Question at position 29 The value of ∆H° = –416 kJ and the value of ∆S° = –209 J/K for the reaction: P4O10(s) + 6H2O(l) → 4H3PO4(s) Calculate ∆G° for this reaction at 15 °C. –413 kJ2720 kJ–476 kJ–356 kJ单项选择题
A
–413 kJ
B
2720 kJ
C
–476 kJ
D
–356 kJ

Question

Question at position 29 The value of ∆H° = –416 kJ and the value of ∆S° = –209 J/K for the reaction: P4O10(s) + 6H2O(l) → 4H3PO4(s) Calculate ∆G° for this reaction at 15 °C. –413 kJ2720 kJ–476 kJ–356 kJ单项选择题

A

–413 kJ

B

2720 kJ

C

–476 kJ

D

–356 kJ

BlackTom AI · Accepted Answer

ΔG° = ΔH° − TΔS°, so compute using consistent units (kJ and K).

1. Relevant concepts
- Standard Gibbs free energy: ΔG° = ΔH° − TΔS°.
- Temperature must be in kelvin; entropy units must match enthalpy units (convert J to kJ).

2. Step-by-step reasoning
- Given: ΔH° = −416 kJ, ΔS° = −209 J·K−1, T = 15 °C.
- Convert temperature: T = 15 + 273.15 = 288.15 K.
- Convert entropy to kJ·K−1: ΔS° = −209 J·K−1 × (1 kJ / 1000 J) = −0.209 kJ·K−1.
- Compute TΔS°: TΔS° = 288.15 K × (−0.209 kJ·K−1).

3. Calculations (all intermediate steps)
- 0.209 × 288.15 = (0.200 × 288.15) + (0.009 × 288.15)
  = 57.63 + 2.59335 = 60.22335 kJ.
- Because ΔS° is negative, TΔS° = −60.22335 kJ.
- ΔG° = ΔH° − TΔS° = (−416.00000 kJ) − (−60.22335 kJ)
  = −416.00000 kJ + 60.22335 kJ = −355.77665 kJ.

4. Why the correct answer is correct
- Rounding to three significant figures (matching given data) gives ΔG° ≈ −356 kJ. The negative ΔG° indicates the reaction is spontaneous under standard conditions at 15 °C. Because ΔS° is negative, the −TΔS term is positive, making ΔG° less negative than ΔH° (i.e., −356 kJ is less negative than −416 kJ).

5. Why other options are incorrect
- −413 kJ: close but ignores correct magnitude of the TΔS correction (would correspond to a much smaller entropy effect).
- 2720 kJ: wrong sign and magnitude (entropy/temperature sign would not produce large positive value).
- −476 kJ: more negative than ΔH°, which would require a positive ΔS° (not the case here); thus incorrect.

6. Conclusion
Final calculated ΔG° ≈ −355.78 kJ, rounded to −356 kJ, which matches the provided answer: −356 kJ.

Question at position 29 The value of ∆H° = –416 kJ and the value of ∆S° = –209 J/K for the reaction: P4O10(s) + 6H2O(l) → 4H3PO4(s) Calculate ∆G° for this reaction at 15 °C. –413 kJ2720 kJ–476 kJ–356 kJ单项选择题

类似问题

Question at position 28 A reaction proceeds toward equilibrium. At a certain point in the reaction, the concentrations of reactants and products are such that ΔG = ΔG°. What conclusion can reasonably be drawn about the reaction at this time? Q = 1K > QK = QK = 1K < Q

Question at position 26 The equilibrium constant for a reaction is 0.48 at 25 ºC. What is the value of ΔGº (kJ/mol) at this temperature? –4.2 kJ/mol 4.2 kJ/mol1.5 x 102 kJ/mol1.8 kJ/molMore information is needed

Question at position 25 A reaction has a Gibbs free energy change of –55 kJ/mol at 80 °C. What must be true? The reaction is at equilibrium.The reaction is non-spontaneous at all temperatures.The reaction is spontaneous at 80 °C.The reaction occurs rapidly.

Question at position 13 Consider the following reaction: 3NO2(g) + H2O(l) ⇌ 2HNO3(l) + NO(g) ∆G° = 9.4 kJ Calculate ∆G for this reaction at 25 °C under the following conditions: partial pressure of NO2 = 0.115 atm and partial pressure of NO = 9.70 atm. 20.4 kJ43.2 kJ11.2 kJ8.2 kJ31.1 kJ

Question at position 7 Which of the following does not have ∆Gof = 0 H2(g)Ca(s)N2(l)O2(g)Hg(l)

Question at position 4 For a reaction to be spontaneous at all temperatures, the signs of ΔHº and ΔSº must be and , respectively. negative, negativepositive, negativenegative, positivepositive, positivepositive, zero

Question at position 28 Predict the signs of ∆G, ∆H, and ∆S for the spontaneous combustion of octane. C8H14(l) + 23 O2(g) → 16 CO2(g) + 14 H2O(g) ∆G < 0, ∆H > 0, ∆S > 0 ∆G < 0, ∆H < 0, ∆S > 0 ∆G > 0, ∆H > 0, ∆S < 0 ∆G < 0, ∆H < 0, ∆S < 0 ∆G > 0, ∆H > 0, ∆S > 0

更多留学生实用工具

智能学习助手

风格化写作助手

论文查重助手

文献引用助手

课堂转译助手

课堂笔记助手

Quiz搜索助手

学校历年真题

智能刷题助手

智能匹配练习

希望你的学习变得更简单

Question at position 29 The value of ∆H° = –416 kJ and the value of ∆S° = –209 J/K for the reaction: P4O10(s) + 6H2O(l) → 4H3PO4(s) Calculate ∆G° for this reaction at 15 °C. –413 kJ2720 kJ–476 kJ–356 kJ单项选择题

类似问题

Question at position 28 A reaction proceeds toward equilibrium. At a certain point in the reaction, the concentrations of reactants and products are such that ΔG = ΔG°. What conclusion can reasonably be drawn about the reaction at this time? Q = 1K > QK = QK = 1K < Q

Question at position 26 The equilibrium constant for a reaction is 0.48 at 25 ºC. What is the value of ΔGº (kJ/mol) at this temperature? –4.2 kJ/mol 4.2 kJ/mol1.5 x 102 kJ/mol1.8 kJ/molMore information is needed

Question at position 25 A reaction has a Gibbs free energy change of –55 kJ/mol at 80 °C. What must be true? The reaction is at equilibrium.The reaction is non-spontaneous at all temperatures.The reaction is spontaneous at 80 °C.The reaction occurs rapidly.

Question at position 13 Consider the following reaction: 3NO2(g) + H2O(l) ⇌ 2HNO3(l) + NO(g) ∆G° = 9.4 kJ Calculate ∆G for this reaction at 25 °C under the following conditions: partial pressure of NO2 = 0.115 atm and partial pressure of NO = 9.70 atm. 20.4 kJ43.2 kJ11.2 kJ8.2 kJ31.1 kJ

Question at position 7 Which of the following does not have ∆Gof = 0 H2(g)Ca(s)N2(l)O2(g)Hg(l)

Question at position 4 For a reaction to be spontaneous at all temperatures, the signs of ΔHº and ΔSº must be__________ and __________, respectively. negative, negativepositive, negativenegative, positivepositive, positivepositive, zero

Question at position 28 Predict the signs of ∆G, ∆H, and ∆S for the spontaneous combustion of octane. C8H14(l) + 23 O2(g) → 16 CO2(g) + 14 H2O(g) ∆G < 0, ∆H > 0, ∆S > 0 ∆G < 0, ∆H < 0, ∆S > 0 ∆G > 0, ∆H > 0, ∆S < 0 ∆G < 0, ∆H < 0, ∆S < 0 ∆G > 0, ∆H > 0, ∆S > 0

更多留学生实用工具

希望你的学习变得更简单

Question at position 4 For a reaction to be spontaneous at all temperatures, the signs of ΔHº and ΔSº must be and , respectively. negative, negativepositive, negativenegative, positivepositive, positivepositive, zero