Question

1. C2H4(g) + H2O(g)<->CH3CH2OH(g)

Using standard thermodynamic data at 298K, calculate the free energy change when 2.000 moles of C2H4(g) react at standard conditions.

2. 2BrF₃(g)Br₂(g) + 3F₂(g)

Using standard thermodynamic data at 298K, calculate the free energy change when 2.27 moles of BrF₃(g) react at standard conditions.

Answer 1

1)

The reaction is

     C2H4 (g) + H2O (g) CH3CH2OH (g)

G rxn = Gf products - Gf reactants

Gf CH3CH2OH (g) = -168.3 KJ / mol

Gf H2O (g) = - 228.4 KJ / mol

Gf C2H4 (g) = 68.2 KJ / mol

Therefore

G rxn = Gf products - Gf reactants

              = -168.3 - [ -228.4 + 68.2)

             = -168.3 + 160.2

             = -8.1 KJ

This is per mole of C2H4.

Therefore for 2 moles of C2H4, the free energy change = 2 mol x -8.1 KJ / mol

                                                                             = - 16.2 KJ

Thus the change in free energy = - 16.2 KJ.

----------------------------------------------------------------------------------------------------------------

2)

The reaction is

2 BrF3 (g) Br2 (g) + 3 F2 (g)

G rxn = Gf products - Gf reactants

Gf Br2 (g) = 3.3 KJ / mol

Gf F2 (g) =0

Gf BrF3 (g) = -288.7 KJ / mol

Therefore

G rxn = Gf products - Gf reactants

              = 3.3 - [ 2 * -288.7)

             = 3.3 + 577.4

             = 580.7 KJ

This is for 2 moles of BrF3.

Therefore for 2.27 moles of BrF3 the change in free energy = 580.7 x 2.27 / 2 = 659 KJ

Thus the change in free energy = 659 KJ.

-------------------------------------------------------------------------------------