Chem 351 Fall 2011 MT: Thermodynamics

Part 6: THERMODYNAMICS

This should have been a reasonably straight forward calculation, but care needs to be taken to get the right structures and the right molecular formula, then do the math correctly.

a. CH₄ + Br₂--> CH₃Br + HBr

b. Remember that breaking bonds requires energy (energy in) and making bonds releases energy (energy out)

Bonds broken = Br-Br and C-H = 413 + 193 = 606 kJmol^-1 (or 98 + 46 = 144 kcalmol^-1 )

Bonds made = C-Br and H-Br = 288 + 366 = 654 kJmol^-1 (or 69 + 87 = 156 kcalmol^-1 )

Overall = bonds broken - bonds made = -48 kJmol^-1 (or -12 kcalmol^-1 )

c. Based on the results of the calculation shown above, the reaction is calculated to be exothermic.

d. In step 2A, you break C-H and make H-Br, so the DH = 413 - 366 = + 47 kJmol^-1 (or 98 - 87 = +11 kcalmol^-1 )

e. Two possible explanations, either:

(i)	Step 2B is more endothermic than step 2A, so it will be a slower rate of reaction for this step 2. In step 2B, you break C-H and make C-Br, so the DH = 413 - 288 = + 125 kJmol^-1 (or 98 - 69 = +29 kcalmol^-1 )
or
(ii)	An orbital based rationale where the Br radical can more readily approach and overlap with the H orbital to abstract the C-H than is can get access to the C to abstract the C.

f. Using an excess of bromine would lead to the reaction of bromomethane with more bromine to give polybromination products such as dibromomethane.