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.

The correct option is;

b) This is an endothermic reaction

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: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :