Just giving it a try:
Part (a)The enthalpy formation of CH
3OH, •CH
2OH and H•
Part (b)Reaction: CH
3OH + Br• <=> •CH
2OH + HBr
K = k
1/k
2Combining :delta: G = -RT ln K and :delta: G= :delta: H-T :delta: S gives :delta: H=T :delta: S-RT ln K
Assume :delta: H and :delta: S are constant for T=300K and 350K,
300 :delta: S - 8.31451 x 300 ln (10
6/10
7) = 350 :delta: S - 8.31451 x 350 ln (4 x 10
8/7.7 x 10
8)
:delta: S = 1/50 { 8.31451 x [350 ln (4x10
8/7.7x10
8) - 300 ln (1x10
6/1x10
7)] } = 126 J mol
-1 K
-1 :delta: H (reaction) = 300 :delta: S - 8.31451 x 300 ln (1x10
6/1x10
7) = 43.5 kJ mol
-1 :delta: H (reaction)
= :delta: H (•CH
2OH) + :delta: H (HBr) - :delta: H (CH
3OH) - :delta: H (•Br)
= :delta: H (•CH
2OH) + (-36) - (-201.6) - 111.9
= 43.5 kJ mol
-1:delta: H (•CH
2OH) = -10.2 kJ mol
-1CH
3OH
•CH
2OH + •H
Bond strength of C-H in CH
3OH = :delta: H (•CH
2OH) + :delta: H (•H) - :delta: H (CH
3OH) = -10.2 + 218 - (-201.6) = 409.4 kJ mol
-1