Question at position 13 Which of the following results in a decrease in the entropy of the system? N2(g) at 25 °C → N2(aq) at 25 °CHe(g) at 2.5 atm → He(g) 1.1 atmO2(g) at 300 K → O2(g) at 400 K2H2O2(g) → 2H2O(g) + O2(g)H2O(s) at 0 °C → H2O(l) at 0 °CNH3(l) at −34.5 °C → NH3(g) at −34.5 °C

Question

BlackTom AI · Accepted Answer

Entropy measures the number of accessible microstates (disorder); processes that restrict translational or configurational freedom produce a negative ΔS. Gas dissolution into a solvent strongly confines molecules and therefore lowers entropy.

1) Relevant concepts
- Entropy (S) increases with more accessible microstates: higher temperature, phase change to gas, increase in moles of gas, or expansion typically give ΔS > 0.
- For ideal gases under pressure change: ΔS = nR ln(Pi/Pf) (or nR ln(Vf/Vi)).
- For temperature changes at constant pressure: ΔS = nCp ln(Tf/Ti).
- For phase changes: ΔS = ΔHtransition / T (positive for melting/vaporization).
- Dissolution of a gas into a liquid usually decreases entropy because gas molecules lose translational freedom and become solvated.

2) Step-by-step reasoning for each option
- N2(g) at 25 °C → N2(aq) at 25 °C: Gas molecules become solvated and confined; number of accessible microstates decreases → ΔS < 0.
- He(g) at 2.5 atm → He(g) 1.1 atm: Pressure decreases (expansion). Per mole ΔS = R ln(Pi/Pf) = 8.314 J·mol−1·K−1 × ln(2.5/1.1) ≈ 8.314 × 0.821 ≈ +6.8 J·K−1 (positive).
- O2(g) 300 K → 400 K: Temperature rise increases entropy. Using Cp,m ≈ (7/2)R ≈ 29.1 J·mol−1·K−1, ΔS = Cp ln(400/300) ≈ 29.1 × ln(1.333) ≈ 29.1 × 0.288 ≈ +8.4 J·K−1.
- 2H2O2(g) → 2H2O(g) + O2(g): Gas moles increase from 2 to 3 → more microstates; ΔS > 0 (rough estimate per reaction: R ln(3/2) > 0).
- H2O(s) 0 °C → H2O(l) 0 °C: Melting increases entropy. ΔS = ΔHfus/T ≈ 6.01 kJ·mol−1 / 273.15 K ≈ +22.0 J·K−1.
- NH3(l) −34.5 °C → NH3(g) −34.5 °C: Vaporization at boiling point increases entropy (liquid → gas increases disorder) → ΔS > 0.

3) Calculations/intermediate steps shown above for He and O2 and melting to demonstrate sign and magnitudes.

4) Why the correct answer is correct
Dissolving N2(g) into water localizes and solvates N2 molecules, drastically reducing translational freedom and accessible microstates compared with the gaseous state. That confinement produces a negative ΔS.

5) Why other options are incorrect (brief)
Each other process either increases temperature, expands a gas (lower pressure), increases gas moles, or converts solid/liquid to gas/liquid, all of which increase disorder and give ΔS > 0.

6) Conclusion
Therefore the only process listed that results in a decrease in system entropy is N2(g) at 25 °C → N2(aq) at 25 °C, which matches the provided correct answer.

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 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 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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