Question

For a gaseous reaction, standard conditions are 298 K and a partial pressure of 1 atm for all species. For the reaction N 2 ( g ) + 3 H 2 ( g ) − ⇀ ↽ − 2 NH 3 ( g ) the standard change in Gibbs free energy is Δ G ° = − 32.8 kJ/mol . What is ΔG for this reaction at 298 K when the partial pressures are P N 2 = 0.100 atm , P H 2 = 0.150 atm , and P NH 3 = 0.650 atm ? Δ G = kJ / mol

Answer 1

N2(g) + 3H2(g) ----> 2NH3(g)

Reaction Quotient

Q = (PNH3)2 / ( PN2 × (PH2)3 )

Q = (0.650)2 / ( 0.100 × (0.150)3 )

Q = 1251.85

Standard change in free energy

Δ G° = - 32.8 KJ/mole  

Gas constant = R = 0.008314 KJ/K.mole

Temperature = T = 298 K

Change in free energy

Δ G = Δ G° + RT ln(Q)

Δ G = - 32.8 + 0.008314 × 298 × ln(1251.85)

Δ G =  - 15.129 KJ / mole