Question

Consider the reaction:

2 NO₂ (g) ⇌ N₂O₄ (g)

The ΔGf° for NO₂ (g) = 52 kJ/mol and for N₂O₄ (g) = 98 kJ/mol at 298 K.

Calculate the ΔG at the following values.

0.8 M NO₂ (g) & 3.0 M N₂O₄ (g)

4.0 M NO₂ (g) & 0.9 M N₂O₄ (g)

2.2 M NO₂ (g) & 2.4 M N₂O₄ (g)

2.4 M NO₂ (g) & 2.2 M N₂O₄ (g)

Answer 1

G = G^o +RTlnQ

G^o = standard free energy change. This can be calculated as follow:

G^o = G^oproduct - G^oreactant

         = 98 - 2x52 = 98 - 104 = -6 kJ/mol = -6000 J/mol

Q = Reaction quotient = The product of the concentrations of the products of a reaction divided by the product of the concentrations.

(1) 0.8 M NO₂ (g) & 3.0 M N₂O₄ (g)

Q = [N₂O₄] / [NO₂]²

    = 3 / 0.64 = 4.68

G = G^o +RTlnQ = -6 + RT ln4.68

      = -6000 + 8.314 x 298 ln4.68 = -2176 J/mol = -2.176 KJ/mol

(2) 4.0 M NO₂ (g) & 0.9 M N₂O₄ (g)

Q = 0.9 / 16 = 0.056

G = -6000 + 8.314 x 298 ln0.056 = -13141.3 J/mol = -13.141 KJ/mol

(3) 2.2 M NO₂ (g) & 2.4 M N₂O₄ (g)

Q = 2.4 / 4.84 = 0.495

G = -6000 + 8.314 x 298 ln0.495 = - 7742 J/mol = -7.742 KJ/mol

(4) 2.4 M NO₂ (g) & 2.2 M N₂O₄ (g)

Q = 2.2 / 5.76 = 0.381

G = -6000 + 8.314 x 298 ln0.381 = - 8390.7 J/mol =-8.3 KJ /mol