(1)Gaseous chlorine reacts with ethylene to give 1,2-dichloroethane: Cl2(g) C2H4

Question

(1)Gaseous chlorine reacts with ethylene to give 1,2-dichloroethane: Cl2(g) C2H4(g) ClCH2CH2Cl(g) The above reaction is carried out in a special steel reaction vessel that is designed to maintain a constant pressure of 1 bar. 10.00 g chlorine and excess ethylene are introduced into separate compartments of the vessel, which is immersed in a well- insulated water bath containing 1000 g water. The initial temperature of the water is 25.00 °C. Then the chlorine is allowed to react with the ethylene; after the reaction is complete and the system has reached thermal equilibrium, the temperature of the water bath is observed to have risen to 31.14 °C (a) Calculate AH° for the reaction given in equation (1), in kJ/mol. (You may neglect the heat capacity of the steel reaction vessel and its contents in this calculation.) (b)Given the bond dissociation enthalpies below and the AH" for reaction (1), estimate the BDE of a C-Cl bond. (You will not need to use all the given BDE's Bond C-C C=C BDE (kJ/mol 348 612 837 Bond CI-CI H-Cl C-H BDE (kJ/mol 243 432 (c)Another possible reaction of Cl2 with ethylene is the formation of vinyl chloride and HCl, shown below. Estimate AH° for this reaction Cl2 (g) + C2H4 (g) CH2-CHCI g) + HCl (g) (d)Which reaction would you expect to have the more positive AS°, (1) or (2)? Explain your answer.

Explanation / Answer

Solution:- (a) mass of water = 1000 g

change in temperature of water = 31.14 - 25.00 = 6.14 0C

q = m c delta T

where, q is heat energy, mass is mass in grams, c is specific heat in J/g.0C and delta T is change in temperature.

q = 1000 g x 4.184 J/g.0C x 6.14 0C

q = 25689.76 J

heat given = heat taken

so, if 25689.76 J of heat is taken by water then same amount of heat would be lost by the reaction of Cl2 with ethylene. So, q lost by Cl2 reaction = -25689.76 J

mass of Cl2 = 10.00 g

moles of Cl2 = 10.00 g x (1mol/70.9 g) = 0.1410 mol

delta H0 = -25689.76 J/0.1410 mol x (1kJ/1000 J)

delta H0 = -182.2 kJ/mol

(b) delta H0 = sum of bond energy of reactants - sum of bond energy of products

For the given equation, one Cl-Cl bond and one C=C bond is breaking and two C-Cl bonds and one C-C bond is forming. So....

-182.2 = [(Cl-Cl) + (C=C)] - [2(C-Cl) + (C-H)]

-182.2 = [243 + 612] - [2(X) + 348]

-182.2 = 855 - 2X - 348

-182.2 = 507 - 2X

2X = 507 + 182.2

2X = 689.2

X = 689.2/2

X = 344.6

So, BDE for C-Cl bond is 344.6 kJ/mol

(c) delta H0 = [(Cl-Cl) + (C-H)] - [(C-Cl) + (H-Cl)]

delta H0 = [(243) + (412)] - [(344.6) + (432)]

delta H0 = 655 - 776.6

delta H0 = -121.6 kJ/mol

(d) Theoretically, delta S0 would be begative for the first reaction as the number of gaseous species is decreasing. For second reaction it would be zero as we have equal number of gaseous species on both sides.

delta S0 = delta H0/T

for the first reaction,

delta S0 = -182.2kJ/mol)/298K

delta S0 = -0.611 kJ/mol.K or -611 J/mol.K

for second reaction..

delta S0 = -121.6 kJ/mol)/298K

delta S0 = -0.408 kJ/mol.K or -408 J/mol.K

From calculations, delta S0 is less negaive than the value we get for first reaction it means the value would be more positive for second reaction as compared to the first reaction.

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