Question

Answer (a) and (b) (a) For the decomposition of ethane (C2H6) into two ethyl radicals the heat of reaction (AHx) is 368 kJ mol!. C2H6 – 2CH In addition, the heat of formation AHF of CH is -85 kJ mol! Calculate the heat of formation of a methyl radical. [10 marks] (b) The reaction trans - cis perfluorobut-2-ene was studied kinetically and the forward rate constant was determined to be: kr = 3.16 x 1013 exp{-31,030/T) s! Given that AH=+3.42 kJ moland AS constant (kr) at 750 K. 2.03 JK-mol-!. Determine the reverse rate [15 marks]

Answer 1

The image below shoes the calculation neededfor question 1), and 2) :

Cat6 2CH3 Hoxn = 368k5/mol BHp og ez HG = -85 kg) mol. & Hoon not produd - ohnen where, na moles, of product = 2x 0 Hproduct - D Horeavant 2x H product = strent a treactant- = 368kJ/mol + (-85 kg/mol) • 283 kJ/mol Dit product = 283 k3 /mol DH product = 141.5 kJ/mol, The enthalpy of formation of Methyl radical is 141.5 kJ/mol trans ses cis perfluorobut-zene. kb Rf = 3.168 lose 4033 Ryo Ky — DH = 3.42 kJ/mol sua DH-TOS 05 = - 2003 165/mol.kl also, Dr RT In Reg. T- 750K R= 8.314 I/molok I = RT lnko a DH-TDS lutling the values, in kog = 3.42 RJ 3.42 kJ/mol 1-2.035/ml k} (8.314 J/molok) (50K) 8.314 5/mulok = 0.55 – 60.248 Inky = 0.79 Keq = 0.79

Keq = 0.79 = 2.203 1 Kf = 3.16 x 1013 {- 31030} s- = 3:16 410's oe s! = 3.16 X10% (0.107 210-17) & putting in (A) keq = K = kg = Kt koa. 0.34 X10'S 2.203 kg = 0.15 x1043) Ans.