Final answer:
To calculate the standard entropy change for the reaction, the absolute entropy values of the reactants and products are needed. Using the given equation and the Thermodynamic properties of pure substance table, the standard entropy change for the reaction 2Al(s) + 3CuO(s) → Al₂O₃(s) + 3Cu(s) at 25.0 °C can be calculated as 13.6 J/(mol·K).
Step-by-step explanation:
To calculate the standard entropy change for the reaction, we can use the products minus reactants rule and the given absolute entropies of the products and reactants. The standard entropy change (ΔS°) can be calculated using the equation: ΔS° = ΣnS°(products) - ΣmS°(reactants). For the given reaction 2Al(s) + 3CuO(s) → Al₂O₃(s) + 3Cu(s), we can use the absolute entropies of the substances involved from the Thermodynamic properties of pure substance table. Let's calculate it step by step:
- Find the absolute entropy (S°) values for each reactant and product from the table.
- Sum the product entropy values and subtract the sum of the reactant entropy values to calculate ΔS°.
- Apply the correct stoichiometric coefficients to the entropy values to account for the balanced equation.
From the table, we have:
S°(Al(s)) = 50.9 J/(mol·K)
S°(CuO(s)) = 48.8 J/(mol·K)
S°(Al₂O₃(s)) = 50.9 J/(mol·K)
S°(Cu(s)) = 33.2 J/(mol·K)
Now, let's calculate the entropy change (ΔS°):
ΔS° = [2S°(Al₂O₃(s)) + 3S°(Cu(s))] - [2S°(Al(s)) + 3S°(CuO(s))]
ΔS° = [2 × 50.9 J/(mol·K) + 3 × 33.2 J/(mol·K)] - [2 × 50.9 J/(mol·K) + 3 × 48.8 J/(mol·K)]
Calculating the values gives us:
ΔS° = 13.6 J/(mol·K)
Therefore, the standard entropy change for the given reaction at 25.0 °C is 13.6 J/(mol·K).