Question

Calculate ΔG∘ (in kJ/mol) for the following reaction at 1 atm and 25 °C:

C₂H₆ (g) + O₂ (g)  → CO₂ (g) + H₂O (l) (unbalanced)

ΔH_f C₂H₆ (g) = -84.7 kJ/mol; S C₂H₆ (g) = 229.5 J/K⋅mol;
ΔH_f ∘ CO₂ (g) = -393.5 kJ/mol; S CO₂ (g) = 213.6 J/K⋅mol;
ΔH_f H₂O (l) = -285.8 kJ/mol; SH₂O (l) = 69.9 J/K⋅mol;
SO₂ (g) = 205.0 J/K⋅mol

Answer 1

First, balance:

C2H6 (g) + O2 (g)  → CO2 (g) + H2O (l)

C2H6 (g) + 7/2O2 (g)  → 2CO2 (g) + 3H2O (l)

now, substitute data

dG = dH - T*dS

dH = Hprod - Hreact = [2CO2 (g) + 3H2O (l) ] -[C2H6 (g) + 7/2O2 (g) ] = (2*-393.5 + 3*-285.8) - (-84.7 + 7/2*0) = -1559.7

dS = Sprod - Sreact = [2CO2 (g) + 3H2O (l) ] -[C2H6 (g) + 7/2O2 (g) ] = (2*213.6 + 3*69.9) - (229.5 + 7/2*205) = -310.1

now..

dG = -1559.7 - 298*(-310.1/1000)

dG = -1467.2 kJ/mol (this is based per 1 mol of C2H6)

if whole reaction is needed:

2C2H6 (g) + 7 O2 (g)  → 4CO2 (g) + 6H2O (l)

GRXN = 2*dG = 2*-1467.2 = -2934.4kJ