Chemistry Thermodynamics Math Question

Question

Calculate the change in Gibbs free energy when 50 mg of liquid bromine dissolves in aqueous solution at 41 C.

SubstanceΔfH0	kJ/molS0	J/mol K
Br2(l)	0.0	152.2
Br2(aq)	-2.6	130

Answer 1

Your table is a little "messed up", but extrapolating what I think it should be we have the following:

Br₂(l) ===> Br₂(aq)

∆G = ∆H - T∆S

For Br₂(l): ∆G = 0 - (314K x 152.2) = -47791 J = 47.8 kJ/mol

For Br₂(aq): ∆G = (-2.6 kJ/mol) - (314K x 0.130 kJ/mol-K) = -2.6 - 40.8 = -43.4 kJ/mol

Br₂(l) ===> Br₂(aq)

∆G_rxn = ∆G_products - ∆G_reactants = -43.3 kJ/mol - 47.8 kJ/mol = -91.1 kJ/mol

50 mg Br₂ x 1 g/1000 mg x 1 mol/159.8 g x -91.1 kJ/mol = 0.0285 kJ = 28.5 J