Question

Consider this reaction occurring at 298 K: N2O(g) + NO2(g) 2 3 NO(g) a. Show that the reaction is not spontaneous under standard conditions by calculating A Grx b. If a reaction mixture contains only N20 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? c. Can the reaction be made more spontaneous by an increase or decrease in temperature? If so, what temperature is required to make the reaction spontaneous under standard conditions?

Answer 1

Answer:

A.) ΔG = 107.8 kJ/mol

ΔG rxn is positive implies that the reaction is non spontaneous.

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

C) T > 923.4 K, therefore, the temperature should be 924K in order to make the reaction spontaneous.

Explanation:

N2O(g) + NO2(g) - 3NO(g)

A) ΔG = ΔG products - ΔG reactants

 ΔGf (N2O) = 103.7 kJ/mol

ΔGf (NO2) = 51.3 kJ/mol

ΔGf (NO) = 87.6 kJ/mol

ΔG rxn = 3*87.60 - (103.7 + 51.3)

= 107.8 kJ/mol

ΔG rxn is positive implies that the reaction is non spontaneous.

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

ΔG rxn = ΔG°rxn + RTln(K)

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

When the reaction ceases to be spontaneous, then ΔG 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, ΔG rxn should be negative.

It is known that:

ΔG= ΔH - TΔS

ΔG= ΔH - TΔS < 0

ΔH< TΔS

T < ΔH / ΔD

For the reaction:

ΔH = 3×NO - (N2O + NO2)

= 3×(91.3) - (81.6 + 33.2)

= 159.1 kJ/mol

ΔS = 3×210.8 - (220.0 + 240.1)

= 172.3 J/mol.K

= 0.1723 kJ/mol.K

From the above expression.

T > 159.1 / 0.1723

T > 923.4 K

Therefore, the temperature should be 924K in order to make the reaction spontaneous.