Using the table of bond dissociation enthalpies, calculate ? H ° for the reactio
ID: 967417 • Letter: U
Question
Using the table of bond dissociation enthalpies, calculate ?H° for the reaction A.
(Use the entries for propene and propenyl halides for the calculation of cyclohexene and cyclohexenyl halides, respectively. Pay attention to the algebraic sign of your answer!)
Using the table of bond dissociation enthalpies, calculate AF° for the reaction A. (Use the entries for propene and propenyl halides for the calculation of cyclohexene and cyclohexenyl halides, respectively. Pay attention to the algebraic sign of your answer!) Table of bond dissociation enthalpies C-H bond BDE (kJ/mol)C-X bondBDE (kJ/mol) X2 and HX bonds BDE (kJ/mol) CH3-H C2H5-H (CH3)2CH-H 414 (CH3)3C-H 405 CH2-CHCH2-H 372 CH2-CH-H 351 355 C1-CI Br-Br 439 CH3-Cl C2H5-Cl (CH3)2CH-C 355 (CH3)3C-Ci CH2CHCH2-Cl 288 247 192 151 431 368 297 422 H-Cl H-Br H-I 355 464 kJ/mol The following are the propagation steps of reaction A st Calculate AH° for each propagation step AH for the 1St step is AHo for the 2"d step is kJ/mol kJ/mol hich propagation step is rate-determining?Explanation / Answer
Solution:
1) You have to put energy into a bond (any bond) to break it. Bond breaking is endothermic. Let's break all the bonds of the reactants:
one C-H +372 kJ
one ClCl +247 kJ
The sum is 372+247=+619 kJ
2) You get energy out when a bond (any bond) forms. Bond making is exothermic. Let's make all the bonds of the one product:
one CCl 288 kJ
one ClH 431 KJ
The sum is= -288+(-431)= 719 kJ
3) H = the energies required to break bonds (positive sign) plus the energies required to make bonds (negative sign):
+619 + (719) = 100 kJ/mol
B) H for the first step: one C-H +372 kJ breaking and one H-Cl 431 KJ is making
H = Ebonds broken + Ebonds formed = +372=(-431) = -59 KJ/mol
H for the 2nd step: one Cl-Cl +247 kJ breaking and one C-Cl 288 KJ is making
H = Ebonds broken + Ebonds formed = +247=(-288) = -41 KJ/mol
Since 2nd step is less enthalpy hence rate determining
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