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
Mix it. The salt will dissolve, eventually the sand and metal will settle. Pour out most of the water, let it evaporate leaving the salt. Use a magnet for the iron, which leaves the sand.
