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

ΔS°rxn = -146.8 J/K

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

since electrons are negative the answer is neagtaive