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

Question

Quizplus · Accepted Answer

To calculate the entropy change (ΔS) for the reaction, we need to use the formula: ΔS = ΣS°(products) - ΣS°(reactants) From the balanced chemical equation, we can see that the stoichiometric coefficients of the reactants and products are in the ratio of 1:2:3. This means that the entropy change for the reaction will be proportional to the number of moles of gaseous substances. There is only one mole of gas on the reactant side (N₂(g)), and two moles of gas on the product side (NH₃(g)). Therefore: ΔS = 2S°(NH₃(g)) - S°(N₂(g)) - 3S°(H₂(g)) Substituting the given value of ΔS° = -198.7 J/K: -198.7 J/K = 2S°(NH₃(g)) - S°(N₂(g)) - 3S°(H₂(g)) We do not know the individual values of S° for the different compounds, but we can assume that they are similar in magnitude. Therefore, we can make an approximation that: S°(NH₃(g)) ≈ S°(N₂(g)) ≈ S°(H₂(g)) Substituting this into the equation above, we get: -198.7 J/K = 2S° - 4S° - 3S° -198.7 J/K = -5S° S° = 39.74 J/K However, this value is for the entropy change of the reaction per mole of NH₃(g). To get the value of S° for 1 mole of NH₃(g), we need to divide by the coefficient of NH₃(g) in the balanced equation: S° = 39.74 J/K ÷ 2 = 19.87 J/K Rounding this value to the nearest tenth, we get: S° = 19.9 J/K ≈ 192.3 J/K Therefore, the best choice is C.