Find the enthalpy of reaction for CH4 + 2C12-->CH2CI2 + 2HCI from bond enthalpies. Average Bond Enthalpies (kJ/mol)Single BondsC-H 413C-C 348C-N 293C-O 358C-F 485C-CI 328C-Br 276C-I 240C-S259Si-H323Si-Si226Si-C 301Si-O 368Multiple BondsC=C 614C=C 839C=N 615C=N 891C=O 799C=0 1072N-HN-NN-ON-FN-CIN-Br391163201272200243436H-HH-F 567H-Cl 431H-Br 366H-I 299N=N 418N=N 941O-H 4630-0 146O-F 1900-CI 203O-I234S-HS-F339327S-Cl253218S-BrS-S 266sS=OS=S495523418F-F 155Cl-F 253CI-CI 242Br-F237Br-Cl 218Br-Br 193I-CI 208175151I-BrI-I Table 9.5 Bond Enthalpies (in kJ/mol)*HCNOSFEΗBrIH436413391463339567431366299C348393358259485328276240N163201272200243614 C=N839C=N418N=O9450 0HHH-C + H+Single BondsS0146190203266327253218F234Multiple BondsC=C615C=C891N=N607N=N498*Data are taken from http://wiki.chemedd1.org/index.php/15.4_Bond_Enthalpies.159253237HCIC=0C=0S=OS=S242218208Br193175IYou can imagine that the reaction takes place in steps involving the breaking andforming of bonds. Starting with the reactants, you suppose that one C-H bond andthe Cl-Cl bond break.151HH-C-H+CI-CI - H-C + H+ Cl + ClHHThe enthalpy change is BE(C-H) + BE(C1-CI). Now you reassemble the fragmentsto give the products.804 (in CO₂)10763234189.11 Bom

Answer:Here, B.E. is used to bond energy and ai and bi are the number of bonds of each type present…

Answered: Find the enthalpy of reaction for CH4 +…

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Chemistry

Find the enthalpy of reaction for CH4 + 2C12--> CH2CI2 + 2HCI from bond enthalpies.

Chemistry: The Molecular Science

5th Edition

ISBN:9781285199047

Author:John W. Moore, Conrad L. Stanitski

Publisher:John W. Moore, Conrad L. Stanitski

Chapter6: Covalent Bonding

Section6.6: Bond Properties: Bond Length; Bond Energy

Problem 6.6CE

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Question

Find the enthalpy of reaction for CH4 + 2C12--> CH2CI2 + 2HCI from bond enthalpies.

Average Bond Enthalpies (kJ/mol)
Single Bonds
C-H 413
C-C 348
C-N 293
C-O 358
C-F 485
C-CI 328
C-Br 276
C-I 240
C-S
259
Si-H
323
Si-Si
226
Si-C 301
Si-O 368
Multiple Bonds
C=C 614
C=C 839
C=N 615
C=N 891
C=O 799
C=0 1072
N-H
N-N
N-O
N-F
N-CI
N-Br
391
163
201
272
200
243
436
H-H
H-F 567
H-Cl 431
H-Br 366
H-I 299
N=N 418
N=N 941
O-H 463
0-0 146
O-F 190
0-CI 203
O-I
234
S-H
S-F
339
327
S-Cl
253
218
S-Br
S-S 266
s
S=O
S=S
495
523
418
F-F 155
Cl-F 253
CI-CI 242
Br-F
237
Br-Cl 218
Br-Br 193
I-CI 208
175
151
I-Br
I-I

Table 9.5 Bond Enthalpies (in kJ/mol)*
HCNOSFE
Η
Br
I
H
436
413
391
463
339
567
431
366
299
C
348
393
358
259
485
328
276
240
N
163
201
272
200
243
614 C=N
839
C=N
418
N=O
945
0 0
H
H
H-C + H+
Single Bonds
S
0
146
190
203
266
327
253
218
F
234
Multiple Bonds
C=C
615
C=C
891
N=N
607
N=N
498
*Data are taken from http://wiki.chemedd1.org/index.php/15.4_Bond_Enthalpies.
159
253
237
H
CI
C=0
C=0
S=O
S=S
242
218
208
Br
193
175
I
You can imagine that the reaction takes place in steps involving the breaking and
forming of bonds. Starting with the reactants, you suppose that one C-H bond and
the Cl-Cl bond break.
151
H
H-C-H+CI-CI - H-C + H+ Cl + Cl
H
H
The enthalpy change is BE(C-H) + BE(C1-CI). Now you reassemble the fragments
to give the products.
804 (in CO₂)
1076
323
418
9.11 Bom

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