H C H H C CH2 HaC CH3 H C CH3 2 H2 (1) Calculate the enthalpy change (2) Assurmi
ID: 553204 • Letter: H
Question
H C H H C CH2 HaC CH3 H C CH3 2 H2 (1) Calculate the enthalpy change (2) Assurming the entropy change is zero, calculate the equilibrium constant at 298 K TABLE 4-2 Bond-Dissociation Enthalpies for Homolvtic Cleavages A:B A-+-B Enthalpy Enthalpy keal/mol Bond kJ/mol kcal/molBond H X bonds and X X bonds Bonds to secondary cuhons 95 106 104 CHo)bCH H 106 38 (CH) CH C 58 (CHs) CH Br 435 CH3)2CHF 335 159 242 192 151 F-F Br Br I-I (CH,)2CH-I 36 381 136 CH CH OH 103 Bonds o wetiary carbons 88CH)CH 431 91 106 381 H Br (CH)C F 331 119(CH)C C HO H HO OH 498 213 (CH C Br (CH)CIExplanation / Answer
In the given reaction the bonds that are broken C-C(Secondary) and C-H(tertiary carbon)
Energy absorbed for breaking bonds = C-C(Secondary) + C-H(tertiary carbon) = 356+381 = 737 KJ
The bonds that are formed C-C(tertiary) and C-H(primary carbon)
Energy released in formation of bonds = 339+410 = -749 KJ
dH, Enthalpy change = Energy for breaking of bonds-energy for formation of bonds
dH = 737 - 749 = -12 KJ
The free energy equation is given by, G = H-TS
If entropy change = 0, then TS =0
G = H = -12 KJ
Now, the equilirium constant is given by G = -RTlnK
lnK = -(G/RT)
K = e^-(G/RT)
K = e^-(-12KJ/8.314*10^-3KJ*298) = 127
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