Physical Chemistry Quantum Chemistry and Molecular Interactions 1st Edition Andrew Cooksy Solutions by TBtutors

Type:

Solution Manual

Physical Chemistry Quantum Resource: Chemistry and Molecular Interactions Edition:

1st Edition

Author(s):

Andrew Cooksy

Physical Chemistry Quantum Mechanics, Spectroscopy, and Molecular Interactions Solutions Manual by Andrew Cooksy

February 4, 2014

Contents Contents

Objectives Review Questions

Chapter Problems Notes on Maple and Mathematica commands . . . . . . . . . . . . . . . . . . . . . . . . . . . Chapter A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Chapter 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Chapter 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Chapter 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Chapter 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Chapter 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Chapter 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Chapter 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Chapter 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Chapter 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Chapter 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Chapter 11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Chapter 12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Chapter 13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

11 11 11 23 39 63 95 134 154 184 205 229 249 263 274 282

Sources and Acknowledgments

293

Bibliography

297

Objectives Review Questions Chapter 1 1.1 We use Eq. 1.1 to convert wavelength to frequency (taking advantage of the constant speed of light c) and we use Eq. 1.2 to convert frequency to photon energy: ν=

c 2.998 · 108 m s−1 = = 3.0 · 1011 s−1 λ 1.0 · 10−3 m

by Eq. 1.1

Ephoton = hν = (6.626 · 10−34 J s)(3.0 · 1011 s−1 ) = 2.0 · 10−22 J.

Eq. 1.2

1.2 The de Broglie wavelength (Eq. 1.3) is our measure of the degree of quantum character in our system. We calculate λdB and compare it to the domain to determine if we need quantum mechanics to describe the physics. In this case, to find the de Broglie wavelength we need to calculate the momentum p from the kinetic energy, but we can do that: p2 mv 2 = 2mp √2 p = 2mK = 2(1.008 amu)(1.661 · 10−27 kg amu−1 )(4.0 · 10−21 J) = 3.66 · 10−24 kg m s−1

λdB =

h 6.626 · 10−34 J s = 1.8 · 10−10 m = 1.8 Å. = p 3.66 · 10−24 kg m s−1

Because 1.8 · 10−10 m ≪ 1.0µm = 1.0 · 10−6 m, it is unlikely that quantum effects arising from this motion will be significant. 1.3 The atom is in an n = 2 state, and we can use the Bohr model of the atom to calculate the correct values of the energies. From Eq. 1.15 we can calculate the total energy, and from Eq. 3.7 we can calculate the potential energy. The question does not specify units, and the most convenient units for the total energy are Eh : Z2 22 En = − 2 Eh = − Eh = −0.5 Eh. 2n 2(22 ) The potential energy depends on the radius of the electron orbit in the Bohr model, rn =

22 n2 a0 = a0 = 2a0 , Z 2

which gives us U =−

Ze2 2e2 e2 =− =− = −1.00 Eh. 4πǫ0 r (4πǫ0 )(2a0 ) (4πǫ0 )a0 )

Chapter 2 1

2.1 We apply the operator to the function, and see if we can find the original function again afterward: 1 1 d 3x e2x = 3e2x + 3x(2e2x ) α̂f (x) = x dx x 1 2x 2 1 2x = 3e + 2 · 3e = f (x). + x x2 x The result is equal to f (x) times another function of x, so this is not an eigenvalue equation. 2 1 d 2x2 1 = 3e 3(4x)e2x x dx x 2 = 4 · 3e2x = 4f (x).

α̂g(x) =

But g(x) is an eigenfunction of α̂, because the result is the original function g(x) times the eigenvalue 4. 2.2 We use the average value theorem, Eq. 2.10, integrating between 0 and a. The integral over x4 sin(cx) can be found using a symbolic math program. Setting c = 2π/a for now to simplify the notation, we have: Z Z a 2 a 2 sin (cx) x4 dx = 0.176a4. ψ ∗ (x4 ) ψ dx = a 0 o 2.3 To write the Schrödinger equation we need the Hamiltonian, which consists of the kinetic energy operator −(h̄2 /2m)∂ 2 /∂x2 , and the potential energy function described in the problem. In this case, the potential energy is given by the formula for a line, to which we assign a slope U0 . We can also add a constant, but it will have no effect on the relative energies or the wavefunctions, so we may as well set it equal to zero. Our potential energy function therefore is U0 x, and the Schrödinger equation becomes h̄2 ∂ 2 + U x ψ = Eψ. − 0 2m ∂x2 2.4 We use Eq. 2.41 to calculate the energy, with a mass mp and the volume given: h2 (n2 + n2y + n2z ) 8mV 2/3 x (6.626 · 10−34 J s)2 E100,1,1 = (1002 + 12 + 12 ) 8(1.673 · 10−27 kg)(1.0 · 10−18 m3 )2/3

Enx ,ny ,nz =

Eq. 2.41

= 3.28 · 10−25 J.

Chapter 3 3.1 We combine the radial and angular parts of the wavefunction as dictated by the quantum numbers, and also substitute Z = 3 for lithium: ψ3,1,−1 (r, θ, φ) = R3,1 (r) Y1−1 (θ, φ) r √ 3/2 r 3r 3 3 4 2 −r/a0 1− e sin θ e−iφ . = √ a a 2a 8π 27 3 0 0 0 3.2 We are using an integral to find an average value, so we use the average value theorem (Eq. 2.10), where the operator is r (the distance from the nucleus) and the wavefunction is given by ψ3,1,−1 (r, θφ) with Z = 3 for lithium: 3 Z ∞ 2 2 32 3 r 3r 1 − e−2r/a0 r3 dr = 25a0 /3. 3 7 a0 a 2a 0 0 0 2

3.3 The number of angular nodes is given by l, which is 1 for a p orbital, and the number of radial nodes is equal to n − l − 1 = 3 − 1 − 1 = 1: 1 angular node, 1 radial node. Chapter 4 4.1 This ion has three electrons and an atomic number Z = 4. We need one kinetic energy term for each electron, three terms for the attraction of each electron for the nucleus, and then three terms for electron–electron repulsions, one for each distinct pairing of the electrons: 1 and 2, 2 and 3, and 1 and 3:

h̄2 4e2 4e2 e2 4e2 − − − + ∇(1)2 + ∇(2)2 + ∇(3)2 − 2me 4πǫ0 r1 4πǫ0 r2 4πǫ0 r3 4πǫ0

1 1 1 + + r12 r23 r13

4.2 Neutral beryllium has 4 electrons, so Be+ has 3 electrons, which we place in the lowest energy subshells 1s and 2s for an electron configuration 1s2 2s1 . The zero-order energy is then the sum of the energies we would calculate if each electron were alone in that subshell. That one-electron energy is −Z 2 Eh /(2n2 ). We have two n = 1 electrons and one n = 2 electron in the configuration, and Z = 4 for Be, so we arrive at 42 1 1 1 E0 = − Eh + + 2 12 12 22 16 9 = −18 Eh. = 2 4 4.3 We use Eq. 4.30, which calculates the effective atomic number by treating the electron as though it were a single electron in an atom with a variable atomic number: 1/2 1/2 2ǫ(i)n2 2(0.182) Eh(32 ) Zeff = − = = 1.81. Eh Eh 4.4 We reverse the labels 1 and 2 in the function and then check to see whether the function has changed sign: Then we find that P̂21 ψ(1, 2) = cos(−x2 ) cos(y1 ) − cos(−x1 ) cos(y2 ) = − cos(−x1 ) cos(y2 ) + cos(−x2 ) cos(y1 ) = −ψ(1, 2). Therefore, the function is antisymmetric. 4.5 According to the arrow diagrams, we have 1s1

ml = 0 ↑ ↑ ↑ ↑ ↑ ↑ ↓ ↓ ↓ ↓ ↓ ↓

ml 0 0 0 0 0 0 0 0 0 0 0 0

ms +1/2 2p1 +1/2 +1/2 +1/2 +1/2 +1/2 −1/2 −1/2 −1/2 −1/2 −1/2 −1/2

ml = −1 ↑ ↓ ↑ ↓

0 ↑ ↓ ↑ ↓

↑ ↓ ↑ ↓

ml −1 0 +1 −1 0 +1 −1 0 +1 −1 0 +1

ms +1/2 +1/2 +1/2 −1/2 −1/2 −1/2 +1/2 +1/2 +1/2 −1/2 −1/2 −1/2

ML −1 0 +1 −1 0 +1 −1 0 +1 −1 0 +1

MS +1 +1 +1 0 0 0 0 0 0 −1 −1 −1

P P 3 P 3 P 3 P 3 P 1 P 1 P 1 P 3 P 3 P 3 P 3

where we found maximum values of L = 1 and S = 1 initially (based on the largest values of ML and MS ), and then after assigning the 9 3 P states, we were left with three MS = 0 states, which gave L = 1, S = 0 for the 1 P term. Breaking the 3 P into its component J values from L−S = 0 to L+S = 2, and ordering according to Hund’s rules, the final list of states is 3 P0 , 3 P1 , 3 P2 , 1 P1 . Chapter 5 5.1 This molecular ion has 2 nuclei with ZLi = 3 and ZH = 1 and 3 electrons, for a total of 5 particles. That means we should have 5 kinetic energy terms (one for each particle), 6 electron–nucleus attraction terms, 1 nucleus–nucleus repulsion term, and 3 electron–electron repulsion terms. (The total number of potential energy terms for N particles is always N (N − 1)/2, in this case 5 · 4/2 = 10, which gives the number of distinct pairs of particles. There is a potential energy term for each pair of interacting particles.) Using the standard form of the kinetic energy operator for each particle and the Coulomb potential for each pair of particles, we end up with the following: h̄2 1 3 1 3 1 e2 3 Ĥ = − − − − − − ∇(1)2 + ∇(2)2 + ∇(3)2 + − 2me 4πǫ0 rLi1 rH1 rLi2 rH2 rLi3 rH3 h̄2 1 1 3 h̄2 1 − + + + ∇(Li)2 − ∇(H)2 . + r12 r23 r13 RAB 2mLi 2mH 5.2 The orbital we’re constructing combines an s orbital (spherical) with a p orbital lying along the bond axis. If we keep the same orientation of nuclei A and B with respect to the z axis direction that is used elsewhere in the chapter, then the s and p orbitals have the same phase where they overlap, so we will get constructive interference between the two nuclei. However, we expect a node (where the new wavefunction will change sign) somewhere to the +z side of nucleus B, where the negative phase of the p orbital cancels the positive phase of the exponentially decaying s orbital.

5.3 The problem describes a curve such as Fig. 5.14, but a with minimum at R = 1.5 Å where the potential energy reaches a value U = −200 kJ mol−1.

E (kJ mol −1)

1.5A 0

−200

5.4 We can deduce from the orientation of the orbitals that (i) only the s, px , and py atomic orbitals are involved (because the orbitals lie in the xy plane) and (ii) orbital 1 consists of only s and px character (because it points along the x axis). All of the original px orbital density must be distributed somewhere among all three hybrid orbitals, so if we increase the amount of px in orbital 1, then the px character of orbitals 2 and 3 must decrease. The px orbital character tends to elongate the hybrid orbital along the x axis. By removing that character from orbitals 2 and 3, we elongate them more along the y axis instead, which will increase the angle between orbitals 2 and 3. (That angle approaches 180◦ in the limit that only s and py character remains, because then you have an sp hybrid, rather than an sp2 4

hybrid.) As the angle between orbitals 2 and 3 diverges, the angles between 1 and 2 and between 1 and 3 decrease. 5.5 For the proton NMR, we have two chemical shifts, one at about δ = 3.0 for the protons adjacent to the Br and another at about δ = 3.5 for the protons adjacent to the Cl. Each of these is then split into a triplet by the interaction with the two protons at the neighboring carbon atom. The actual chemical shifts turn out to be shifted downfield (to higher δ) because we have two electronegative atoms relatively close together: 1 H: 2 triplets, equal intensity, δ ≈ 3.55, 3.75. For the 13 C, we get two chemical shifts (one for each carbon atom). The J-coupling to the protons is usually very large and potentially a source of confusion, so 13 C spectra are usually purposely gathered under conditions that eliminate the J-coupling, resulting in singlets: 13 C: 2 singlets, equal intensity, δ ≈ 25, 35. Chapter 6 6.1 For this one, we can use the mathematical approach: ˆ Iˆσ̂xy ψ(x, y, z) = Iψ(x, y, −z) = ψ(−x, −y, z) = Ĉ2 (z)ψ(x, y, z). Therefore, Iˆσ̂xy = Ĉ2 (z). 6.2

H C

C H

This molecule has a Ĉ2 symmetry axis along the C C bond and two vertical mirror planes that contain that bond. Those and the identity are the only symmetry elements, so the point group is C2v and the symmetry elements are Ê, Ĉ2 , σ̂xz , σ̂yz . 6.3 1,1-Dibromoethene is in the point group C2v . We evaluate the results of the direct product Γi ⊗ Γµ , where Γµ may be any of A1 , B1 , and B2 for electric dipole transitions (because these correspond to the functions x, y, and z), or A1 , A2 , B1 , and B2 for Raman transitions (because each of these corresponds to some quadratic function such as x2 or xy). Therefore, the possible upper states for an electric dipole transition would be 1 A1 , 1 B1 , 1 B2 , and for a Raman transition would be 1 A1 , 1 B1 , 1 B2 , 1 A2 . 6.4 The molecule is in the point group C2h . The π bond lies perpendicular to the plane of the nuclei— the xy plane. That π orbital must be symmetric under the Ĉ2 rotation of the molecule (which doesn’t change what lies above or below the xy plane) but changes sign under inversion and under reflection through the xy plane. The representation is therefore au . Chapter 7 7.1 The molecule has 7 electrons, but 4 of them are in the 1s core orbitals and are not expected to contribute to the bonding. In the ground state, we would put 2 of the remaining 3 electrons into the 2σg bonding orbital and the last electron in the 2σu antibonding orbital, predicting a bond order of (2 − 1)/2 = 1/2. 7.2 The molecule has 7 electrons, and we would predict the MO configuration 1σg2 1σu2 2σg2 2σu1 . 5

7.3 Because the superscripts are not the same, the spins of the two states are not the same and the transition is forbidden by the spin selection rule ∆S = 0. 7.4 The transition would be an emission transition, forbidden by the spin selection rule (because ∆S 6= 0). Therefore, the transition would occur by the process of phosphorescence. Chapter 8 8.1 According to Eq. 8.17, 1 Ev = v + ωe = (1450 cm−1)(2.5) = 3625 cm−1. 2 The wavefunction for the v = 2 state we can assemble from the normalization constant A2 , the Hermite 2 polynomial H2 (y), and the exponential function e−y /2 (Eq. 8.16):

kµ h̄2

1/8

1 √ 8 π

1/2

(4y 2 − 2)e−y /2

where y = (R − Re )(kµ/h̄2 )1/4 . 8.2 The spacing increases with E (because the walls are steeper than in the harmonic oscillator potential), but more slowly than particle in box (because the walls are not infinitely steep). The wavefunctions are similar to those in the harmonic oscillator, but with less variation in amplitude from the center to the walls (because the bottom of the well is flatter). And because the walls are steeper than in the harmonic oscillator, the tunneling does not extend as far.

8.3 The reduced mass µ of a diatomic is given by mA mB /(mA + mB ), and if mA = mB (i.e., any homonuclear diatomic), then µ = mA /2. For 39 K2 , where each atom has a mass of 39.098 amu, the reduced mass is µ=19.55 amu. To estimate the vibrational constant, we need a guess of the force constant. Choosing k ≈ 12 N m−1 , halfway between the values of 17 N m−1 for Na2 and 7 N m−1 for Cs2 in Table 8.2, we predict s r k ( N m−1 ) 12 −1 ωe ( cm ) = 130.28 ≈ (130.28) = 102 cm−1. µ ( amu) 19.55 8.4 The point group is D∞h . There are two equivalent C H bonds. In group theory, we consider all the equivalent bonds at the same time, so we can either have the two bonds move in phase to get the symmetric stretch, which has σg symmetry, or they can move exactly out of phase to get the antisymmetric stretch, which has symmetry σu . Checking the functions for whether these correspond to functions for an IR active mode (x, y, or z) or a Raman active mode (any quadratic function of x, y, and z), we obtain the following results, in summary: σg (Raman active), σu (IR active). Chapter 9 6

9.1 We use Eq. 9.5, combining the reduced mass of 7.55 amu and the equilibrium bond length to get Be ( cm−1 ) =

16.858 16.858 = 1.755 cm−1 . = 2 (7.55)(1.128)2 µ ( amu) Re ( Å)

9.2 From Table 9.2 we obtain the values A = B = 9.94 cm−1 and C = 6.30 cm−1, which corresponds to an oblate top with energy levels given by Eq. 9.22: Erot = Kc2 (C − B) + BJ(J + 1) = (12 )(6.30 − 9.94) + (9.94)(2)(3) cm−1 = 56.0 cm−1 .

9.3 The rotational constant will increase, according to Eq. 9.16, as the moment of inertia decreases. Of the three molecules, 12 C16 O has the lowest moment of inertia, because it has only two atoms (of the same masses as the atoms that make up the CO2 molecules). Between the two CO2 isotopologues, the 18 O-substituted molecule has the greater isotopic mass, so the greater moment of inertia. The ordering is therefore 12 C18 O2 , 12 C16 O2 , 12 C16 O. 9.4 We fold the values of the rotational constants from Table 9.1 into Eqs. 9.8 and 9.9, setting v = 1, to find the rotational energies for J = 2 and J = 3. The difference between those energies is the transition energy ∆E that we’re looking for: 1 αe = [20.9557 − (1.5) (0.798)] cm−1 = 19.759 cm−1 Bv=1 = Be − v + 2 Erot = Bv J(J + 1) − Dv [J(J + 1)]

Erot (J = 2) = (19.759 cm−1)(2)(3) − (2.15 · 10−3 cm−1 ) [(2)(3)]2

Erot (J = 3) = (19.759 cm−1)(3)(4) − (2.15 · 10−3 cm−1 ) [(3)(4)]

∆E = Erot (J = 3) − Erot (J = 2) = 118.32 cm−1 . Chapter 10 10.1 Repulsion applies to all examples, and is expected to be roughly proportional to e−aR , although other forms are used to model this (such as the R−12 repulsive term in the Lennard–Jones potential). (a) dispersion: U (R) ∝ αA αB /R6 (by Eq. 10.38). (b) H-bonding: U (R) ∝ µ2A µ2B /R6 (by Eq. 10.19), although a case can be made that hydrogen bonding is strong enough that the structures cannot be treated as freely rotating at typical temperatures, in which case the µA µB /R3 of Eq. 10.16 would be more appropriate. In at least one popular model potential used to predict biochemical structure, the distance-dependence of hydrogen bonding is given using either a Lennard–Jones 6-12 potential (as in our Eq. 10.44) or a 10-12 potential (in which the distance dependence is a much more quickly decaying R−10 [1]). (c) dipole–dipole, dipole–induced dipole: U (R) ∝ µ2A µ2B /R6 , µA αB /R6 (Eqs. 10.19, 10.23). The dipole–induced dipole is important here because CO is only very weakly polar, but has a relatively high polarizability. Note the important distinction between these two terms: polar means the positive and negative charges in the molecule are already well separated to create a permanent dipole moment, while polarizable means that an external electric field can easily separate the charges, whether or not they are already well separated. (d) dipole–dipole (non-rotating): U (R) ∝ µ2A αB /R3 (by Eq. 10.16). You could make a good case that dipole–induced dipole is important here as well. I have only left it off because the interaction between non-rotating dipoles, varying as R3 , will tend to be more important if only because it decays so much more slowly than the dipole–induced dipole interaction. ′

10.2 From Eq. 10.17, the interaction between a multipole of 2k charges and another of 2k charges has ′ a potential energy proportional to R−k−k −1 . A quadrupole is formed by an arrangement of 4 charges, for which k = 2, so the interaction between two quadrupoles will vary in proportion to R−2−2−1 = R−5 . 7

10.3 The dipole moment of HI is 0.45 D and the polarizability of N2 is 0.20 Å . Plugging these values and the distance of 3.90 Å into Eq. 10.23 yields the following: 4µ2A α (4πǫ0 )R6 4(0.45 D)2 (3.3356 · 10−30 C m/D)2 (0.20 · 10−30 m3 ) =− (1.113 · 10−10 C2 J−1 m−1 )(3.90 · 10−10 m)6

u2−2∗ (R) = −

= −4.60 · 10−24 J = −0.00277 kJ mol−1.

This interaction energy is quite weak, partly because the N2 polarizability is so small. 10.4 We can expect dispersion to be most important, because the bromine atoms are so large (and consequently have high polarizabilities). The molecule should also be quite polar, given the electronegativity of the Br atom, and so dipole–dipole interactions will also run high. The dipole–induced dipole attraction will also be present, but is usually weaker than the direct dipole–dipole interaction: b<a<c. 10.5 The well should extend from 3.61 Å to 5.42 Å and be 190 K (132 cm−1 ) deep, as in the following graph.

10.6 The degrees of freedom that couple most strongly are usually those that have the most similar energy spacing. In this case, there is a transfer of energy away from vibrations in molecule A. Rotational energy spacings tend to be at least two orders of magnitude smaller than the vibrational spacings, and electronic transitions are typically an order of magnitude greater. Therefore, the likeliest place for vibrational energy to go is into other vibrations. In this case, if A has lost vibrational energy, it is most likely to have gone into vibrational excitation of B. Chapter 11 11.1 The dispersion force binds CO2 molecules together in a cluster, because there is no monopole (ionic charge) or dipole moment to bind them. As the cluster adds more units, the average binding energy per molecule will generally decrease when the cluster is very small. For small clusters, removing one unit reduces the overall binding significantly. But the effect depends on the cluster and the cluster size. For larger clusters, the loss of one unit has little effect on the remaining cluster. We expect the molecules at the surface of the cluster to be the most weakly bound, because they are not completely surrounded by the stabilizing neighbors. As the cluster size increases, the ratio of the volume (proportional to the cluster size N ) to the surface area (proportional to N 2/3 ) steadily increases, assuming the cluster shape remains roughly spherical, so on average for large clusters the binding energy may increase slightly with N . 11.2 The weak bonding interactions will be modeled by the van der Waals term (which includes dispersion forces) and the electrostatic potential energy term (which accounts for charge-charge interactions across distances longer than the typical chemical bond): UvdW and Uelectrostatic . 8

Chapter 12 12.1 The pair correlation function for this system should resemble the curves shown for water in Figs. 12.3g–i. The oscillations should become smoother as we approach the boiling point. low temp

near boiling point

12.2 Yes. Molecules of the two substances will attract one another, principally through dispersion. Chapter 13 13.1 The unit cell has two square faces opposite one another, and the rest of the faces are equivalent rectangles. There is one Ĉ4 principal rotation axis, a horizontal mirror plane, and four vertical mirror planes. The crystallographic point group is therefore D4h . 13.2 No. The reason is that no charge separation can exist in a regular monatomic crystal, and polarity is one of the requirements for piezoelectricity.

Chapter Problems Notes on Maple and Mathematica commands Several solutions in this manual have added notes on how to set up expressions under the symbolic mathematics programs Maple (Waterloo Maple, Inc.) and Mathematica (Wolfram Research) to assist in solving the problem. Please note, however, that a symbolic math program is useful only after we have overcome the conceptual challenges of the problem. Where these programs have their greatest use is when we need to solve simultaneous or transcendental equations, or obtain numerical values or algebraic expressions for integrals. We still have to know how to set up an integral, and we have to decide where and how to apply approximations (which turns out to be more important than one might think). Once we have taken these steps, the program may be able to take over. Even then, however, the math in the majority of our problems is limited to relatively straightforward algebraic manipulations, however complex the concepts may be. For most of the cases where the Maple and Mathematica syntax is not shown, little would differ from the written mathematics already appearing in the solution. This is largely the goal of symbolic math programs in the first place: to accept input in the form that one would write the problem on paper. The following problems in this volume give specific commands for use with Maple or Mathematica: A.10 A.18 3.29 3.31

1.9 1.20 2.3 2.4 2.10 2.35 3.47 4.26 9.19

Chapter A A.1 This problem uses a common manipulation, one of the features of logarithms that makes them so useful: pKa = − log10 Ka = − log10 e−∆G/(RT ) ∆G =− − log10 e RT ∆G = (0.434). RT

log xa = a log x

This shows, if you don’t mind us getting ahead of ourselves a little, that the pKa is directly proportional to the free energy of dissociation, ∆G, and inversely proportional to the temperature, T . A.2 The idea here is that, even if we think at first we have no idea what the number ought to be, a closer look at the available choices makes it clear that we can spot some potentially ridiculous answers: a. 2 · 1010 m s−1 is faster than the speed of light. 11

b. 2 · 105 m s−1 has no obvious objections. c. 2 m s−1 is the speed of a slow walk, and would imply, for example, that you could send an e-mail message over a cable connection to a friend half a mile away, and then run the half-mile to arrive and deliver the message in person before the e-mail finishes traveling through the wires. When we have calculations that toss around factors of 10−34 , for one example, this is a significant skill. The correct answer is 2 · 105 m s−1 . 3

A.3 The volume is roughly 125 Å , which we can show is not big enough to hold more than about 15 atoms. Chemical bonds, formed between overlapping atoms, are roughly 1 Å long, and so typical atomic 3 diameters are roughly 2 Å or more, and occupy a volume on the order of (2 Å)3 = 8 Å . A volume of 3 125 Å , therefore, cannot hold more than about 125/8 = 15.6 atoms. Among the choices, the only reasonable value is 8. A.4

a. Chemical bond lengths in molecules are always in the range 0.6–4.0 Å, or 0.6 · 10−10 to 4.0 · 10−10 m. 25 · 10−8 m is much too large for a bond length. no b. Six carbon atoms have a mass of 6 · 12 = 72 amu. With the added mass of a few hydrogen atoms at 1 amu each, 78 amu is a reasonable value. yes

A.5

a. The derivatives d[A] and dt have the same units as the parameters [A] and t, respectively. Both sides of the equation should therefore have units of mol L−1 s−1 . That means that k needs to provide the units of s−1 and cancel one factor of concentration units on the righthand side. k has units of L s−1 mol−1 . b. The argument of the exponential function must be unitless, so kB must cancel units of energy (J) in the numerator and temperature (K) in the denominator. The correct units are J K−1 . c. The units all cancel, and Keq is unitless. d. Squaring both sides of the equation, we can solve for k: µω 2 = k. k must therefore have units of kg s−2 . 2

A.6 There are two factors on the lefthand side, (2x+ 1)2 and e−ax . For the product to be zero, at least one of these factors must be zero. If (2x + 1) = 0, then All three are valid solutions.

x = − 12 . If e−ax = 0, then

x→ ± ∞.

A.7 In general, for any complex number (a + ib), the complex conjugate is (a + b)∗ = a − ib. We look for the imaginary component and and invert its sign: a. x − iy : a = x b = −y, x + iy. b. ix2 y 2 : a = 0 b = x2 y 2 ,

−ix2 y 2 .

c. xy(x + iy + z) : a = x2 y + xyz d. a = x/z

b = y/z,

b = xy 2 , x2 y + xyz − ixy 2 ,

xy(x − iy + z).

(x − iy)/z .

e. eix = 1 + ix − x2 − ix3 + x4 + ix5 − . . . a = 1 − x2 + x4 − . . .

b = x − x3 + x5 − . . .

a − ib = 1 − ix − x2 + ix3 + x4 − ix5 − . . . = e−ix . 12

f. 54.3: a = 54.3 b = 0, 54.3. A.8 This problem tests a few algebraic operations involving vectors, particularly useful to know when we look at angular momentum and (often related) magnetic field effects. √ ~ = 02 + 22 + 12 = a. The length of a vector is calculated using the Pythagorean theorem: |C| √ 5. ~ +B ~ = (1 + 1, 0 + 0, 0 + 1) = (2, 0, 1). b. We add vectors one coordinate at a time: A c. The dot product of two vectors multiplies the values for each coordinate of the two vectors and ~ · B ~ = (1 · 1) + (0 · 0) + (0 · 1) = 1. sums the results: A ~ · C ~ = (1 · 0) + (0 · 2) + (0 · 1) = 0. d. In the case of perpendicular vectors, this gives us zero: A e. The cross product involves a little more work, and yields a new vector, perpendicular to the two ~ × B ~ = (0 · 1 − 0 · 0, 0 · 1 − 1 · 1, 1 · 0 − 0 · 1) = (0, −1, 0). original vectors: A A.9 If we accept that the Taylor series expansion is exact if we take it to infinite order, then the Euler formula can be proven by the expansions of ex (Eq. A.25), sin x (Eq. A.26), and cos x (Eq. A.27): eix =

∞ X 1 (ix)n n! n=0

1 i 5 i x − ... = 1 + ix − 12 x2 − x3 + x4 + 6 24 120 1 1 5 1 = (1 − 12 x2 + x4 − . . .) + i(x − x3 + x − . . .) 24 6 120 = cos x + i sin x . This equation is of practical importance to us, and is famous among mathematicians for tying together three fundamental mathematical values—π, i, and e—in one equation: eiπ = 1. A.10

• Maple: We can have Maple solve the equation P − Va2 (Vm − b) m

directly using the solve command. After checking that all of the units are indeed consistent, enter the Maple command solve((1.000-(3.716/Vˆ2))*(V-0.0408)/(0.083145*298.15)=1,V); The resulting solution, 24.98, is in the same units as b, so our final value to three significant digits is 25.0 L mol−1 . • Mathematica: In Mathematica we use the Solve command to find the value of a variable in an algebraic equation: Solve[(1.000 - (3.716/Vˆ2))*(V - 0.0408)/(0.083145*298.15) == 1, V] Note that we need to use a double equals sign in the command. This command gives 24.979, which rounds to 25.0 L mol−1 . 13

• Successive approximation: There are several ways to solve this, corresponding to different forms of the equation that leaves Vm on one side. One way to set up the equation quickly is to recognize that (Vm − b) will vary rapidly compared to P − (a/Vm2 ), so we can isolate Vm as follows: P − Va2 (Vm − b) m =1 RT a P − 2 (Vm − b) = RT Vm RT Vm − b = P − Va2 m

Vm =

P − Va2

+ b.

Substituting in the values for P , a, b, R, and T (making sure that the units are all compatible), we can reduce the equation to the following: Vm ( L mol−1 ) =

24.790 (0.083145)(298.15) + 0.0408 = + 0.0408. 1 − 3.716 1 − 3.716 V2 V2 m

Guessing an initial value of 1 L mol

−1

yields the following series of approximations: 24.790 + 0.0408 = −9.0864 Vm = 1 + 3.716 12 24.790 Vm = 3.716 + 0.0408 = 25.999 1 + 5.330 2 24.790 Vm = 3.716 + 0.0408 = 24.967 1 + 22.099 2 24.790 Vm = 3.716 + 0.0408 = 24.979 1 + 24.796 2 24.790 Vm = 3.716 + 0.0408 = 24.979. 1 + 24.834 2

The series has converged to the three significant digits requested. The final value for Vm is 25.0 L mol−1 . A.11 Here we apply the rules of differentiation summarized in Table A.3. a. f (x) = (x + 1)1/2 df = 12 (x + 1)−1/2 . dx b. f (x) = [x/(x + 1)]1/2 −1/2 x df 1 dx 1 d 1 = 2 +x dx x+1 x + 1 dx dx x + 1 −1/2 x 1 x . − = 21 x+1 x + 1 (x + 1)2 14