Chemistry 65.100 A and V
April 16, 1998
Final Exam Answers

Part A.

1. DG⁰ is the free energy change under standard conditions (25^oC, 1 atm, all reactants and products in their standard states). R is the gas constant, T is the absolute temperature and K_eq is the equilibrium constant at temperature T.

2. To answer this, you need to draw the Lewis structures of each, and determine whether there are any lone pairs on the central atom (tetrahedral structures will have none).

CH₄ has 4 + 4(1) = 8 valence electrons, all of which are used as bonding electrons, and thus there are no lone pairs, and thus CH₄ is tetrahedral.
SH₄ has 6 + 4(1) = 10 valence electrons, and thus has one lone pair on the S atom, and thus is not tetrahedral.
XeF₄ has 8 + 4(7) = 36 valence electrons. Eight are around each F atom, leaving two lone pairs on the Xe. The structure therefore is not tetrahedral.

3. The bonding in graphite consists of sp² hybrid obritals in the plane, holding each C atom to three others. The bonding between planes is via overlap of the extra p-orbital remaining after sp² hybridization. The bonding in diamond is all overlap of sp³ hybrids. This bonding occurs in all three dimensions, and the structure is thus much stronger and harder.

4. Ar_(g) has only dispersion forces between the atoms, whereas HCl_(g) molecules will also have stronger dipole-dipole forces holding the molecules to one another. Thus, Ar will more closely obey the ideal gas law, which assumes no interactions between neighbouring species in the gas phase.

5. These substances dissolve because the entropy of solution is positive enough that the free energy of solution becomes negative.

6. A catalyst changes the mechanism of the reaction to one having a lower activation energy, hence a faster rate.

7. The configuration [Ar] 4s¹ 3d¹⁰ is more stable because it has one half-filled orbital and one filled orbital.

8. Cathodic protection works by bringing the metal to be protected into electrical contact with a metal that is more easily oxidized. This second metal oxidizes preferentially, releasing electrons which are at a high enough potential to prevent electrons from being released by the protected metal. To cathodically protect iron, the metal in question must oxidize more easily than iron. Referring to the reduction potentials table, we see that Mg are below Fe, and so will protect Fe. Cu and Pb are above Fe, and so can not protect Fe.

Part B.

1a. Oxidation reaction: H_2(g) à 2 H⁺_{(aq, xM)} + 2 e^-
Reduction reaction: 2 H⁺_{(aq, 1 M)} +2 e^- à H_2(g)
Overall reaction: 2 H⁺_{(aq, 1 M)} à 2 H⁺_{(aq, xM)}

b. Under standard state conditions, the oxidation and reduction reactions cancel out, since [H_(aq)] would always be 1 M at standard state conditions. Thus, E^o = 0.000 V. Under non-standard conditions (i.e. when x ¹ 1 M), we use the Nernst equation to find the cell potential:

E = E^o – [RT/(nF)] ln (Q)

here, E = 0.108 V, E^o = 0.000 V, R = 8.314 J K^-1 mol^-1, T = 50^oC = 323 K, n=2, F=96500 C mol^-1

For the overall reaction as written above, Q = x²/1²

Thus, 0.108 = 0 – 8.314(323)/(2(96500)) ln(x²)
recognizing that ln (x²) = 2 ln(x), we have:
0.108 = -0.0278 ln(x)
solving, x = 0.021 M

2a. PV = nRT, or, P = nRT/V
here, n = 0.580 kg/0.0709 kg mol^-1 = 8.18 mol
R = 0.082 L atm K^-1 mol^-1
T = 200^oC = 473 K
V = 15 L

Thus, P = 8.18(0.082)(473)/15.0 = 21.2 atm

b. The vdW equation is P = (nRT)/(V-nb) –a(n/V)²
here, a=6.49 atm L² mol^-2 and b=0.0562 L mol^-1
Thus, P = (8.18)(0.082)(473)/(15.0-8.18(0.0562)) – 6.49(8.18/15.0)² = 19.9 atm

c. The vbW pressure is lower because it takes into account the interactions among the Cl₂ molecules, which tend to lower the pressure.

3a. DH^o = DH_f^o(CO) + 2DH_f^o(H₂) – DH_f^o(CH₃OH)
= -110.5 + 2(0) – (-238.7) = 128.2 kJ mol^-1
DS^o = S^o(CO) + 2S^o(H₂) – S^o(CH₃OH)
= 197.6 + 2(130.6) – 127 = 331.8 J K^-1 mol^-1

DG^o = DH^o - TDS^o Note that the superscript o means "standard state", i.e. at 298 K.

Thus, DG^o = 128200 – 298(331.8) = +29,300 J mol^-1
Note that this value indicates that there will be very little products at equilibrium!

b. The equilibrium constant is found from:

K_eq = exp[-DG^o/(RT)]
= exp (-29300/(8.314(673))
= 0.00530

c.

At equilibrium, we thus have:

K_eq = [CO][H₂]²/[CH₃OH]

since the volume is the same for all three (all are in the same container), we can use the number of moles instead of the concentrations. Thus,

K_eq = 0.00530 = x(2x)²/(1-x)

Since K_eq is so small, we expect very little product. Thus 1-x » 1. Thus the expression simplifies to 0.0053 = 4x³. Solving, x = 0.11. (Note that assumption is not perfect, since x is as large as it is.)

Thus, at equilibrium, we will have 1-x = 0.89 mol CH₃OH, 0.11 mol CO, 0.22 H₂

4a. pH = pK_a + log₁₀{[base]/[acid]}
here, [base] = [HCOO^-] = 0.63 M, [acid] = [HCOOH] = 0.55 M
Thus, pH = 3.74 + log₁₀(.63/.55) = 3.80

b. The equilibrium reaction is: HCOOH + H₂O ¾ H₃O⁺ + HCOO^-

Before adding the acid, the pH was 3.80, i.e. [H₃O⁺] = 1.589 x 10^-4 M, and [HCOO^-] = 0.63 M. Adding 0.020 mol HNO₃ will initially cause the [H₃O⁺] to jump to 1.589 x 10^-4 + 0.020/1000 = 1.789 x 10^-4, but then the acid dissociation equilibrium will shift left, consuming some of the added acid, and maintaining the pH almost constant. Here, K_a = 10^-pKa = 10^-3.74 = 1.82 x 10^-4.

Thus, at equilibrium we have 1.82 x 10^-4 = (1.789 x 10^-4 – x)(0.63-x)/(0.55+x)
Assuming x is small (we would not have much of a buffer if this were not so), we can simplify the above to:

1.82 x 10^-4 = (1.789 x 10^-4 – x)(.63/.55)
solving, x = 2.01 x 10^-5
Thus, at equilibrium, we have[H₃O⁺] = 1.789 x 10^-4 – 2.01 x 10^-5 = 1.588 x 10^-4. Thus, pH = -log₁₀(1.588 x 10^-4) = -3.799 (i.e. a pretty good buffer)

5a. The unit cell edge length = l =383.3 pm. Since this is a face centred cubic unit cell, the diagonal of one face = 4r. Using pythagoras’ theorem, we have l² + l² = (4r)². Thus, r² = l²/8. Solving, r = 135.5 pm.

b. The FCC unit cell contains 4 atoms in total. Thus, its mass =
4(192.22 g mol^-1)/6.02 x 10²³ mol^-1 = 1.277 x 10^-21 g

The edge length = 383.3 pm = 3.833 x 10^-8 cm. Volume = l³ = (3.833 x 10^-8 cm)³ = 5.631 x 10^-23 cm³

Thus, density = m/V = 1.277 x 10^-21 g / 5.631 x 10^-23 cm³ = 22.7 g cm^-3

6a. t_1/2 = 32 min.
k = -ln(2)/t_1/2 = 0.693/32 min = 0.0217 min^-1

b. [PAN] = [PAN]₀ e^-kt
= 5.0 x 10¹⁴ molecules L^-1 e^{-0.0217 min-1(90 min)} = 7.12 x 10¹³ molecules L^-1

c. At 30^oC (303 K), k = 0.0217 min^-1
At 40^oC (313 K), k = 0.693/16 min = 0.0433 min^-1
Plotting ln(k) vs. 1/T gives a straight line having a slope = -E_a/R. The slope of the line is [ln(.0217) – ln(.0433)] / (1/303 – 1/313) = -6552 K
Thus, E_a = -R(slope) = -8.314 J K^-1 mol^-1 (-6552 K) = 54,500 J mol^-1 = 54.5 kJ mol^-1

7a. ³²₁₅P à ³²₁₆S + ⁰_-1e
b. k = -ln(2)/t_1/2 = 0.693/4.5day = 0.154 day^-1
c. [⁴⁷Ca] = [⁴⁷Ca]₀ e^-kt
= (1.0 mg) e^{-0.154 day-1(19.0 day)} = 0.0536 mg

8a i. p-amino methyl benzene
or, 1-amino-4-methyl benzene
or, p-amino toluene
or, p-methyl aniline

ii. 3-fluoro-3-hexene

iii. 5-bromopentanoic acid

iv diaminomethane

8b i. 1,2-diiodopropane

ii. o-dichlorobenzene and m- dichlorobenzene and p-dichlorobenzene

iii. diethylamine hydrochloride

iv methyl butanoate

v N-ethylpentamide

9. PCl₆^- valence electrons = 5+6(7)+1 = 48. Arranging all 48 around the Cl atoms leaves no lone pairs. The structure is thus of the form AX₆ and must be octahedral

SiCl₃^- valence electrons = 4 + 3(7)+1 = 26. This will leave 1 lone pair on the Si, and the structure is thus of the form AX₃E. This will be trigonal pyramidal (like ammonia)

NH₄⁺ valence electrons = 5 + 4(1)-1 = 8. This is AX₄, and is tetrahedral.

H₃PO₄. You must first recognize this as an oxo-acid, wehre the H atoms are bound to the O atoms. Around the P, there are thus only the four O atoms. Valence electrons = 5+4(6)+3(1) = 32. The structure around the P is AX₄, i.e. tetrahedral

10a. Bond order of B₂ = (4-2)/2 = 1. The molecule is thus stable.
b. To make the molecule unstable by adding electrons, we would have to add enough so that the number of bonding electrons = number of antibonding electrons. You would thus have to fill all of the orbitals shown, at which point you would have 8 bonding and 8 antibonding, i.e. you would have to add 10 electrons! You would only have to remove 2 to make it unstable, at which point there would be 2 bonding and 2 antibonding.
c. BO(B₂^-) = (5-2)/2 = 1.5
BO(B₂) = (4-2)/2 = 1
BO(B₂⁺) = (3-2)/2 = 0.5
The bond length is greater if the bond order is lower, since the bond is weaker. Thus, in increasing order of bond length are B₂^-, B₂ and B₂⁺

d. All three are paramagnetic, since all three would have at least one unpaired electron.