Question at position 13 For which of the following reactions will ΔS be negative? 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g) 2H2S(g) + 3O2(g) → 2H2O(l) + 2SO2(g) Na2CO3(s) + 2 HCl(aq) → 2 NaCl(aq) + H2O(l) + CO2(g) CaSO3(s) → CaO(s) + SO2(g) 2NH3(g) → N2(g) + 3H2(g)

Question

BlackTom AI · Accepted Answer

Entropy generally increases with more gas molecules and with phase changes that produce more dispersed states (solid → liquid → gas). For each reaction, compare numbers of gas moles and note phase changes to decide sign of ΔS.

1) Relevant concepts
- Entropy (S) increases with the number of microstates: more gas molecules and formation of aqueous/ gaseous species from solids/liquids usually raises S.
- Phase changes: gas >> aqueous > liquid > solid in entropy magnitude.
- A decrease in total gas moles or formation of liquids/solids from gases tends to give ΔS < 0.

2) Step-by-step reasoning (count gases and phases)
Reaction A: 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)
- Reactants: solids + liquid, products: aqueous ions + 1 mol gas.
- Change: creation of dissolved ions and a gas increases dispersal → ΔS positive.

Reaction B: 2H2S(g) + 3O2(g) → 2H2O(l) + 2SO2(g)
- Reactants: 2 + 3 = 5 mol gas.
- Products: 2 mol SO2(g) + 2 mol H2O(l) (liquid).
- Change: gas moles drop from 5 → 2 and gaseous H2S and O2 become liquid H2O (large entropy loss) → ΔS negative.

Reaction C: Na2CO3(s) + 2HCl(aq) → 2NaCl(aq) + H2O(l) + CO2(g)
- Reactants: no gas; Products: 1 mol gas (CO2) + aqueous species.
- Change: formation of a gas and dissolution generally raises entropy → ΔS positive.

Reaction D: CaSO3(s) → CaO(s) + SO2(g)
- Reactants: solid; Products: solid + 1 mol gas → increased disorder → ΔS positive.

Reaction E: 2NH3(g) → N2(g) + 3H2(g)
- Reactants: 2 mol gas → Products: 4 mol gas (increase in gas molecules) → ΔS positive.

3) Calculations / intermediates: gas mole counts shown above (Reactants vs Products): only B shows a decrease (5 → 2).

4) Why the correct answer is correct
Reaction B produces fewer gas molecules and converts gaseous reactants into liquid water, causing a significant decrease in molecular disorder and hence a negative ΔS.

5) Why other options are incorrect
A, C, D, E all either produce gas, produce dissolved ions, or increase the number of gas molecules—each increases disorder so ΔS is positive.

6) Conclusion
Thus the only reaction with ΔS negative is 2H2S(g) + 3O2(g) → 2H2O(l) + 2SO2(g), matching the provided answer.

Question at position 13 For which of the following reactions will ΔS be negative? 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g) 2H2S(g) + 3O2(g) → 2H2O(l) + 2SO2(g) Na2CO3(s) + 2 HCl(aq) → 2 NaCl(aq) + H2O(l) + CO2(g) CaSO3(s) → CaO(s) + SO2(g) 2NH3(g) → N2(g) + 3H2(g)单项选择题

类似问题

Question at position 10 For which of the following processes will ∆S be negative? 2NH3(g) → N2(g) + 3H2(g)CaSO3(s) → CaO(s) + SO2(g)Br2(solid) → Br2(liquid)2H2S(g) + 3O2(g) → 2H2O(l) + 2SO2(g)

Question at position 12 Calculate ΔS° for the reaction 4Cr(s) + 3O2(g) → 2Cr2O3(s) [table] Substance | Cr(s) | O2(g) | Cr2O3(s) S°(J/K·mol) | 23.77 | 205.138 | 81.2 [/table] 147.7 J/(K mol)−147.7 J/(K mol)548.1 J/(K mol)−548.1 J/(K mol)310.1 J/(K mol)

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)

Calculate the value of S° for the reaction shown: 2 NH3(g) N2(g) + 3 H2(g) At 25°C the values of entropy in J K−1 mol−1 are ammonia, 192.77; nitrogen, 191.61; and hydrogen, 130.68.

Propanol (C3H8O) has a boiling point of 97.0 ºC and ΔHvap =42.27 kJ mol-1. Calculate ΔS for the process given below. CH3CH2CH2OH (l) → CH3CH2CH2OH (g)

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