TABLE 9.4 Bond Energies (kJ/mol)* Single Bonds C-H 414 C-C 348 C-N 293 C-O 351 C

Question

TABLE 9.4 Bond Energies (kJ/mol)* Single Bonds C-H 414 C-C 348 C-N 293 C-O 351 C-F 439 C-CI 328 C-Br 276 C-1 238 C-S 259 N-H 889O H 463 FF 159 N-N 163OO 146 C F 253 N-O 201O-F 190 CCI 242 N-F 272 O-C203Br-F 237 N-CI 200 O-I234Br-Cl 218 N-Br 243 Br-Br 193 S-H 339 -C 298 H-H 436 S-F327 I-Br180 H-F 569 S-C 251 1-I H-CI 431 S-Br 218 H-Br 368 S-S 266 H-1 297 131 Si-H 293 Si-Si 226 Si-C 801 Si-O 368 Multiple Bonds C-C 611 CC 837 C=N 615 498 N-N 418 N N 946 C=() 799 CEO 07 S O 523 S=S 418 Te bd eegies for the diatomie molecules ca be measured directly; the other num- bers are average bond energies

Explanation / Answer

1)

Using Table 9.4, calculate an approximate enthalpy (in kJ) for the reaction of 2.09 g gaseous methanol (CH3OH) in excess molecular oxygen to form gaseous carbon dioxide and gaseous water. (Hint, remember to first write the balanced equation.) bond energies

m = 2.09 g of CH3OH

mol = mass/MW = 2.09/32 = 0.0653 mol of methanol

CH3OH + O2 = CO2 + H2O

Balance:

CH3OH + 3/2O2 = CO2 + 2H2O

Hrxn = Hprod - Hreact

C-H = 3*414 = 1242

C-O = 351

O-H = 463

O=O = 3/2*498 = 747

Hreact= 1242 + 351 + 463 + 747 = 2803 kJ/mol

C=O = 2*799 = 1598

O-H = 2*465 = 930

Hprod= 930+1598 = 2528

Hrxn = 2528 - 2803 = -275 kJ/mol

We have n = 0.0653 mol of Methanol

E = Hrxn * n = 0.0653 * (-275 ) = -17.9575 kJ/mol

2)

E = 2760 kJ

n = E/Hrxn = 2760 / 275 kJ/mol = 10.04 mol are needed

PV = nRt

V = nRT/P = (10.04*0.082)(273+18)/(5.14) = 46.6098 L

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