Question

Consider the following reaction occurring at 298 K: N2O(g)+NO2(g)⇌3NO(g) Part A Show that the reaction is not spontaneous under standard conditions by calculating ΔG∘rxn.

Part B If a reaction mixture contains only N2O and NO2 at partial pressures of 1.0 atm each, the reaction will be spontaneous until some NO forms in the mixture. What maximum partial pressure of NO builds up before the reaction ceases to be spontaneous?

Part C What temperature is required to make the reaction spontaneous under standard conditions?

Answer 1

N2O(g) + NO2(g)         3NO(g)

A) Grxn = G products - G reactants

Gf (N2O) = 103.7 kJ/mol

Gf (NO2) = 51.3 kJ/mol

Gf (NO) = 87.6 kJ/mol

G rxn = 3*87.60 - (103.7 + 51.3)

= 107.8 kJ/mol

since G rxn is possitive, the reaction is non spontaneous.

(B). P (N2O) = P (NO2) = 1 atm

G rxn = G°rxn + RT*ln(K)

G rxn = G°rxn + RT*ln (P NO)^3 / [P(N2O) * P(NO2)]

When the reaction ceases to be spontaneous, G rxn = 0.

0 = 107.8 * 10^3 + 8.314 * 298 * ln (P NO)^3 / (1 * 1)

107.8 * 10^3 = - 8.314 * 298 * ln (P NO)^3

ln (P NO)^3 = - 43.51

3 * ln (P NO) = - 43.51

ln (P NO) = -14.50

P (NO) = e^(-14.50)

P (NO) = 5.0 x 10^-7 atm

(C). For a reaction to be spontaneous, G rxn should be negative.

We know that:

G = H - T S

H - T S < 0

H < T S

T < H / S

For the reaction:

H = 3*NO - (N2O + NO2)

= 3*(91.3) - (81.6 + 33.2)

= 159.1 kJ/mol

S = 3*210.8 - (220.0 + 240.1)

= 172.3 J/mol.K

= 0.1723 kJ/mol.K

Putting in above expression.

T > 159.1 / 0.1723

T > 923.4 K

Hence, temperature should be 924 K to make the reaction spontaneous.