The balanced equation for the reaction of ammonia and oxygen is the following. 4 nh3(g) + 5 o2(g) → 4 no(g) + 6 h2o(g) the standard molar entropies for the products and reactants are listed below. calculate the change in standard molar entropy for this reaction at 298.0 k and standard pressure, in j/mol·k. (enter your answer to the tenths place. include the sign of the value in your answer.)

ΔS° = 180.5 J/mol.K

Explanation:

Let's consider the following reaction.

4 NH₃(g) + 5 O₂(g) → 4 NO(g) + 6 H₂O(g)

The standard molar entropy of the reaction (ΔS°) can be calculated using the following expression.

ΔS° = ∑np × S°p - ∑nr × S°r

where,

ni are the moles of reactants and products

S°i are the standard molar entropies of reactants and products

ΔS° = 4 mol × S°(NO(g)) + 6 × S°(H₂O(g)) - 4 mol × S°(NH₃(g)) - 5 mol × S°(O₂(g))

ΔS° = 4 mol × 210.8 J/K.mol + 6 × 188.8 j/K.mol - 4 mol × 192.5 J/K.mol - 5 mol × 205.1 J/K.mol

ΔS° = 180.5 J/K

This is the change in the entropy per mole of reaction.

they are made of gases.

they have a spherical shape.

they make energy through fusion.

i'm wiling to try and you : )