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) sectio

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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

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2022-04-20T06:36:16+00:00 1 Answer 0

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  1. Myna
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    2022-04-20T06:37:48+00:00

    Consider the halogenation of ethene is as follows:
    CHu2082=CHu2082(g) + Xu2082(g)u2192 Hu2082CX-CHu2082X(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
    u0394 H = + 611 – 1104 = – 493 kJ

    2C-Cl formed = (2 * -339) = – 678 kJ
    u0394H = + 611 – 678 = -67 kJ

    2 C-Br formed = (2 * -280) = -560 kJ
    u0394H = + 611 – 560 = + 51 kJ

    2 C-I Formed = (2 * -209) = -418 kJ
    u0394H = + 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

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