Final answer:
The total entropy change for the reaction H₂ + 1/2 O₂ -> H₂O at 298 K was calculated to be -123.7 J/mol·K by using the provided molar entropy values and the 'products minus reactants' rule. However, since the question's answer choices are all positive and do not match the calculated value, there may be a mistake in the provided values or the question itself.
Step-by-step explanation:
To calculate the total entropy change (ΔS) for the reaction
H₂(g) + ½ O₂(g) → H₂O(l), we can use the standard molar entropy values of the products and reactants at 25°C (298 K). Each entropy value must be multiplied by the respective stoichiometric coefficient from the balanced chemical equation:
- S°(H₂O(l)) = 109.6 J/(mol·K)
- S°(H₂(g)) = 130.7 J/(mol·K)
- S°(O₂(g)) = 205.2 J/(mol·K)
The total entropy change is the sum of the entropies of the products minus the sum of the entropies of the reactants:
ΔS = S°(products) - S°(reactants)
ΔS = [1 mol × S°(H₂O(l))] - [1 mol × S°(H₂(g)) + ½ mol × S°(O₂(g))]
ΔS = [1 × 109.6 J/(mol·K)] - [1 × 130.7 J/(mol·K) + ½ × 205.2 J/(mol·K)]
ΔS = 109.6 - (130.7 + ½ × 205.2)
ΔS = 109.6 - (130.7 + 102.6)
ΔS = 109.6 - 233.3
ΔS = -123.7 J/mol·K
Since the question provides only positive values for the total entropy, it appears there's confusion with the sign or the given values.
Additionally, when considering the enthalpy change (ΔH) of the reaction, which is -286 kJ/mol at 298 K, you can relate it to the entropy change and temperature to calculate the Gibbs free energy change (ΔG) using the equation:
ΔG = ΔH - TΔS
This is important because a spontaneous reaction is characterized by a negative ΔG. However, since the answer choices provided do not match the calculated ΔS, there may be an issue with the figures provided in the question or a misunderstanding about the expected answer.