ΔS° = -198.7 J/K for the reaction shown below.Calculate S° for NH₃(g) .
N₂(g) + 3 H₂(g) → 2 NH₃(g)

Question

ΔS° = -198.7 J/K for the reaction shown below.Calculate S° for NH₃(g). N₂(g)+ 3 H₂(g)→ 2 NH₃(g)

Quizplus · Accepted Answer

We can use the equation ΔS° = ΣS°(products) - ΣS°(reactants) to calculate the standard molar entropy of NH₃(g). ΔS° = -198.7 J/K = 2S°(NH₃(g)) - (S°(N₂(g)) + 3S°(H₂(g))) We are asked to find S°(NH₃(g)), so we can rearrange the equation and substitute the given values: S°(NH₃(g)) = (ΔS° + ΣS°(reactants)) / 2 S°(NH₃(g)) = (-198.7 J/K + 3S°(H₂(g)) + S°(N₂(g))) / 2 We can look up the standard molar entropy values for H₂(g) and N₂(g) in a table, or use the fact that they are gases and likely have similar standard molar entropy values to diatomic gases (e.g. S°(N2(g)) = 191.5 J/K∙mol). Let's assume that S°(H₂(g)) and S°(N₂(g)) are both about 200 J/K∙mol. Then: S°(NH₃(g)) = (-198.7 J/K + 3(200 J/K∙mol) + 1(200 J/K∙mol)) / 2 S°(NH₃(g)) = 192.3 J/K∙mol Therefore, the best choice is C, with a standard molar entropy of 192.3 J/K∙mol.