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

2C-Cl formed = (2 * - 339) = - 678 kJ

ΔH = + 611 - 678 = - 67 kJ

2 C-Br formed = (2 * - 280) = - 560 kJ

ΔH = + 611 - 560 = + 51 kJ

2 C-I Formed = (2 * - 209) = - 418 kJ

ΔH = + 611 - 418 = + 193 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