Question

Look at this reaction:

3NO(g) ⇆ N2O(g) + NO2(g)
Keq = 1.8×10^(18) (at 298 K),
△H° = -155 kJ/mol.
Which of the following statements about this reaction is not correct?

a) ΔS < 0 because there is more order in the products than in the reactants.
b) This is an endothermic reaction.
c) The products are heavily favored at room temperature.
d) At a higher temperature, products would be less favored than at room temperature.

Answer 1

The correct option is;

b) This is an endothermic reaction

Step-by-step explanation:

To answer the question, we analyze each of the options as follows;

a) ΔS < 0 because there is more order in the products than in the reactants.

The above statement is correct as there are fewer molecules in the product than in the reactants

b) This is an endothermic reaction.

The heat of the reaction is △H° = -155 kJ/mol which is negative, suggestive of an exothermic reaction, therefore the above equation is not correct

c) The products are heavily favored at room temperature.

Based on the high value of keq at 298 K the above statement is correct

d) At a higher temperature, products would be less favored than at room temperature.

The above statement is true as the amount of heat produced will be returned bck to the system tending to reverse the reaction.

Answer 2

statements about this reaction is not correct:

b) this is an endothermic reaction.

Step-by-step explanation:

reaction:

• 3NO(g) ↔ N2O(g) + NO2(g)

∴ Keq = 1.8 E18 (298 K)

∴ ΔH° = - 155 KJ/mol

• Ln K = - ΔG°m/RT

⇒ - ΔG°m/RT = Ln(1.8 E18) = 42.034

⇒ - ΔG°m = (42.034)(RT)

∴ R = 8.314 J/mol.K

⇒ - ΔG°m = (42.0349((8.314 J/mol.K)(298 K))

⇒ ΔG°m = - 104143.05 J/mol = - 104.143 KJ/mol

• ΔG°m = ΔH° - TΔS°

⇒ - TΔS° = ΔG°m - ΔH° = - 104.143 - ( -155) = 50.86 KJ/mol

⇒ ΔS° = - (50.86)/(298 K)

⇒ ΔS° = - 0.1707 KJ/mol.K

we have:

∴ ΔG° ≈ ΔH° < 0....... the reaction is exothermic