Question

9) Ethylene, C2H,reacts with elemental halogens in the gas phase according to the following reaction to form 1,2-dihaloethanes: НН + X-X (g) HH (9) 1,2-dihaloethane, C2H2 Ethylene, C2H Elemental halogen Here, "X-X" is shorthand for Cl2. Br2.or 2 a) Use the enthalpies of formation and Third Law entropies in the table below to calculate Gibbs free energies of reaction for the formation of each dihaloethane. Although you're going to do 3the same calculation three times, you only need to show one sample calculation for how you produce each answer. Fill your answers into the blanks in the table following this problem. b) Which reaction favor products most strongly? c) What is the major contributor to the equilibrium position of each reaction? That is, are the reactions driven primarily by a favorable enthalpy change or a favorable entropy change? d) Of the three reaction, which one would show the largest effect from a temperature change. What would be the effect of a large decrease in reaction temperature? AreactionG H° So formation Ethene (9) 52.47 kJ/mole 219.32 J/mole K 1,2-dichloroethane (g) -129.79 kJ/mole 208.53 J/mole K 1,2-dibromoethane (9) -38.33 kJ/mole 331.1 J/mole K 1,2-diiodoethane (g) 66.5 kJ/mole 348.2 J/mole K Chlorine, C2 (g) 0.00 kJ/mole223.066 J/mole K Bromine, Br2 (9) 30.907 kJ/mole 245.463 J/mole K lodine, 12 (g) 62.42 kJ/mole 260.687 J/mole K
Can someone help

Answer 1

Solution :-

Part a) First we need to calculate the delta H rxn and delta S rxn and then using them we can find the delta G rxn

Lets calculate for the 1,2 dichloroethane

Delta H rxn = sum of delta Hof product – sum of delta Hof reactant

                  =[C2H4Cl2*1] – [(C2H4*1)+(Cl2*1)]

                  = [-129.79*1]-[(52.47*1)+(0.00*1)]

                  = -182.26 kJ

Now lets calculate the entropy change

Delta S rxn = sum of delta S product – sum of delta S reactant

                  =[C2H4Cl2*1] – [(C2H4*1)+(Cl2*1)]

                  = [208.53*1]-[(219.32*1)+(223.066*1)]

                  = -233.856 JK

Delta G= delta H – (T*delta S)

              = -182.26 kJ – (-0.233856 kJ/K * 298 K)

              = -112.57 kJ

Calculating the delta G for the other two products using the same formula by using the values of respective reactant and product

Lets calculate for the 1,2 dibromoethane

Delta H rxn = sum of delta Hof product – sum of delta Hof reactant

                  =[C2H4Br2*1] – [(C2H4*1)+(Br2*1)]

                  = [-38.33*1]-[(52.47*1)+(30.907*1)]

                  = -121.707 kJ

Now lets calculate the entropy change

Delta S rxn = sum of delta S product – sum of delta S reactant

                  =[C2H4Br2*1] – [(C2H4*1)+(Br2*1)]

                  = [331.1*1]-[(219.32*1)+(245.463*1)]

                  = -133.683 JK

Delta G= delta H – (T*delta S)

              = -121.707 kJ – (-0.133683 kJ/K * 298 K)

              = -81.9 kJ

Lets calculate for the 1,2 diiodoethane

Delta H rxn = sum of delta Hof product – sum of delta Hof reactant

                  =[C2H4I2*1] – [(C2H4*1)+(I2*1)]

                  = [66.5*1]-[(52.47*1)+(62.42*1)]

                  =   -48.39 kJ

Now lets calculate the entropy change

Delta S rxn = sum of delta S product – sum of delta S reactant

                  =[C2H4I2*1] – [(C2H4*1)+(I2*1)]

                  = [348.2*1]-[(219.32*1)+(260.687*1)]

                  = -131.807 JK

Delta G= delta H – (T*delta S)

              = -48.39 kJ – (-0.131807 kJ/K * 298 K)

              = -9.11 kJ

Part b) the reaction of the formation of 1,2-dibromoethane has the highest free energy change among the given reactions therefore the the products are produced strongly in the formation of the 1,2-dibromoethane

Part c) Ethalpy change is the major contributor in determining the spontaneity of the reaction

Favourable enthalpy change that means when the reaction gives out the energy (exothermic reaction) is more favourable because the products of the reaction are more stable than the reactants

Part d) Since the reaction of the chlorine has the highest enthalpy change value that is -182.26 kJ/mol

So this reaction is the most exothermic among the three so when the temperature is reduced largely then this will show the major increase in the formation of the product for the 1,2-dichloroethane.

Therefore the reaction of the 1,2-dichloroethane will show the largest effect of large temperature decrease and it will produce more product.