65.100 A and V
Christmas Exam 1998
Answers

Part A.

1. The oxidation number of the Mn atom in MnO₄^- is +7, which is very high. Thus, the Mn atom will be easily reduced, i.e. it is a strong oxidant.

2. A 4s orbital has three (3) nodes.

3. The effective nuclear charge decreases going down a group because electrons are being added to higher and higher principle quantum shells, and so are farther from the nucleus. In other words, the electrons in the inner shells shield those in the outer shells from the nucleus.

4. Example 1: paramagnetism: MO theory predicts paramagnetism in species containing unpaired electrons
Example 2: Correlation of bond length with ionization
Example 3: Correlation of bond energy with ionization
Example 4: bond order (predicts existence or not of a species)

5. K_eq = exp(-DG^o/RT). Thus, if DG^o is small and positive, K_eq is just less than 1, indicating that at equilibrium, there will be more reactants than products.

6. The reaction will not be spontaneous at any temperature.

7. At high pressure, the molecules are close together, and so V<<V_container. Also, molecules interact with one another, resulting in a lower pressure than that predicted by the ideal gas law.

8. CH₄ has more atoms, hence more vibrational modes than CO₂. Thus, CH₄ can absorb IR radiation at a greater number of frequencies that CO₂.

9. One of the bonds is a sigma (s) bond, resulting from the overlap of sp hybrids on each carbon. The other two bonds are pi (p) bonds, formed by sideways overlap of parallel p-orbitals on the carbon atoms.

10. The system can exchange heat, and can exchange work with the system.

11. The system will shift in the exothermic direction (right!)

12. Q = p²_NOCl/(p²_NO p_Cl2) = 122/(.1²(1.0)) = 14400. We see that Q>Kp. Thus, the system must shift to the left so as to decrease Q until Q=Kp.

Part B

1. Fe⁺³ + SCN^- --> FeSCN⁺²

 

Thus, at equilibrium, we have:

[FeSCN⁺²] / ([Fe⁺³] [SCN^-]) = 1100
x/ ((0.02 - x)(0.10 - x)) = 1100
x = 1100 (.002 - .02x - .1x + x²)
1100 x² - 133 x + 2.2 = 0

Using the quadratic formula, x = 0.101 or 0.0198

We reject x=0.101, since this would lead to negative concentrations. Thus, x = 0.0198

Thus, [Fe⁺³] = 0.0200 - x = 0.0002 M
[SCN^-] = 0.1000 - x = 0.0802 M
[FeSCN⁺²] = x = 0.0198 M

2. DH^o = DH^o_f(Cu₂O_(s)) - 2DH^o_f(Cu_(s)) — ½ DH^o_f(O_2(g))
= -170 - 0 - 0
= -170 kJ mol^-1

DS^o = S^o(Cu₂O_(s)) - 2S^o(Cu_(s)) - ½ S^o(O_2(g))
= 93 - 2(33) - ½ (205)
= -75.5 J K^-1 mol^-1

DG^o = DH^o - TDS^o
= -170000 - 298(-75.5)
= =147500 J mol^-1

K_eq = exp (-DG^o/(RT)) = exp (+147500 J mol^-1 / (8.314 J K^-1mol^-1(298 K))
= e^59.5
= 6.9 x 10²⁵

3. (a) 1/l = 0.01097 nm^-1 [1/3² — 1/4²]
solving, l = 1875 nm

(b) 1/l = 0.01097 nm^-1 [1/3² — 1/¥²]
solving, l = 820 nm

E = hn = hc/l = 6.63 x 10^-34 J s (3.00 x 10⁸ m s^-1) / (820 x 10^-9 m)
= 2.42 x 10^-19 J (per photon)
x 6.02 x 10²³ mol^-1 = 146000 J / mol photons
= 146 kJ/mol photons

(c) Total energy absorbed = 250 g x 4.18 J g^{-1 o}C^-1 x (100-20)^oC = 83600 J

Energy per mole photons = N_avhc/l
= 6.02 x 10²³ mol^-1 x 6.63 x 10^-34 J s (3.00 x 10⁸ m s^-1) / (0.0155 m)
= 7.7 J/mol photons

Thus, moles photons required = 83600 J / (7.7 J/mol photons) = 10818 mol photons (= 6.5 x 10²⁷ photons)

4. (a) N₂O contains 16 valence electrons. Using the rules for drawing Lewis structures:

::N=N=O::

:NºN-O:::

:::N-NºO:

We see that the third structure has a formal charge of -2 on the N atom. Thus, the other structures two are far more likely.

(b) (i) F is more electronegative than C, thus each dipole points from the C to the F

(ii) Using VSEPR, we predict CF₄ to be of the form AX₄, which is tetrahedral.

(iii) The molecule is symmetrical. Thus, all four dipolar bonds cancel out leaving no net dipole!

5. (a) S₂O₈^-2 --> SO₄^-2
S₂O₈^-2 --> 2 SO₄^-2
S₂O₈^-2 + 2 e^- --> 2 SO₄^-2

Cr⁺³ --> Cr₂O₇^-2
2 Cr⁺³ --> Cr₂O₇^-2
2 Cr⁺³ + 7 H₂O --> Cr₂O₇^-2
2 Cr⁺³ + 7 H₂O --> Cr₂O₇^-2 + 14 H⁺
2 Cr⁺³ + 7 H₂O --> Cr₂O₇^-2 + 14 H⁺ + 6 e^-

Take three times the reduction reaction, and add it to the oxidation reaction:

3 S₂O₈^-2 + 6 e^- --> 6 SO₄^-2
2 Cr⁺³ + 7 H₂O --> Cr₂O₇^-2 + 14 H⁺ + 6 e^{-
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2 Cr⁺³ + 3 S₂O₈^-2 + 7 H₂O --> Cr₂O₇^-2 + 6 SO₄^-2 + 14 H⁺

(b) (0.001 Cr⁺³ / 52 g mol^-1) x (14 mol H⁺ / 2 mol Cr⁺³) = 1.35 x 10^-4 mol H⁺
[H⁺] = 1.35 x 10^-4 mol / 1.000 L = 1.35 x 10^-4 mol/L
pH = -log₁₀[H⁺] = -log₁₀(1.35 x 10^-4) = 3.87

6. (a) P = nRT/V = 0.5 mol(0.082 L atm K^-1 mol^-1)(298 K)/ 1.00 L = 12.2 atm

(b) P = (nRT)/(V-nb) - a(n²/V²)

= 0.5(0.082)(298)/(1.00 - 0.5(0.0391)) - 1.39(0.52/12)

= 12.11 atm

(c) The answers are different because the ideal gas law does not work well at high pressures

(d) P = nRT/V = 0.5 mol(0.082 L atm K^-1 mol^-1)(298 K)/ 10.00 L = 1.22 atm

(e) P = (nRT)/(V-nb) - a(n²/V²)
= 0.5(0.082)(298)/(10.00 - 0.5(0.0391)) - 1.39(0.52/102)
= 1.22 atm

(f) The values are very close because the ideal gas law works well at such low pressures.