

Answer: The mass of the hydroiodic acid is 128. g/mol. Therefore

distance traveled by NH3/ distance traveled by HI = (MHI/MNH3)1/2 = (128./17.0)1/2 = 2.74

If the tube is 122.5 cm long, the reaction will occur roughly 33 cm from the hydroiodic acid end of the tube, and 89.5 cm from the ammonia end of the tube.

If the temperature of the tube is lowered from 25 C to -30 C, there will be no change in the point at which the reaction is first observed. The average rate of diffusion for each molecule will be decreased by the same proportion. So while it will take longer to observe the reaction, the reaction will first be observed to occur at the same point in the tube.

Identifying the molar mass of a gas by effusion

Answer: The rate of gaseous effusion of an atom or molecule is proportional to the average speed. The average speed is proportional to the inverse square root of the mass of the atom or molecule. That means that

rate of effusion UNKNOWN / rate of effusion O2 = (MO2/MUNKNOWN)1/2

The rate of effusion is inversely propotional to the time required for the escape of the gas. The higher the rate of effusion, the shorter the time required to escape. That means that

escape time for UNKNOWN/ escape time for O2 = (MUNKNOWN/MO2)1/2

The ratio of the escape times is 95.8 seconds/47.3 seconds. We known the molar mass of dioxygen is 31.998 g/mol. So the molar mass of the unknown gas is (95.8/47.3)2 x 31.998 g/mol = 131. g/mol. Inspection of the Periodic Table tells us that the unknwon gas is xenon!

Osmotic pressure and ion selective membranes

Answer: No detectable concentration of chloride ions will be created in the surrounding solution. The random diffusion of ions will not create a measurable separation of charge on the macroscopic scale. Why not? The cost in energy to separate the charges will simply be too great.

The migration of chloride ions out of the the "sausage" must be accompanied by either (1) the migration of a similar number of positively charged particles out of the membrane or (2) the migration of a similar number of negatively charged particles into the membrane, so that each domain remains electrically neutral.

Osmotic pressure and the molar mass of a protein

Answer: 6.8 x 104 g/mol

How much work is done in blowing up a balloon?

Answer: The pressure is taken to be the constant atmospheric pressure of 750 torr. The work done ON the system is

w = - PextΔV

The work done BY the system to inflate the balloon is PextΔV = 1.5 L atm = 0.15 kJ. The system does a positive amount of work to inflate the balloon.

Using the relation that H = E + PV, we know that ΔH - ΔE = Δ(PV). As the pressure is constant, Δ(PV) = P ΔV, which is exactly the work done BY the system to inflate the balloon.

Deeper consideration of the heats of solution

Answer: The heat of solution of both sodium chloride and potassium nitrate is endothermic. This results from the fact that it takes more energy to separate the ions in the solid than is recovered in the solvation of the charged ions by the dipolar water molecules.

Why is it that the dissolution of sodium chloride was observed to be endothermic while the dissolution of lithium chloride is exothermic? Both sodium and lithium are alkali metals and exist as monovalent cations. In each case, they are paired with chloride ions. What gives?

It is the size of the cations that matters! The lithium ion, with a diameter of 1.80 Å, is substantially smaller than the sodium ion, which has a diameter of 2.32 Å. The charge-charge attraction is inversely proportional to the distance between the charges. The smaller lithium ion has a stronger attraction to the dipolar water molecules. More energy is released on solvating the smaller lithium ion than the larger sodium ion. As in the case of the dissolution of calcium chloride, the dissolution of lithium chloride salt in water is highly exothermic.

Measuring heats of reaction at constant pressure

Answer: We are told that the heat of reaction is -26.29 kJ for the reaction of one mole of barium nitrate with one mole of iron sulfate in solution. So 26.29 kJ of energy will be added to the surroundings

qsys = -qsurr

For our experiment we take the surroundings to be the calorimeter so that

qsurr = qcalorimeter = C ΔT

where C is the calorimeter constant - the heat required to raise the temperature of the calorimeter by one degree celsius.

We assume that the heat that is added to the surroundings is entirely used to heat the water in the calorimeter. That is, we assume that the solution is perfectly insulated, and that no heat is absorbed by the container holding the solution.

There is 2.0 liters of water with a mass of 2.0 kg. The specific heat of the solution is cS=4.18 J/g C. The calorimeter constant is assumed to be C=m cS where m is the mass of the solution.

The heat added to the calorimeter is

qcalorimeter = 26.29 kJ = C ΔT = m cS Δ T.

The change in the temperature of the solution is computed to be ΔT=3.1C.

Rules of thumb regarding the thermodynamics of the dissolution of salts

Answer: For the dissolution of anhydrous salts, the energy cost to overcome the charge-charge attractions and separate the ions in the lattice of the ionic solid is more than compensated for by the favorable charge-dipole interactions in the solvation of the ions by water. The fact that the favorable interaction energy between the salt ions and water more than compensates for the cost of separating the ions in the solid salt makes the dissolution of most salts exothermic.

To make an anhydrous salt from a hydrate, there is an energy cost to overcome the favorable charge-dipole interactions between the salt ions and the dipolar water molecules. That makes the process endothermic.

Reversible and irreversible isothermal expansions of an ideal gas

Answer: For the irreversible, three-step isothermal expansion: q=16.0 Latm, w=-16.0 Latm, ΔE=0, and ΔH=0. For the reversible isothermal expansion: q=23.0 Latm, w=-23.0 Latm, ΔE=0, and ΔH=0.

The work done on the gas in each case is negative; as it is expanding, the gas does a positive amount of work. The heat added to the system in each case is positive; as the gas expands it does work, and to maintain the constant temperature (and energy) heat must be added to the gas. In agreement with our expectations, wirrev > wrev. In the reversible expansion, the gas does more work.

Variations on the thermite reaction

Answer: We find for the reaction of cobalt(II) oxide with aluminum

2 Al(s) + 3 CoO(s) → 3 Co(s) + Al2O3(s)

Another metal oxide that can be reacted with aluminum in a variation on the thermite reaction is cadmium(II) oxide

2 Al(s) + 3 CdO(s) → 3 Cd(s) + Al2O3(s)

Stability of metallic aluminum in an aerobic atmosphere

Answer: The surface of aluminum metal is readily oxidized by atmospheric oxygen. Why it is that the reaction does not proceed in a highly exothermic manner, leading to the complete oxidation of the aluminum to corundum? A coat of aluminum oxide forms over the surface of the aluminum metal. The size of the aluminum oxide is such that it can organize itself on the surface, sealing the surface to further reaction of the underlying aluminum metal with the atmospheric oxygen.

In the case of rusting iron, the size of the iron oxide is such that the surface oxide does not adhere to the underlying iron metal and tends to flake. The flaking exposes the underlying iron to oxygen, causing it to rust. Eventually, the iron will rust through in a near complete conversion of the metallic iron to iron oxide.

Constant heats of reaction

Answer: The molar enthalpy of reaction for the dissolution of magnesium sulfate in water at standard state and 25C is

ΔHo = ( ΔHfo[Mg2+(aq)] + ΔHfo[SO42-(aq)] ) - ΔHfo[MgSO4(s)] =(-446.85 kJ - 909.27 kJ) - (-1284.9 kJ) = -71.22 kJ

As in the case of the dissolution of calcium chloride, the reaction is exothermic.

To liberate the same heat as a thermite reaction using 1.0 g of aluminum and 3.0 g of ferric oxide

qthermite = (mAl/2 MAl) ΔHothermite = qdissolution = (mMgSO4/MMgSO4) ΔHodissolution

Using MMgSO4=120.366 g/mol, we would need to use 26.7 g of magnesium sulfate.

What about table salt? The heat of solution for sodium chloride is +3.87 kJ. It's endothermic! So no amount of sodium chloride can be dissolved to add a positive quantity of heat to the surroundings.

Dehydration of sugar in water

Answer: Rather than waiting a minute to see the slow growth of a column of porous carbon, there will be an immediate reaction where great volumes of bitter sweet and smouldering black foam will pour forth and over the beaker and surroundings.

Why does this reaction occur more rapidly than the previous version where the solid sugar is mixed with the acid solution?

Thinking at the level of atoms and molecules, in the previous protocol the reaction can only occur at the surface of the solid sugar crystals. In this version, using the dissolved sugar, the reaction can occur everywhere and at once in the liquid solution.

The heat of reaction for neutralization of a strong acid with a strong base

Answer: The complete balanced equation includes spectator ions as well as reactive species. To a first approximation, the spectator ions play no role in the chemistry of the reaction. They exist in the same state in the solution of reactants and in the solution of products. Therefore, they do not contribute to the overall change in enthalpy of the solution or the heat of reaction.

Intermolecular interactions and the vaporization of liquids

Answer: Thinking at the molecular level, for a molecule of water from the liquid to escape the liquid and enter the gas phase, the force of intermolecular attraction to other molecules in the liquid must be overcome. A water molecule at the surface of liquid water will be attracted by dipole-dipole and hydrogen bonding attractions to the neighboring water molecules. There is an energetic cost to remove a molecule of water from the liquid state and add it to the gaseous state.

There will also be an energetic cost to remove an ethanol molecule from liquid ethanol, as we will need to overcome the attractive interactions due to hydrogen bonding and dispersion. However, the hydrogen bonding is less effective. There are at most three hydrogen bonds to a given ethanol molecule, rather than four to a given water molecule.

In the case of cyclohexane, we will need to overcome the intermolecular attractions due to dispersion, which are relatively weak compared with the hydrogen bonding interactions in ethanol and water.

The heat of vaporization of water

Answer: This is a bit of trick question, as we are comparing apples and oranges. The heat that must be added to a mole of water at 25C to convert it to water vapor at 25C is different from the standard listing of the heat of vaporization, ΔHvap. In the standard listing, the heat of vaporization is the amount of heat that must be added to one mole of liquid water at its normal boiling temperature to convert it to water vapor at its normal boiling temperature.

What we have computed is instead the heat of vaporization at 25C. We computed the heat that must be added to convert a mole of liquid water at 25C to a mole of water vapor at 25C, using the enthalpies of formation of liquid water and gaseous water.

We can also compute the result using a thermodynamic cycle. Start with liquid water at 25C, heat it to the boiling temperature, vaporize it, then cool the water vapor to 25C. The heat added to the system in that three-step process is

qP = ncP(l)(100C-25C) + ΔHvap + ncP(g)(25C-100C)

Using the heat capacities cP(l)=75.4 J/K and cP(g)=36.0 J/K, and ΔHvap=40.7 kJ/mol, we find qP=44 kJ. That result is in close agreement with the result of our calculation of the enthalpy of reaction using enthalpies of formation.

Waiting for the improbable

Answer: If we have 50 yellow gum balls and 50 purple gum balls, we find Ωsegregated=(50!)2=9.25017065 x 10128 while Ωmixed = (2 x 50)!= 9.33262154 x 10157 so that there are

Ωmixed/Ωsegregated=1.00891345 x 1029

times the number of ways of arranging the mixed products.

If we visit a new arrangement every 2 seconds, it would take on average

2(1.0 x 1029)(2 seconds) = 4.0 x 1029 seconds = 1.3 x 1022 years

which is far longer than the age of the universe! Note that we multiplied by two since we were asked to find the yellow gum balls on top. As Spock would say, the event is "possible, but very improbable."

The entropy of disorder in low temperature crystals

Answer: Each carbon monoxide molecule can access two states: forward or backward arrangements. If we take a mole of carbon monoxide molecules in the crystal, there would be Ωcrystal=2N0 possible configurations of the distinguishable CO molecules in the crystal, where N0 is Avogadro's number. Note that the molecules are distinguishable because they are fixed at their sites in the solid crystal lattice.

Using the Boltzmann entropy formula, we find

Scrystal = kB ln[Ωcrystal] = kB ln[2N0] = N0kB ln[2] = R ln[2]

which is 5.76 J/K.

Quantitative determination of the free energy of reaction

Answer: Using a standard table of thermodynamic data collected at standard state and 25C, we find for the reaction

NH3(g) → NH3(aq)

that ΔHo=-80.29 kJ - (-46.11 kJ) =-34.18 kJ, that ΔSo= 111.3 J/K - (192.34 J/K) =-81.04 J/K, and ΔGo= ΔHo - TΔSo =-34.18 kJ - (298.15K)(-81.04 J/K) = -10.01 kJ.

Note that the enthalpy favors the dissolved state due to the strong polar attractions between the ammonia and water molecules. The entropy favors the gaseous state where there are many more possible configurations of the ammonia molecules. At a temperature of 25C, we find that ΔGo < 0 and the dissolution of ammonia gas in water is favorable.

Computing the solubility of ammonia in water

Answer: Given the change in the Gibbs free energy at standard state, ΔGo, we find

K = exp[-ΔGo/RT]=exp[-(-10.01 x 103 J)/((8.314 J/(mol K)) (298.15K))] = 56.7

We also know that K=([NH3]/PNH3)(1 atm/1M). Since the pressure of ammonia gas is taken to be 1 atmosphere, the solution would be approximately 56.7 M. But pure water is roughly 55M, so the ammonia solution would have approximately a 1:1 ratio of ammonia to water molecules.

Does this result make sense? The formula that we have used to compute the equilibrium constant is based on thermodynamic data derived for relatively dilute solutions. This application extends the use of that data to highly concentrated solutions where the results cannot be expected to be accurate.

Heavy lifting with the ammonia fountain

Answer: For 1.0L of ammonia gas at 1.0 atm pressure, the number of moles of ammonia gas is

nNH3 = PV/RT = (1.0 atm)(1.0 L)/((0.08205 L atm/(mol K)) (298.15K)) = 0.041 moles

So the volume of water required to form a 56.7M solution of ammonia from the 1.0L of ammonia gas would be VH2=nNH3/56.7 mol/L = 0.72 mL.

Now if -ΔG=mgh, we find

- nNH3 ΔGo = mgh

but since m=nH2O MH2O, the moles of water times the molar mass of water, we find

h = -nNH3 ΔGo/(nH2O MH2O g)

Using g=9.8 m/s2, MH2O=18 g/mol, and the fact that for the concentrated solution the ratio of ammonia to water is approximately 1:1, we find

h = -(-10.01 x 103 J/mol)/((18 x 10-3 kg/mol)(9.8 m/s2)) = 57 x 103 m = 57 km

Wow! The water could be raised 57km. That is well on its way out of the Earth's atmosphere! Note that the result is independent of the moles of ammonia gas used.

Exploring the chemistry of other flash compounds

Answer: The balanced equations for the chemistry of the various flash powders are

4 KNO3(s) + 10 Mg(s) → 2 K2O(s) + 2 N2(g) + 10 MgO(s)

KClO3(s) + 2 Al(s) → KCl(s) + Al2O3(s)

3 KClO4(s) + 8 Al(s) → 3 KCl(s) + 4 Al2O3(s)

In each case, a source of oxygen is combined with a metal that is readily oxidized, leading to a violently exothermic reaction.

Balancing redox reactions

Answer: The balanced half-reaction for the oxidation of carbon in methane is

2 H2O(l) + CH4(g) → CO2(g) + 8 H+(aq) + 8 e-

and the balanced half-reaction for the reduction of oxygen is

8 e- + 8 H+(aq) + 2 O2(g) → 4 H2O(l)

The overall reaction is

CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)

You can imagine two electrochemical half-cells, one a standard oxygen half-cell, with oxygen gas at 1 atmosphere bubbled over a platinum electrode in a 1M aqueous solution of acid, and the other a similarly constructed standard methane half-cell.

Computing the charge on a single electron

Answer: Using the method of Stoney, we first compute the moles of hydrogen and convert that to moles of electrons and then number of electrons. We find that the number of moles of hydrogen gas corresponding to 28.7mL at 30.11inHg and 21.9C is nH2 = 1.19x10-3 mol. It follows that the total number of electrons used to reduce that many moles of dihydrogen is

Ne- = N0 (2 mol e-/1 mol H2) nH2 = 1.44 x 1021

where N0 is Avogadro's number. The total charge that passes through the wire during the reaction is

Q = It = (0.25 C/s)(15.0 min)(60 s/min) = 225 C

The charge on the electron can be computed from the ratio of the total charge to the number of electrons

e = Q/Ne- = 225 C/1.44 x 1021 = 1.56 x 10-19 C

Correcting the calculation by accounting for the vapor pressure of water

Answer: If the total pressure is 764.8 mmHg, and 19.8 mmHg is contributed by water vapor, the partial pressure of hydrogen gas is actually 745 mmHg. From that corrected pressure, we find that nH2 = 1.16 x 10-3 mol.

Continuing the calculation, we find that the total number of electrons is 1.40 x 1021 and that the revised estimate of the charge on the electron is 1.60 x 10-19 C. The result is in agreement with the accepted value of the charge on the electron to within the number of significant figures allowed by the data.

If you are fond of your fingers, read this problem

Answer: The cyanate ion, OCN-, has a lowest energy Lewis structure with a single bond between the carbon and oxygen, and a triple bond between carbon and nitrogen. Each atom has a filled valence shell. The partial charges on carbon and nitrogen are zero, and the partial charge on oxygen is -1. The partial charges agree with the differences in electronegativity. By these measures, the cyanate ion is expected to be stable.

The fulminate ion, CNO-, has a lowest energy Lewis structure with a double bond between the carbon and nitrogen, and a single bond between nitrogen and oxygen. The partial charges on the carbon and nitrogen are zero, and the partial charge on oxygen is -1. While the Lewis structure minimizes the formal charges, and the formal charges agree with the differences in electronegativity, the carbon does not have a filled valence shell, making the compound unstable.

The balanced equation for the explosion of silver fulminate is

Ag2C2N2O2(s) → 2 Ag(s) + 2 CO(g) + N2(g)

where one mole of silver fulminate produces three moles of gas.

Direct measures of relative reduction potentials

Answer: The standard reduction potential of silver EoAg+|Ag = 0.7996V while that of copper is EoCu2+|Cu = 0.3402V. The silver bar will remain as is. No silver will be oxidized, and there is no silver ion in solution to be reduced. Therefore the copper ion will remain in solution, and no reaction is expected to occur.

Predicting the change in cell voltage as a function of solution concentrations

Answer: The standard reduction potential of cobalt EoCo2+|Co = -0.28V while that of nickel is EoNi2+|Ni = -0.23V. The nickel has the higher standard reduction potential so the nickel will be reduced and the cobalt will be oxidized. The standard state cell voltage is EoCo|Co2+||Ni2+|Ni = -0.23V - (-0.28V) = 0.05V.

The overall reaction is

Co(s) + Ni2+(aq) → Co2+(aq) + Ni(s)

where Q=[Co2+]/[Ni2+], the number of electrons involved in the reaction is n=2, and the temperature is 25C. We find

ECo|Co2+||Ni2+|Ni = Eo Co|Co2+||Ni2+|Ni - 0.0296V log10 [Co2+]/[Ni2+].

More simply put, the cell voltage is predicted by the Nernst equation to be

ECo|Co2+||Ni2+|Ni = 0.05V - 0.0296V log10 [Co 2+]/[Ni2+].

For the various concentrations listed, we find 0.05V, 0.08V, 0.1V, 0.05V, and 0.0V, in order. Note that the last set of concentrations leads to a cell voltage that is approximately zero, meaning that the system is close to equilibrium.

Predicting concentration cell voltages

Answer: The standard state cell voltage for any concentration cell is zero. Therefore EoCu|Cu2+||Cu2+|Cu = 0V. The overall reaction is

Cu(s) + Cu2+(aq)[left] → Cu2+(aq)[right] + Cu(s)

where Q=[Cu2+]R/[Cu2+]L, the number of electrons involved in the reaction is n=2, and the temperature is 25C. We find

ECu|Cu2+||Cu2+|Cu = - 0.0296V log10 [Cu2+]R/[Cu2+]L.

For the various concentrations listed, we find 0.0V, 0.03V, 0.06V, 0.0V, and -0.06V, in order.

Using concentration cells to determine unknown solution concentrations

Answer: The standard state cell voltage for any concentration cell is zero. Therefore EoCu|Cu2+||Cu2+|Cu = 0V. The overall reaction is

Cu(s) + Cu2+(aq)[unknown] → Cu2+(aq)[0.01M] + Cu(s)

where Q=[0.01M]/[Cu2+]unknown, the number of electrons involved in the reaction is n=2, and the temperature is 25C. We find

ECu|Cu2+||Cu2+|Cu = 0.058V = - 0.0296V log10 [0.01M]/[Cu2+]unknown.

. Solving for the unknown concentration we find [Cu2+]unknown = 0.91M. The concentration is higher than 0.01M, as we expected.

The principal driving force in the reaction is to make the concentration of copper ion in each beaker equal. The voltage will cause electrons to flow toward the half-cell of higher concentration, reducing the concentration of copper ion.

The formation of colloidal sulfur and the point of equilibrium

Answer: The addition of acid will convert sulfite ion to hydrogen sulfite

SO32-(aq) + H+(aq) → HSO3-(aq)

Therefore, the addition of acid will remove sulfite ion, shifting the equilibrium

S2O32-(aq) → S(s) + SO32-(aq)

to the right, in favor of products.

Temperature dependence of chemical reaction rates

Answer: To determine the activation energy, we use the result

ln[ k(T2)/k(T1)] = (Eact/R) (1/T1 - 1/T2)

where T1=293.15, T2=318.15K, and k(T2)/k(T1)=2. We find that Eact = 21.5 kJ/mol.

Rates of catalysis are influenced by the form of the catalyst

Answer: For the reaction to be catalyzed, the reactant must bind to or otherwise interact with the catalyst. The rate of the catalyzed reaction will depend on the concentration of reactant interacting with the catalyst, and that concentration will depend upon the number of catalytic binding sites available to the reactant.

When the potassium permanganate is in crystalline form, the binding sites will be limited to the surface of the crystal. When the crystals are macroscopic in size, that vast majority of the permanganate will be found on the interior of the crystal and unavailable for binding. When the permanganate is dissolved in solution, there is a great increase in the number of permanganate molecules accessible to the reactant, and the rate of reaction is dramatically increased.

The optimal form of a catalyst is crucial to the optimization of an catalytic process, and an important consideration for engineers working to design catalytic reaction processes on an industrial scale.

Enhancements in the rate of reaction

Answer: To determine the change in the activation energy induced by interaction between the reactant and the catalyst, we use the result

ΔEact = Eactuncat - Eactcat = RT ln[ kcat(T)/kuncat(T)]

where we are told that kcat/kuncat = 4 x 108. We find that ΔEact = 51.2 kJ/mol = 12.2 kcal/mol.

The Boltzmann probability and black body radiation

Answer: Using the fundamental relation between the energy of a photon and the wavelength of light

Ephoton = hc/λ

we find that for red light of wavelength 650 nm, the photon energy is 3.04 x 10-19 J = 183 kJ/mol. The equivalent temperature can be found from the relation

E = kBT

where kB = 1.38 x 10-23 J/K is Boltzmann's constant. The photon energy is equivalent to a temperature of 22.0 x 103K. For yellow light of wavelength 570 nm, the photon energy is 3.47 x 10-19 J = 209 kJ/mol equivalent to a temperature of 25.1 x 103K. At last, for violet light of wavelength 400 nm, the photon energy is 4.95 x 10-19 J = 298 kJ/mol equivalent to a temperature of 35.8 x 103K.

The photoelectric effect and the ionization of atoms

Answer: The binding energy for the electron is 13.6 electron-volts (or eV). To convert to Joules

13.6 eV = 13.6 electron V (1.60 x 10-19 C/electron) = 2.18 x 10-18 CV = 2.18 x 10-18 J

The equivalent frequency can be found using Einstein's relation for the energy of a photon

ν = E/h = 2.18 x 10-18 J / 6.626 x 10-34 Js = 3.28 x 1015 Hz

These numbers are already familiar, from our study of the ionization energy and line spectra of the hydrogen atom.

Exploring the Lyman, Balmer, and Paschen spectral series

Answer: Our general result for the allowed photon energies for emissions from the hydrogen atom are

Ephoton = -ΔEatom = -(Efinal - Einitial) = 2.18 x 10-18 J (1/nf2 - 1/ni2)

In analyzing the Lyman spectral series, we make use of our result for the allowed frequencies of emission in the hydrogen atom, where the initial state is any state and the final state is the ground electronic state

Ephoton = -ΔEH atom = 2.18 x 10-18 J (1/nf2 - 1/ni2) = Ry (1 - 1/ni2)

The longest wavelength of light emitted would correspond to the smallest possible energy of transition, between the ni=2 and nf=1 states

Ephoton = hc/λ = Ry (1 - 1/22) = 3Ry/4

That means that λshort = 4hc/3Ry = 121 nm. The shortest wavelength of light emitted would correspond to the largest possible energy of transition, between the ni=∞ and nf=1 states. The corresponding wavelength is λlong = hc/Ry = 90.9 nm. The results fall largely in the ultraviolet region of the spectrum (12.5-375 nm).

The Balmer series consists of the emission lines that correspond to transition to the final nf=2 electronic state from any higher energy electronic state. For those transitions λshort = 4hc/Ry = 364 nm and λlong = 36hc/5Ry = 654 nm. The spectral emissions fall largely in the visible region of the spectrum (400-700 nm).

The Paschen series consists of the emission lines that correspond to transition to the final nf=3 electronic state from any higher energy electronic state. For those transitions λshort = 9hc/Ry = 818 nm and λlong = 144hc/7Ry = 1870 nm. The spectral emissions fall largely in the infrared region of the spectrum (780 nm - 1 mm).

The "F-center" of an ionic crystal modeled as a particle-in-a-box

The energy of the absorbed photon must be equal to the energy difference between the ground and first excited states of the electron in the F-center of the potassium chloride crystal. Modeling the electron as a particle in a cubic box, the ground state will be E111 and the first excited state E211=E121=E112. We find

Ephoton = hc/λ = E211 - E111 = 3h2/(8 m L2)

Solving for L=6.83 x 10-10m. That box length is slightly larger than the ionic diameter of a chloride ion, estimated by Linus Pauling to be 3.62Å.

The bromide ion is expected to be larger than the chloride ion, so the F-center vacancy will be larger in potassium bromide. The larger "box" will lead to a longer wavelength of absorption. Since the potassium chloride appears magenta, the F-center absorbs green light. The potassium bromide absorbs at a longer wave length, so it might absorb yellow and appear blue. We identify the blue crystals as potassium bromide.

Modeling the "solvated electron" as a particle-in-a-box

The energy of the absorbed photon is equal to the energy difference between the ground and first excited states of the electron in the solvated electron. We model the electron as being confined to a cubic box, with ground state E111 and first excited state E211=E121=E112. We find

Ephoton = hc/λ = E211 - E111 = 3h2/(8 m L2)

where λ=9.8 x 10-7nm. That leads to L=9.4 x 10-10m. Note that the electron is more delocalized than in the case of the confinement to the F-center of potassium bromide. The 980nm light that is absorbed falls in the infrared region of the spectrum (780 nm - 1 mm).

Ionization energy and effective nuclear charge

The ionization energies of the first ten elements can be used to estimate the effective nuclear charge "felt" by the ionized valence electron. Our estimate of the energy of an electron in the multielectron atom is

En = -Ry Zeff2/ n2

If we assume that the final ionized state corresponds to n=∞, we can estimate the ionization energy to be

I.E. = En=∞ - En = Zeff2Ry /n2

where n is the principal quantum number of the outermost valence electron that will be ionized, and

Zeff = n(I.E./Ry)½

The first ionization energies for the first ten neutral atoms are H (13.5984 eV), He (24.5874 eV), Li (5.3917 eV), Be (9.3227 eV), B (8.2980 eV), C (11.2603 eV), N (14.5341 eV), O (13.6181 eV), F (17.4228 eV), and Ne (21.5645 eV). We can compute Zeff for those same ten atoms and find H (1.000), He (1.345), Li (1.259), Be (1.656), B (1.562), C (1.820), N (2.068), O (2.001), F (2.264), and Ne (2.518).

Space oddity

There are six frequencies of light that can be emitted as a result of transitions in the He+ ion initially in its third excited state. One example is

νHe+4→1 = (15/4) Ry/h = (15/4) (3.29 x 1015 Hz) = 1.23 x 1016 Hz

The other possible frequencies are νHe+4→2 = (3/4) Ry/h, νHe+4→3 = (7/36) Ry/h, νHe+3→1 = (32/9) Ry/h, νHe+3→2 = (5/9) Ry/h, and νHe+2→1 = 3 Ry/h.

The possible frequencies of light that could be absorbed by a H atom initially in its ground electronic state include

νH1→2 = (3/4) Ry/h = (3/4) (3.29 x 1015 Hz) = 2.57 x 1015 Hz

as well as νH1→3 = (8/9) Ry/h, and νH1→4 = (15/16) Ry/h. It happens that the transition He+(n=4 → n=2) is resonant with the transition H(n=1 → n=2) as

νH1→2 = (3/4) Ry/h = νHe+4→2

Light emitted by the He+ ion initially in its third excited state can be absorbed by a H atom in its ground electronic state.

Modeling electronic transitions in multielectron atoms

The energy associated with a transition in the one-electron Cu+28 ion between its inital ground state (n=1) and final first excited state (n=2) is

ΔECu+28 = -Ry Z2 (1/nfinal2 - 1/ninitial2) = -2.18 x 10-18 J (292) (1/22 - 1/12) = 1.37 x 10-15 J

For light of wavelength 154. x 10-12 m in the x-ray spectrum, the energy of a corresponding photon is

Ephoton=ΔECu=hc/λ=1.29x10-15J.

The transition energy in the neutral copper atom, Cu, is less than the transition energy in the one-electron copper ion, Cu+28. In the multielectron atom, the 28 other electrons shield the nucleus's charge from the electron undergoing transition. That electron undergoing transition "feels" a smaller nuclear charge and the transition energy is reduced relative to that in the one-electron ion.

Agreement between theory and experiment does not prove that the theory is correct!

The song remains the same. The permanganate remains diamagnetic. The Mn(III) remains more paramagnetic than the Mn(IV). The electron configurations deviate from the configurations predicted by the initial application of the Aufbau Principle, Pauli Principle, and Hund's Rule. An electron is of higher energy when singly or doubly occupying the 4s orbital than when occupying a 3d orbital. The "rule of thumb" that we can apply to the first row of transition metals is that none of the ions of the first row transition metals have 4s electrons.

Trends in bond order in the homonuclear diatomic molecules

The predictions for the bond order are H2 (1), He2 (0), Li2 (1), Be2 (0), B2 (1), C2 (2), N2 (3), O2 (2), F2 (1), and Ne2 (0). Of those possible molecules, we expect that the noble gases He and Ne will not form stable molecules, and they do not. H2, N2, O2, and F2 we have encountered, and the predicted bond orders agree with the predictions of Lewis electron dot structures for those homonuclear diatomic molecules.

Trends in bond order in photoexcited homonuclear diatomic molecules

The predictions for the bond order of the photoexcited homonuclear diatomics are made by considering the electronic configuration of the photoexcited molecule. For example, H2 has a ground state of

σ1s2

with a bond order of 1, and first excited state of

σ1s &sigma1s*

with a bond order of 0. It follows that the bond order of the remaining homonuclear diatomics in their first excited electronic states are He2 (1), Li2 (0), Be2 (1), B2 (1), C2 (2), N2 (2), O2 (2), F2 (1), and Ne2 (1). It appears that it may be possible to observe a stable homonuclear diatomic molecule composed of noble gas atoms if the molecule is in an appropriate excited electronic state.

Note that it is possible to use electronic excitation to change bond order, at times making stable bonds unstable. That is the basis of the field of photochemistry, where light is used to change the electronic states of molecules to carry out bond breaking and formation, in reorganizing atoms and making new molecules.

Taking note of light-sensitive compounds

To make silver chloride, mix elemental silver with hydrochloric acid. To make silver nitrate, mix elemental silver with nitric acid. To make silver sulfate, mix elemental silver with sulfuric acid. Those chemicals have in common that they are all strong acids.

The standard reduction potential for silver(I) is positive

Ag+(aq) + e- → Ag(s)         Eo= 0.7991 V

The oxidation of metallic silver in a solution of pH = 0 is a spontaneous process.

Which salt is most soluble? We know that silver chloride is not soluble, and most sulfates, including silver sulfate, are not soluble. Ah, but most nitrates are soluble. So silver nitrate is the soluble salt of the three.

The positive and the negative in photography

The dageurreotype is the mirror image of the object.

Exploring the colors of dyes using the particle-in-a-box model

We assume that the energy of the absorbed photon must be equal to the energy difference between the HOMO and LUMO of the dye molecule. Modeling the highest lying electron in the molecule as a particle-in-a-box, we take the transition energy to be

Ephoton = hc/λ = ELUMO - EHOMO = 3h2/(8 m L2)

Solving for cyanine, where λmax=523 nm leads to L=6.91 x 10-10m. Solving for pinacyanol, where λmax=604 nm leads to L=7.43 x 10-10m. Solving for dicarbocyanine, where λmax=706 nm leads to L= 8.03 x 10-10m.

The results are quite reasonable! If we take the length of a carbon-carbon bond to be 1.3 Å we find for the three dye molecules boxes of 5.3, 5.7, and 6.2 carbon-carbon bond lengths. Not bad!

Assigning the infrared spectrum of a mixture of gases

We expect that the bond length of HI (1.61 Å) is longer than the bond length of HBr (1.41 Å). Therefore, the bond strength of HI (298 kJ/mol) is expected to be weaker than that of HBr (366 kJ/mol). The transition frequency is proportional to the square root of the force constant

ν ∝ k½

The HI bond has a smaller force constant than the HBr bond and therefore is expected to have a lower frequency of oscillation,

We can make a similar argument in terms of the mass. The reduced mass of the HI bond is larger than the reduced mass of the HBr bond. The transition frequency is inversely proportional to the square root of the reduced mass

ν ∝ 1/μ½

With a larger reduced mass, the HI bond is expected to have a lower frequency of oscillation,

Both arguments suggest that the line at lower frequency in the IR spectrum (6.69 x 1013 Hz) is assigned to HI and the spectral line at higher frequency (7.68 x 1013 Hz) is assigned to HBr.

Assigning the infrared spectrum of a small organic molecule

We can make use of the fact that the transition frequency

ν ∝ (k/μ)½

where k is the force constant of the bond and μ is the bond's reduced mass. The 13C=O bond has a larger reduced mass than the 12C=O bond. The spectral line corresponding to the vibrational excitation of the 13C=O bond in formic acid will be red shifted relative to the 12C=O bond.

Assigning the microwave spectrum of a mixture of gases

The frequency of absorption is due to transitions between rotational quantum states of the molecule. For a transition from the J to J+1 quantum state, the change in energy of the molecule will be

ΔE=EJ+1 - EJ=2hB(J+1)         B = h/(8π2I)

The frequency of transition from the ground (J=0) to first excited (J=1) states in the pure rotational spectrum will correspond to a frequency of

ν = h/(4π2I)

where the moment of inertia I=μreq2. Thinking of the reduced masses, μHCl = 0.9796 amu < μDCl = 1.9044 amu. Therefore, IDCl > IHCl and the frequency of transition in the DCl molecule will be lower than that in the HCl molecule.

How much lower? The frequency of transition is inversely proportional to the moment of inertia, so

νDCl = νHCl (IHCl/IDCl) = νHCl (μHCl/μDCl)

leading us to estimate BDCl = BHCL(μHCl/μDCl) = 0.312x1012Hz (0.9796/1.9044) = 0.160x1012Hz.

The quantum mechanics of a coordination complex

The cobalt(II) ion has an electron configuration [Ar] 3d7. In the CoCl42- complex ion, the four coordinating chloride ligands form a tetrahedral geometry. The localized electrons on the chloride ions donate electron density to each of the four sp3 hybrid atomic orbitals on the central cobalt ion.

The ordering of the d-orbitals in the tetrahedral ligand field is the opposite of that in the octahedral ligand field. The ligand field splitting between the lower lying dz2 and dx2-y2 orbitals and the higher lying dxy, dxz, and dyz orbitals is Δt. The chloride ions are weak field ligands, so that Δt is expected to be small relative to the energy associated with electron-electron repulsion in the individual d-orbitals. The resulting high spin electron configuration is

dz22 dx2-y22 dxy1 dxz1 dyz1

Note that for this compound, the electron configuration of this high spin state is identical to the low spin configuration. The compound is predicted to be paramagnetic.

The quantum mechanics of a silver complex ion

The electron configuration of the Ag+ ion is [Kr] 4d10. The orbital hybridization is sp, consistent with a linear complex. The cyanide ions are strong field ligands. The silver has no choice but to be in a low spin state. The compound is diamagnetic.

Predicting the final reaction product in the diverse chemical soup

The net ionic equation for the reaction is

2 Ag(CN)2-(aq) + S2-(aq) → Ag2S(aq) + 4 CN-(aq)

where the dominant reaction product is a silver sulfide precipitate. In its mineral form, silver sulfide is known as argentite.