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

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)

BlackTom AI · Accepted Answer

Entropy measures the number of accessible microstates (disorder); processes that reduce the number of gas particles or convert gases to condensed phases normally produce a negative ΔS. Use molecular counts and phase changes to evaluate sign of ΔS.

1) 2NH3(g) → N2(g) + 3H2(g)
- Reactant gas molecules = 2
- Product gas molecules = 1 + 3 = 4
- Δn(gas) = 4 − 2 = +2 (increase in gas moles → greater disorder)
- Conclusion: ΔS positive → not the correct choice.

2) CaSO3(s) → CaO(s) + SO2(g)
- Reactant: 1 solid; Products: 1 solid + 1 gas
- A gas is produced from a solid, increasing the number of accessible microstates and disorder.
- Conclusion: ΔS positive → not correct.

3) Br2(solid) → Br2(liquid)
- Solid → liquid increases molecular freedom (liquid is more disordered than solid).
- Conclusion: ΔS positive → not correct.

4) 2H2S(g) + 3O2(g) → 2H2O(l) + 2SO2(g)
- Reactant gas molecules = 2 (H2S) + 3 (O2) = 5
- Product gas molecules = 2 (SO2); plus 2 molecules of liquid H2O (liquids have far fewer translational microstates than gases)
- Δn(gas) = 2 − 5 = −3 (net loss of 3 gas molecules)
- Converting gas molecules to liquid (H2O) further reduces disorder.
- Conclusion: ΔS is negative (significantly decreased entropy).

Why the correct answer is correct: the fourth reaction reduces the total number of gas-phase particles and produces liquids from gases, both of which lower the system’s entropy, giving a negative ΔS. The other options either increase gas moles or change to a more disordered phase, so they yield positive ΔS. Final answer: 2H2S(g) + 3O2(g) → 2H2O(l) + 2SO2(g).

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)单项选择题

类似问题

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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