Using table 8.4, approximate the heat of combustion (kJ/ mol hydrocarbon) for th

Question

Using table 8.4, approximate the heat of combustion (kJ/ mol hydrocarbon) for the complete combustion of methane (CH4) and octane (C8H18). Please show all work thank you!

TABLE 8.4 Average Bond Enthalpies (k/mol) Single Bonds C-H 413 N-H 391 O-H 463 C-C 348 N-N 163 O-O 146 N-O 201 C-N 293 O-F 190 O-Cl 203 C-O 358 272 N-Cl 200 C-F 485 O-I 234 C-Cl 328 N- Br 243 C-Br 276 339 C I 240 H-H 436 S-F 327 S-C1 253 C-S H-F 567 259 H-Cl 431 Br 218 Si-H 323 H-Br 366 S-S 266 H-I 299 226 Si-C 301 Si-O 368 Cl 464 Multiple Bonds C C 614 N-N 418 495 N N 941 839 N o 607 C N 615 S O 523 CEN 891 S S 418 C O 799 C O 1072 F-F 155 Cl-F 253 Cl Cl 242 Br-F 237 Br Cl 218 Br-Br 193 I-C1 208 I-Br 175 I I 151

Explanation / Answer

The combustion reaction is given of methane is given as

CH4 + 2O2 ---> CO2 + 2H2O

so the combustion of 1 mole of methane requires 4 C-H bonds and 2 O=O bonds to be broken and 2 C=O and 4 O=H bonds to be formed

bond enthapies for 4 C-H bonds will be 413 x 4 = 1652 KJ

for 2 O=O, 2 x 495 = 990 kJ

for 2 C=O, 2 x 799 = 1598 KJ

for 2H2O, 4 x 463 = 1852 KJ

so the reaction enthalpy will be ( 1852 + 1598) - ( 1652 + 990) = 808 KJ / mol

so the heat of combustion for methane is -808 kJ / mol

similarly for octane

the reaction is given as,

C8H18 + 25/2 O2 ---> 8CO2 + 9H2O

7 C-C , 18 C-H and 12.5 O=O bonds have to be broken and 16 C=O and 18 O-H bonds have to be formed

so enthalpy = [ ( 16 x 799 ) + ( 18 x463) ] - [ (7x 348) + (18 x 413) + ( 12.5 x 495)]

= 5061 KJ/ mol

so the heat of combustion of octane is -5061 KJ / mol

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