Question

Calculate ΔSrxn∘ for the reaction

2NO(g)+O2(g)→2NO2(g)



Substance S∘ (J/mol⋅K)
NO2 240.0
O2 205.2
NO 210.8

Answer 1

The entropy of the reaction can be calculated alike the enthalpy of the reaction which is equal to the difference between the summation of entropies of the products multiplied by their corresponding stoich coeff. and the summation of the entropies of the reactants multiplied by their corresponding stoich coeff. In this case, the answer is -146.8 J / mol K

Answer 2

ΔS°rxn = -146.8 J/K

Step-by-step explanation:

Let's consider the following reaction.

2 NO(g) + O₂(g) → 2 NO₂(g)

The standard entropy change of the reaction (ΔS°rxn) is equal to the standard entropies of the products times their stoichiometric coefficients minus the standard entropies of the reactants times their stoichiometric coefficients.

ΔS°rxn = 2 mol × S°(NO₂(g)) - 2 mol × S°(NO(g)) - 1 mol × S°(O₂(g))

ΔS°rxn = 2 mol × 240.0 J/mol.K - 2 mol × 210.8 J/mol.K - 1 mol × 205.2 J/mol.K

ΔS°rxn = -146.8 J/K