U9.2 · Absolute Entropy and Entropy ChangePractice · 绝对熵与熵变
Lessons covered:U9.1 Introduction to EntropyU9.2 Absolute Entropy and Entropy Change
- 1Multiple choice
At 25 °C, which of these has the largest standard molar entropy, S°?
ExplanationS° (J mol⁻¹ K⁻¹): NO₂(g) 240.1, NH₃(g) 192.8, NH₄NO₃(s) 151.08, NH₄Cl(s) 94.6. Gases have far more entropy than solids; of the two gases, NO₂ is heavier with more atoms (more ways to vibrate), so its S° is larger. The bigger ionic solid NH₄NO₃ beats NH₄Cl, but both are far below the gases.
- 2Numeric
Calculate ΔS° for the catalytic oxidation of ammonia, 4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g). S° (J mol⁻¹ K⁻¹): NH₃(g) 192.8, O₂(g) 205.2, NO(g) 210.8, H₂O(g) 188.8.
J/KExplanationΔS° = (4 × 210.8 + 6 × 188.8) − (4 × 192.8 + 5 × 205.2) = 1976.0 − 1797.2 = +178.8 J/K. Nine gas molecules become ten, so ΔS° is positive, as predicted.
- 3Numeric
Calculate ΔS° for CO(g) + 2H₂(g) → CH₃OH(l). S° (J mol⁻¹ K⁻¹): CO(g) 197.7, H₂(g) 130.7, CH₃OH(l) 126.8.
J/KExplanationΔS° = 126.8 − (197.7 + 2 × 130.7) = 126.8 − 459.1 = −332.3 J/K. Three gas molecules become one liquid molecule, a big drop in entropy.
- 4Numeric
For N₂(g) + 3H₂(g) → 2NH₃(g), ΔS° = −198.1 J/K. What is the entropy change of the system when 3.40 g of NH₃ (0.200 mol) is made from N₂ and H₂?
J/KExplanationΔS° is for the reaction as written, which makes 2 mol of NH₃. Making 0.200 mol gives −198.1 × 0.200 ÷ 2 = −19.8 J/K. Multiplying by 0.200 directly doubles it.
- 5Multiple choice
Elements have a standard enthalpy of formation of zero, but their standard molar entropy S° is not zero. Why?
ExplanationEnthalpy has no absolute zero, so a reference is chosen: elements in their standard states are set to zero. Entropy does have an absolute zero: a perfect crystal at 0 K (the third law). Above 0 K every substance, including solid elements such as Fe(s) with S° = 27.3 J mol⁻¹ K⁻¹, has positive entropy.
- 6Multiple choice
The graph shows the standard entropy of 1 mol of water against temperature (1 bar). The curve jumps at two temperatures. Which jump is larger, and why?
ExplanationReading the graph: the jump at 273 K is about 22 J/(mol·K) (ΔSfus = 6010 ÷ 273), and at 373 K about 109 J/(mol·K) (ΔSvap = 40,700 ÷ 373), about 5 times as large. Liquid molecules are still touching; as a gas they spread into a far larger volume, so matter and energy become much more dispersed. The smooth rises show that entropy also grows with temperature within one phase, but much less than at a phase change, so “higher temperature always means a bigger rise” isn’t the reason.
Review: U9.2 Absolute Entropy and Entropy ChangeU9.1 Introduction to Entropy
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