Chem 1211 Take H AH and Bond Enthalpes (S pts) values given in the back for the

Question

Chem 1211 Take H AH and Bond Enthalpes (S pts) values given in the back for the combustion propane by using, st the A of the book and then, by using the bond enthalpies in table 8.4. Explain why the two values are different Table 8.4 Average Bond Enthalpies (k/mol) Single Bonds C-H 413 C-C348 C-N 293 C-O 358 N-H 391 N-N 163 N-O 201 N-F272 N CI 200 N-Br243 O H 463 0-0 46 O-E190 o-C 203 0-1 234 15y Cl-CI 485 C-C 328 C-Br276 C-1 C-S H-H 436 H-F 567 H-C 31 HBr366 H-299 S-H 339 S-F 327 S C 253 S-Br218 S-S266 B-Br 19 240 259 Si H 323 Si-Si226 Si-C 301 Si-o 368 Si-C464 Multiple Bonds C 614 839 CN615 1-1 1 N=N NN N=O 418 941 607 S-S 418 0 799 , re CO 1072

Explanation / Answer

a)Via enthalpy of formation

HRXN = Hproducts - HReactants

note --> the H2O(g) value you added, -136.10 is incorrect, the Hf-H2O(g) = -241.8 kJ/mol

HRXN = (3*-393.5 + 4*-241.8) - (-103.85+0)

HRXN = -2043.85

a) Via bond enmthalpy

HRXN = Hformed - Hbroken

bonds fromed:

6x(C=O) = 6*799

8x(O-H) = 8*463

Hformed = 6*799+ 8*46 = 5162

bonds broken

2x(C-C)= 2*348

8x(C-H) = 8*413

5x(O=O) = 5*495

Hbroken = 2*348+ 8*413+ 5*495 = 6475

HRXN = 5162-6475

HRXN = -1313 kJ/mol

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