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avatar+36916 
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First balance the equation

                           C2H4---> 2C+ 2H2

C2H4  +  6 F2 ---------------------------------->  2 CF+  4 HF   

                                                                    Note that we got 4 HF's   Given 2 HF's = - 537kj       - 537 x 2

                                                                                   we got 2 CF4     Given 1 CF4 = - 680 kj       - 680 x 2

                                                                                                                                                      ___________

                                                                                                                                        deltaH = - 2434kj

                                                                                            But I will assume the intermediate reaction occurs too:

                                                                                   C2H4 -----> 2C + 2H2    we have ONE of these reactions  for  - 52.3 kj

        deltaH becomes - 2434 - 52.3 kj =  -2486.3 kj

Apr 18, 2019
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Apr 18, 2019

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