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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We are asked to calculate the standard molar entropy change, ΔS°, for the reaction 2 CH3OH(g) + 3 O2(g) → 2 CO2(g) + 4 H2O(g) using the given S° values.
First, list the S° values and multiply by stoichiometric coefficients for reactants and products.
Reactants: 2 CH3OH(g) with S° = 237.6 J/mol·K gives 2 × 237.6 = 475.2 J/mol·K. For O2(g) with S° = 205.0 J/mol·K, 3 × 205.0 = 615.0 J/mol·K. Total reactant entropy contribution = 475.2 + 615.0 = 1090.2 J/mol·K.
Products: 2 CO2(g) with S° = 213.6 J/mol·K gives 2 × 213.6 = 427.2 J/mol·K. For H2O(g) with S° = 188.83 J/mol·K, 4 × 188.83 = 755.32 J/mol·K. Total product entropy contribution = 427.2 + 755.32 = 1182.52 J/mol·K.
Now compute ΔS° for the reaction using ΔS° = ΣS°(products) − ΣS°(reactants).
So ΔS° = 1182.52 − 1090.2 = 92.32 J/mol·K (approximately 92.3 J/mol·K).
To verify the arithmetic, re-check each multiplication and addition carefully: 2 × 237.6 = 475.2; 3 × 205.0 = 615.0; 2 × 213.6 = 427.2; 4 × 188.83 = 755.32; sum products 427.2 + 755.32 = 1182.52; sum reactants 475.2 + 615.0 = 1090.2; difference 1182.52 − 1090.2 = 92.32.
Consider potential sources of confusion: ensure you are using gas-phase S° values as given, and apply the stoichiometric coefficients to each species correctly; mixing up products and reactants would invert the sign of ΔS°. Also note rounding; 92.32 rounds to about 92.3, which aligns with the computed value.
If you compare with the offered options, you should see a match around +92.3 J·mol⁻¹·K⁻¹, depending on rounding in the intermediate steps.
Incorrect choices would arise from mis-summing the product or reactant contributions, applying wrong signs, or misapplying the stoichiometric coefficients (for example, treating 4 H2O as 2 H2O or forgetting the factor of 2 for CO2, etc.). Misplacing the sign would lead to negative values where the entropy increase is expected for this reaction at standard conditions.
Another common pitfall is ignoring the gas-phase assumption; if a condensed phase value were used for any species, the result would not be comparable to the provided options and would misrepresent the true ΔS° for the gas-phase reaction.

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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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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