Question

< Chapter 8 HW Problem 8.70 - Enhanced - with Feedback < 12 of 15 > Using bond enthalpies in the table, estimate Al for the following gas-phase reactions. un Review | Constants Periodic Table Part A Estimate AH for the reaction. Br-c-H+CI-C1B -C-CI+H-CI 463 146 190 203 234 Express your answer using three significant figures. IVO AL O 2 ? Average Bond Enthalpies (kJ/mol) Single Bonds C-H 413 N-H 391 0-H C-C 348 N-N 163 0-0 C-N 293 N- 0 201 0-F C-0 358 N-F 272 0-0 C-F 485 N- 200 0-1 C-ci 328 N-B: 243 C-Br 276 S-H C-I 240 H-H 436 S-F C-S 259 H-F 567 S- H-ci 431 S-Br ci-c1 242 H-Br 366 S-S Br-Br 193 H-I 299 Multiple Bonds C=C 614 N=N 418 0=0 C=C 839 N=N 941 C=N | 615 N=0 | 607 C=N 891 C=0 799 C=0 1072 339 327 253 218 266 AH- Submit Request Answer Part B 495 Estimate AH for the reaction. H-S-C-c-S-H+2H-Br HH HH Br-C-C-Br + 2 H-S-H HH Express your answer using two significant figures.

Answer 1

PART A

We have, $\Delta$ H ⁰_rxn = Sum of $\Delta$ H ⁰ of reactant bonds - Sum of $\Delta$ H ⁰ of product bonds

$\therefore$$\Delta$H ⁰_rxn = [ 3 $\Delta$ H ⁰ C-Br + $\Delta$ H ⁰ C-H + $\Delta$ H ⁰ Cl-Cl ] - [3 $\Delta$ H ⁰ C-Br + $\Delta$ H ⁰ C-Cl + $\Delta$ H ⁰ H-Cl ]

$\Delta$H ⁰_rxn = $\Delta$ H ⁰ C-H + $\Delta$ H ⁰ Cl-Cl - $\Delta$ H ⁰ C-Cl - $\Delta$ H ⁰ H-Cl

$\Delta$H ⁰_rxn = [ 413 + 242 - 328 - 431 ] kJ / mol

$\Delta$H ⁰_rxn = - 104 kJ / mol

ANSWER : $\Delta$ H ⁰_rxn = - 104 kJ / mol

PART B

We have, $\Delta$ H ⁰_rxn = Sum of $\Delta$ H ⁰ of reactant bonds - Sum of $\Delta$ H ⁰ of product bonds

$\therefore$$\Delta$H ⁰_rxn = [ 4$\Delta$H ⁰ C-H + 2 $\Delta$ H ⁰ H-S +2 $\Delta$ H ⁰ C-S + $\Delta$ H ⁰ C-C + 2 $\Delta$ H ⁰ H-Br ] - [ 4$\Delta$H ⁰ C-H + 2 $\Delta$ H ⁰ C-Br + $\Delta$ H ⁰ C-C + 4 $\Delta$ H ⁰ H-S ]

$\Delta$H ⁰_rxn = 2 $\Delta$ H ⁰ H-S +2 $\Delta$ H ⁰ C-S + 2 $\Delta$ H ⁰ H-Br - 2 $\Delta$ H ⁰ C-Br - 4 $\Delta$ H ⁰ H-S

$\Delta$H ⁰_rxn = [ 2 (339) + 2(259) + 2 (366) - 2(276) - 4 (339) ] kJ / mol

$\Delta$H ⁰_rxn = 20 kJ /mol

PART C

We have, $\Delta$ H ⁰_rxn = Sum of $\Delta$ H ⁰ of reactant bonds - Sum of $\Delta$ H ⁰ of product bonds

$\therefore$$\Delta$H ⁰_rxn = [ 4$\Delta$H ⁰ N-H + $\Delta$ H ⁰ N-N + $\Delta$ H ⁰ Cl-Cl ] - [4$\Delta$H ⁰ N-H + 2 $\Delta$ H ⁰ N-Cl]

$\Delta$H ⁰_rxn = $\Delta$ H ⁰ N-N + $\Delta$ H ⁰ Cl-Cl - 2 $\Delta$ H ⁰ N-Cl

$\Delta$H ⁰_rxn = [163 + 242 - 2(200) ] kJ /mol

$\Delta$H ⁰_rxn = 5 kJ / mol