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单项选择题
A
Q = 1
B
K > Q
C
K = Q
D
K = 1
E
K < Q

Question

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单项选择题

A

Q = 1

B

K > Q

C

K = Q

D

K = 1

E

K < Q

BlackTom AI · Accepted Answer

The thermodynamic relation ΔG = ΔG° + RT ln Q directly applies. Here ΔG equals ΔG°, so substitute and solve:

1) Relevant concepts:
- ΔG is the Gibbs free energy under current conditions.
- ΔG° is the standard Gibbs free energy (when reactants and products are in their standard states).
- Q is the reaction quotient, and RT ln Q quantifies how far the current mixture differs from standard-state conditions.
- Temperature T > 0 and R (gas constant) > 0.

2) Step-by-step algebra:
ΔG = ΔG° + RT ln Q
Given ΔG = ΔG°, set them equal:
ΔG° = ΔG° + RT ln Q
Subtract ΔG° from both sides:
0 = RT ln Q
Divide both sides by RT (nonzero for any physical temperature):
ln Q = 0
Exponentiate both sides:
Q = e^0 = 1

3) Intermediate steps shown above: RT ln Q = 0 ⇒ ln Q = 0 ⇒ Q = 1. No numerical values for R or T are needed because the factor RT is nonzero and cancels.

4) Why this is correct:
The equation shows that the only way ΔG can equal ΔG° is if the RT ln Q term vanishes, which requires Q = 1. This conclusion is independent of the value of ΔG° or the equilibrium constant K.

5) Why the other options are incorrect:
- K > Q or K < Q: These compare K and Q (which determine reaction spontaneity direction relative to equilibrium) but cannot be inferred from ΔG = ΔG°; that equality gives no information about K or equilibrium position.
- K = Q: That would mean the system is at equilibrium (ΔG = 0). But ΔG = ΔG° does not imply ΔG = 0 unless ΔG° = 0; we are not given that.
- K = 1: No information is provided about K; only Q is fixed here.

Conclusion: The correct conclusion is Q = 1, which matches the provided answer.

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单项选择题

类似问题

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

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 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单项选择题

类似问题

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

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