Question

For a gaseous reaction, standard conditions are 298 K and a partial pressure of 1 atm for all species. For the reaction 2 NO ( g ) + O 2 ( g ) − ⇀ ↽ − 2 NO 2 ( g ) the standard change in Gibbs free energy is Δ G ° = − 69.0 kJ/mol . What is ΔG for this reaction at 298 K when the partial pressures are P NO = 0.450 atm , P O 2 = 0.100 atm , and P NO 2 = 0.650 atm ?

Answer 1

Answer:

ΔG = -61.5 kJ/mol (Spontaneous process)

Explanation:

2 NO (g)  +  O₂  (g)   ⇄  2NO₂ (g)

Let's apply the thermodynamic formula to calculate the ΔG

ΔG = ΔG° + R .T . lnQ

We don't know if the gases are at equilibrium, that's why we apply Q (reaction quotient)

ΔG = - 69 kJ/mol + 8.31x10⁻³ kJ/K.mol . 298K . ln Q

How can we know Q? By the partial pressures (Qp)

P NO = 0.450atm

PO₂ = 0.1 atm

PNO₂ = 0.650 atm

Qp = [NO₂]² / [NO]² . [O₂]

Qp = 0.650² / 0.450² . 0.1 = 20.86

ΔG = - 69 kJ/mol + 8.31x10⁻³ kJ/K.mol . 298K . ln 20.86

ΔG = -61.5 kJ/mol (Spontaneous process)