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 neededSingle choice
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 neededSingle choice

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 neededSingle choice

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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 neededSingle choice

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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

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