Consider the halogenation of ethene, where x is a generic halogen: h2c=ch2(g)+x2(g)→h2xc−ch2x(g) you may want to reference (page
412) section 9.10 while completing this problem. part a use bond energies to determine which halogen produces the most exothermic halogenation reaction with ethene. bond bond energy (\rm kj/mol) h−c 414 c−c 347 c=c 611 c−f 552 c−cl 339 c−br 280 c−i 209 f−f 159 cl−cl 243 br−br 193 i−i 151 use bond energies to determine which halogen produces the most exothermic halogenation reaction with ethene. bond bond energy (\rm kj/mol) 414 347 611 552 339 280 209 159 243 193 151 iodine chlorine bromine fluorine
Consider the halogenation of ethene is as follows: CH₂=CH₂(g) + X₂(g) → H₂CX-CH₂X(g) We can expect that this reaction occurring by breaking of a C=C bond and forming of two C-X bonds. When bond break it is endothermic and when bond is formed it is exothermic. So we can calculate the overall enthalpy change as a sum of the required bonds in the products: Part a) C=C break = +611 kJ 2 C-F formed = (2 * - 552) = -1104 kJ Δ H = + 611 - 1104 = - 493 kJ
Part b) As we can see that the highest exothermic bond formed is C-F bond so from bond energies we can found that addition of fluoride is the most exothermic reaction
