Chemistry 65.100 A and V
Mid-Year Examination, December, 2000
ANSWERS

Part A. (5 marks each).

1. Atoms can lower their electronic energy by having half-filled or completely filled valence orbitals. 4s² is filled, whereas 3d⁴ is 4/10 filled. Thus, by shifting an electron from the 4s to the 3d orbital, the configuration would be 4s¹3d⁵, which is two half-filled orbitals, and which is energetically more stable.

2. Since there are three atoms bound to each C atom (two hydrogens and the other carbon), then each carbon must be using sp² hybrid orbitals for bonding to these other atoms. This leaves a p-orbital unchanged on each carbon atom. Each carbon atom thus uses an sp² hybrid to form the (sigma) bond to the other carbon, two sp² hybrids to bond with the two hydrogens, and a p-orbital to form the (pi) bond with the other carbon.

3. The balanced reaction is C₂H_6(g) + 7/2 O_2(g) ¾ 2 CO_2(g) + 3 H₂O_(g). The number of moles of gas thus increases from 4.5 to 5. Since Dn > 0, we would predict that DS^o is positive for this reaction.

4. Elements always have DH^o_f = 0 because the definition of DH^o_f is the enthalpy released or absorbed when the species is formed from the elements (with everything in its standard state). Forming an element from the elements obviously has no enthalpy change associated with it.

5. An endothermic reaction be spontaneous as long as the value of DS is sufficiently positive that DG is negative.

6. A single Cl atom can destroy many ozone molecules in the stratosphere because it is involved in a cyclic destruction process according to one of the following two reaction schemes:

Cl + O₃ ® ClO + O₂
ClO + O ® Cl + O₂

OR

2 Cl + 2 O₃ ® 2 O₂ + 2 ClO
2 ClO ® Cl₂O₂
Cl₂O₂ + hn ® 2 Cl + O₂

7. Starting with NO_(g), ozone is produced in photochemical smog by the following reactions:

NO + 1/2 O₂ ® NO₂
NO₂ + hn ® NO + O
O + O₂ ® O₃

8. For the reaction 2 POCl_3(g) ¾ 2 PCl_3(g) + O_2(g), the equilibrium will shift to the right if the pressure is lowered. According to Le Chatelier's principle, if the pressure is lowered, the equilibrium will shift so as to increase the pressure again, i.e. to the side of the reaction having more moles of gas.

9. For the reaction CH_4(g) + 3 O_2(g) ¾ CO_2(g) + 2 H₂O_(g), the equilibrium shift to the right if the temperature is decreased. First, you must recognize this as an exothermic reaction. Then according to le Chatelier's principle, decreasing the temperature shifts the equilibrium towards the exothermic direction.

10. Dalton's Law of partial pressures simply states that the total pressure in a mixture of gases is equal to the sum of the partial pressures of the individual gases in the mixture.

11. This demonstrates that equilibrium is a dynamic process, since for a large crystal of salt to appear, the NaCl must have been in solution at some point, followed by crystallization. However, the solution is initially saturated, meaning it can accommodate no more NaCl. The only interpretation is that some NaCl crystallized, followed by some dissolution, and so on. In other words, the equilibrium between solid and aqueous phase NaCl is a dynamic process - both the forward and reverse reactions are always occurring.

12. The SbCl₅^-2 does not likely have an exactly square pyramidal shape, since there is a lone pair on the Sb atom in addition to the five bonding pairs. This lone pair repels the bonding pairs more than they repel one another, resulting in bond angles different from the theoretical 90^o.

1. DH^o_f for magnesium fluoride is DH^o for the reaction:

Mg_(s) + F_2(g) ® MgF_2(s)

Mg_(s) ® Mg_(g) +150 kJ
Mg_(g) ® Mg⁺_(g) +735 kJ
Mg⁺_(g) ® Mg⁺²_(g) +1445 kJ
F_2(g) ® 2 F_(g) +154 kJ
2 F_(g) + 2 e^- ® 2 F^-_(g) 2(-328) = -656 kJ
Mg⁺²_(g) + 2 F^-_(g) ® MgF_2(s) -3916 kJ
_____________________________________________________
Mg_(s) + F_2(g) ® MgF_2(s) -2,088 kJ

2. Labelling the oxygen atoms as shown in structure (a), formal charges were worked out as shown in the tables below. Structure (d) has the lowest formal charges, and therefore is the most likely.

3.

At equilibrium at 700 K, [CO_2(g)][H_2(g)] / ([CO_(g)][H₂O_(g)]) = K_c = 5.10

Thus, (1.00 + x)² / (1.00 - x)² = 5.10

Taking the square root of both sides,
(1.00 + x)/(1.00 - x) = 5.10^1/2 = 2.26
1.00 + x = 2.26 - 2.26 x
3.26 x = 1.26
x = 0.39

Thus at equilibrium, [CO_(g)] = [H₂O_(g)] = 1.00 - x = 0.61 mol L^-1
and [CO_2(g)] = [H_2(g)] = 1.00 + x = 1.39 mol L^-1
(To check the answer, (1.00 + x)² / (1.00 - x)² = 1.39²/0.61² = 5.1)

4.(a) To find DG^o, we need values for DH^o and DS^o:

DH^o = DH_f^o(CaO_(s)) + DH_f^o(CO_2(g)) - DH_f^o(CaCO_3(s))
= -635.1 + (-393.5) - (-1206.9) = +178.3 kJ mol^-1
DS^o = S^o(CaO_(s)) + S^o(CO_2(g)) - S^o(CaCO_3(s))
= 38.2 + 213.7 - 92.9 = +159 J K^-1 mol^-1

Making the assumption that DH^o = DH_{1000 C}, and that DS^o = DS_{1000 C}, then

DG₁₀₀₀ = DH^o - TDS^o = +178,300 J mol^-1 - 1273 K (159 J K^-1 mol^-1) = -24,100 J mol^-1 = -24.1 kJ mol^-1
(b) Since DG < 0, the reaction is spontaneous at 1000^oC.
(c) K_p = exp(-DG^o/RT) = exp(+24100 J mol^-1/(8.314 J K^-1 mol^-1 x 1273 K)) = 9.75
(d) For this reaction, K_p = p_CO2, since the other two species are solids. Thus, p_CO2 = 9.75 atm.

5. Using Graham's law, we first find the molecular weight of the unknown substance:
rate_?? / rate_Ar = [MW_Ar/MW_??]^1/2
But in this question, you are given the time it takes for diffusion, which is inversely related to the rate. Thus,
time_Ar / time_?? = [MW_Ar/MW_??]^1/2
rearranging, MW_?? = MW_Ar / [time_Ar/time_??]² = 39.9 g mol^-1 / [4.37/7.73]² = 124.8 g mol^-1

Now assume a 100 g sample:
moles C = 38.4 g / 12.011 g mol^-1 = 3.19 mol C
moles H = 4.82 g / 1.01 g mol^-1 = 4.77 mol C
moles Cl = 56.8 g / 35.5 g mol^-1 = 1.60 mol Cl

Divide each by the lowest to obtain the number relative number of moles of each (i.e. the empirical formula):

moles C / moles Cl = 3.19 / 1.6 = 2
moles H / moles Cl = 4.77 / 1.6 = 3
moles Cl = 1
Thus, the empirical formula is C₂H₃Cl

The molecular weight of this compound is 2x12 + 3 + 35.5 = 62.5, which is 1/2 of the molecular weight of the compound calculated by Graham's law. Thus, multiply the empirical formula by 2 to obtain the molecular formula: C₄H₆Cl₂, which has a molecular weight of 125 g mol^-1.

6(a) P = nRT/V = 2 mol (0.082 L atm K^-1 mol^-1) (298 K) / 1 L = 48.9 atm

(b) P = nRT/(V - nb) - a(n/V)² = 2 mol(0.082 L atm K^-1 mol^-1)(298 K)/(1 L - 2 mol(0.0238 L mol^-1)) - 0.0346 L² atm mol^-1 (2 mol / 1 L)² = 51.3 atm - 0.14 atm = 51.2 atm

(c) P = nRT/(V - nb) - a(n/V)² = 2 mol(0.082 L atm K^-1 mol^-1)(298 K)/(1 L - 2 mol(0.0568 L mol^-1)) - 6.865 L² atm mol^-1 (2 mol / 1 L)² = 55.1 atm - 27.5 atm = 27.7 atm.

(d) The van der Waals equation predicts a higher pressure for He_(g) than the ideal gas law because the He atoms actually take up some space in the container, leading to more frequent collisions and thus a higher pressure. Note that the "a" term, accounting for interatomic interactions, is very small for He and so has a negligible effect on the calculated pressure.

(e) The pressure of the SO_2(g) is less than that He_(g) because SO₂ is a polar molecule, leading to strong intermolecular forces that effectively pull the SO₂ molecules together slightly in the gas phase, resulting in fewer collisions, hence a lower pressure than in He_(g).

7. (a) The bond order of O₂ is (8 - 4)/2 = 2
(b) Oxygen is paramagnetic because of the two unpaired electrons in the p*_2p MO.
(c) Oxygen also diamagnetic because there are several paired electrons in the MOs.
(d) The two p_2p bonding molecular orbitals are formed by sideways overlap of two pairs of p-orbitals, i.e. one pair from each O atom. The s_2p MO is formed by the end-on overlap of the other two p-orbitals, one from each O atm. Because the three atomic p-orbitals are orthogonal to one another, there can be only one end-on interaction.
(e) O₂⁺² would be formed by removal of the two electrons from the p*_2p MO. Then the bond order would be (8 - 2)/2 = 3, which is greater than zero. Thus, O₂⁺² may exist.