阿摩線上測驗
23. 在恆定的溫度(T)和壓力(P)下，根據吉布斯自由能公式，請問以下哪一組熱力學參數組合會 導致反應呈現非自發性行為？
(A) ΔH = +25 kJ, ΔS = +100 J/K, T = 300 K
(B) ΔH = −10 kJ, ΔS = +5.0 J/K, T = 298 K
(C) ΔH = −10 kJ, ΔS = −40 J/K, T = 200 K
(D) ΔH = +25 kJ, ΔS = +5.0 J/K, T = 300 K
答案：登入後查看
統計： A(5), B(14), C(26), D(136), E(0) #3430573
統計： A(5), B(14), C(26), D(136), E(0) #3430573
詳解 (共 1 筆)
陳宗榮
#6601697
布斯自由能公式 =>ΔG = ΔH - TΔS
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ΔG<0 時為自發性反應
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(A) ΔH = +25 kJ, ΔS = +100 J/K =+ 0.1kJ/K, T = 300 K
25 - (300*0.1) = -5 kJ< 0
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(B) ΔH = −10 kJ, ΔS = +5.0 J/K =+ 0.005kJ/K, T = 298 K
-10 - (298*0.005) =-11.49 kJ < 0
(C) ΔH = −10 kJ, ΔS = −40 J/K =- 0.04kJ/K, T = 200 K
-10 - (200*-0.04) = -2 kJ < 0
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(D) ΔH = +25 kJ, ΔS = +5.0 J/K=+ 0.005kJ/K, T = 300 K
25 - (300 * 0.005) = 23.5 > 0
答案 (D)
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