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单项选择题
A
–4.2 kJ/mol
B
4.2 kJ/mol
C
1.5 x 102 kJ/mol
D
1.8 kJ/mol
E
More information is needed

Question

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单项选择题

A

–4.2 kJ/mol

B

4.2 kJ/mol

C

1.5 x 102 kJ/mol

D

1.8 kJ/mol

E

More information is needed

BlackTom AI · Accepted Answer

ΔG° = -RT ln K, so we can compute ΔG° directly from the equilibrium constant.

1) Relevant knowledge points and constants
- Relationship: ΔG° = -RT ln K (R = gas constant, T = absolute temperature).
- Use R = 8.314 J·mol⁻¹·K⁻¹ and T = 25 °C = 298.15 K.
- Convert final answer to kJ·mol⁻¹ (1 kJ = 1000 J).

2) Step-by-step reasoning
- Start with K = 0.48.
- Compute ln K: ln(0.48) = -0.733969 (negative because K < 1).
- Plug into ΔG° = -RT ln K.

3) Calculations with intermediate steps
- R·T = 8.314 J·mol⁻¹·K⁻¹ × 298.15 K = 2478.819 J·mol⁻¹.
- Multiply by ln K (absolute value): 2478.819 × (−0.733969) = −1818.173 J·mol⁻¹.
- Apply the negative sign in formula: ΔG° = −(2478.819 × −0.733969) = +1818.173 J·mol⁻¹.
- Convert to kJ: 1818.173 J·mol⁻¹ ÷ 1000 = 1.818 kJ·mol⁻¹ ≈ 1.8 kJ·mol⁻¹.

4) Why the correct answer is correct
Because K < 1, ln K is negative; the product −RT ln K is therefore positive. The numerical evaluation using standard R and T gives +1.818 kJ·mol⁻¹, which rounds to 1.8 kJ·mol⁻¹. This matches the provided correct choice.

5) Why the other options are incorrect
- −4.2 kJ·mol⁻¹: negative ΔG° would require K > 1 (spontaneous in standard direction); not consistent with K = 0.48.
- 4.2 kJ·mol⁻¹: magnitude about twice the correct value; does not match the calculated product −RT ln K.
- 1.5 × 10^2 kJ·mol⁻¹: far too large given the modest deviation of K from 1.
- "More information is needed": not true, because ΔG° is directly determined by K and T (both given).

6) Conclusion
The calculated standard free energy change is approximately +1.82 kJ·mol⁻¹, so the correct answer is 1.8 kJ/mol, matching the provided answer.

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 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 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

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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 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 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

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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