Using the following So values H2O(g), So = 188.83 J·mol-1·K-1 O2(g), So = 205.0 J·mol-1·K-1 CO2(g), So = 213.6 J·mol-1·K-1 CH3OH(g), So = 237.6 J·mol-1·K-1 Calculate ΔSo in J mol-1 K-1 for the following reaction. 2CH3OH(g) + 3O2(g) → 2CO2(g) + 4H2O(g)

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To tackle ΔS° for this reaction, begin by listing the standard molar entropies (S°) for each species given: H2O(g) = 188.83 J·mol⁻¹·K⁻¹, O2(g) = 205.0 J·mol⁻¹·K⁻¹, CO2(g) = 213.6 J·mol⁻¹·K⁻¹, CH3OH(g) = 237.6 J·mol⁻¹·K⁻¹. The reaction is: 2 CH3OH(g) + 3 O2(g) → 2 CO2(g) + 4 H2O(g).

Step 1: Compute total S° for products. There are 2 moles of CO2 and 4 moles of H2O.
- S° for 2 CO2: 2 × 213.6 = 427.2 J·mol⁻¹·K⁻¹.
- S° for 4 H2O: 4 × 188.83 = 755.32 J·mol⁻¹·K⁻¹.
- Total S°(products) = 427.2 + 755.32 = 1182.52 J·mol⁻¹·K⁻¹.

Step 2: Compute total S° for reactants. There are 2 moles of CH3OH and 3 moles of O2.
- S° for 2 CH3OH: 2 × 237.6 = 475.2 J·mol⁻¹·K⁻¹.
- S° for 3 O2: 3 × 205.0 = 615.0 J·mol⁻¹·K⁻¹.
- Total S°(reactants) = 475.2 + 615.0 = 1090.2 J·mol⁻¹·K⁻¹.

Step 3: Find ΔS° for the reaction using ΔS° = S°(products) − S°(reactants).
- ΔS° = 1182.52 − 1090.2 = 92.32 J·mol⁻¹·K⁻¹.
- Rounding appropriately, ΔS° ≈ 92.3 J·mol⁻¹·K⁻¹.

Option-by-option analysis:
- a. 40.2: This value is not supported by the computed difference between product and reactant entropies. It could arise from miscounting moles or mis-summing S° values, but it does not match the correct total change.
- b. -92.3: The sign would imply a decrease in entropy, which contradicts the calculation here where products have more moles and/or higher per-mole S° than reactants in total.
- c. 92.3: This matches the computed ΔS° value, indicating a net increase in entropy for the reaction under standard conditions.
- d. -40.2: This value is neither the correct magnitude nor the correct sign based on the entropy changes of the species involved and the stoichiometry.

In summary, by summing the product-side entropies and subtracting the sum of the reactant-side entropies, the calculated ΔS° is approximately +92.3 J·mol⁻¹·K⁻¹, reflecting a net increase in disorder for this reaction.

CHM1051 MUM S2 2025 CHM1051 Practice Exam 2

Using the following So values H2O(g), So = 188.83 J·mol-1·K-1 O2(g), So = 205.0 J·mol-1·K-1 CO2(g), So = 213.6 J·mol-1·K-1 CH3OH(g), So = 237.6 J·mol-1·K-1 Calculate ΔSo in J mol-1 K-1 for the following reaction. 2CH3OH(g) + 3O2(g) → 2CO2(g) + 4H2O(g)

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