Question 8.12: Phosgene, a substance used in poisonous gas warfare during W...
Phosgene, a substance used in poisonous gas warfare during World War I, is so named because it was first prepared by the action of sunlight on a mixture of carbon monoxide and chlorine gases. Its name comes from the Greek words phos (light) and genes (born of). Phosgene has the following elemental composition: 12.14% C, 16.17% O, and 71.69% Cl by mass. Its molar mass is 98.9 g/mol. (a) Determine the molecular formula of this compound. (b) Draw three Lewis structures for the molecule that satisfy the octet rule for each atom. (The Cl and O atoms bond to C.) (c) Using formal charges, determine which Lewis structure is the dominant one. (d) Using average bond enthalpies, estimate ΔH for the formation of gaseous phosgene from CO(g) and Cl_{2}(g).
(a) The empirical formula of phosgene can be determined from its elemental composition. Assuming 100 g of the compound and calculating the number of moles of C, O, and Cl in this sample, we have:
\left(12.14 \cancel{g C}\right)\left(\frac{1 mol C}{12.01 \cancel{g C}} \right)= 1.011 mol C
The ratio of the number of moles of each element, obtained by dividing each number of moles by the smallest quantity, indicates that there is one C and one O for each two Cl in the empirical formula, COCl_{2}.
\left(16.17 \cancel{g O}\right)\left(\frac{1 mol O}{16.00 \cancel{g O}} \right)= 1.011 mol O
The molar mass of the empirical formula is 12.01 + 16.00 + 2(35.45) = 98.91 g/mol, the same as the molar mass of the molecule. Thus, COCl_{2} is the molecular formula.
\left(71.69 \cancel{g Cl}\right)\left(\frac{1 mol Cl}{35.45 \cancel{g Cl}} \right)= 2.022 mol Cl
(b) Carbon has four valence electrons, oxygen has six, and chlorine has seven, giving 4 + 6 + 2(7) = 24 electrons for the Lewis structures. Drawing a Lewis structure with all single bonds does not give the central carbon atom an octet. Using multiple bonds, we find that three structures satisfy the octet rule:
(c) Calculating the formal charges on each atom gives: The first structure is expected to be the dominant one because it has the lowest formal charges on each atom. Indeed, the molecule is usually represented by this single Lewis structure.
d) Writing the chemical equation in terms of the Lewis structures of the molecules, we have:
:C≡O: + \overset{\cdot \cdot }{\underset{\cdot \cdot }{:Cl}}— \overset{\cdot \cdot }{\underset{\cdot \cdot }{Cl:}}→\overset{\cdot \cdot }{\underset{\cdot \cdot }{:Cl}}−\overset{\overset{\large{\small{:O:}}}{||} }{C} −\overset{\cdot \cdot }{\underset{\cdot \cdot }{Cl:}}
Thus, the reaction involves breaking a C≡O bond and a Cl—Cl bond and forming a C=O bond and two C—Cl bonds. Using bond enthalpies from Table 8.3,
| TABLE 8.3 Average Bond Enthalpies (kJ/mol) | |||||||
| Single Bonds | |||||||
| C—H | 413 | N—H | 391 | O—H | 463 | F—F | 155 |
| C—C | 348 | N—N | 163 | O—O | 146 | ||
| C—N | 293 | N—O | 201 | O—F | 190 | Cl—F | 253 |
| C—O | 358 | N—F | 272 | O—Cl | 203 | Cl—Cl | 242 |
| C—F | 485 | N—Cl | 200 | O—I | 234 | ||
| C—Cl | 328 | N—Br | 243 | Br—F | 237 | ||
| C—Br | 276 | S—H | 339 | Br—Cl | 218 | ||
| C—I | 240 | H—H | 436 | S—F | 327 | Br—Br | 193 |
| C—S | 259 | H—F | 567 | S—Cl | 253 | ||
| H—Cl | 431 | S—Br | 218 | I—Cl | 208 | ||
| Si—H | 323 | H—Br | 366 | S—S | 266 | I—Br | 175 |
| Si—Si | 226 | H—I | 299 | I—I | 151 | ||
| Si—C | 301 | ||||||
| Si—O | 368 | ||||||
| Si—Cl | 464 | ||||||
| Multiple Bonds | |||||||
| C=C | 614 | N=N | 418 | O=O | 495 | ||
| C≡C | 839 | N≡N | 941 | ||||
| C=N | 615 | N=O | 607 | S=O | 523 | ||
| C≡N | 891 | S=S | 418 | ||||
| C=O | 799 | ||||||
| C≡O | 1072 | ||||||
we have:
ΔH = [D(C≡O)+ D(Cl – Cl)] -[D(C=O) +2D(C—Cl)]
= [1072 kJ + 242 kJ] – [799 kJ + 2(328 kJ)] = -141 kJ
Notice that the reaction is exothermic. Nevertheless, energy is needed from sunlight or another source for the reaction to begin, as is the case for the combustion of H_{2} (g) and O_{2} (g) to form H_{2}O (g)