AP Chemistry · Topic 9.5
Free Energy and Equilibrium Practice
Part of Applications of Thermodynamics.(ENE-5.B)
Practice questions
10
Sample questions
5 of 10 — sign in to practice the rest with adaptive difficulty and mastery tracking.
Sample 1difficulty 2/5
ΔG = 0 corresponds to
- Acheck_circle
Equilibrium
- B
Non-spontaneous
- C
Pure forward
- D
Spontaneous
Why
Equilibrium = no net driving force in either direction.
- A
Sample 2difficulty 3/5
The minimum on a G vs reaction extent plot represents:
- A
Where ΔG° = 0
- B
Where reactant only exists
- Ccheck_circle
The equilibrium position where ΔG = 0
- D
Where K = 0
Why
A reaction proceeds toward minimum free energy; at that minimum, the slope (ΔG with respect to ξ) is zero — equilibrium.
- A
Sample 3difficulty 3/5
From ΔG° = -RT ln K, what is ΔG° when K = 1?
- A
ΔG° < 0
- B
Indeterminate without T
- C
ΔG° > 0
- Dcheck_circle
ΔG° = 0
Why
ln(1) = 0, so ΔG° = -RT(0) = 0 regardless of T.
- A
Sample 4difficulty 4/5
Use ΔG° = -RT ln K with R = 8.314 J/(mol·K).
A reaction has ΔG° = -10. kJ/mol at 298 K. What does this imply for K?
- A
K < 1
- Bcheck_circle
K > 1
- C
K = 1
- D
K = 0
Why
ΔG° = -RT ln K; negative ΔG° gives ln K > 0, so K > 1 and products are favored.
- A
Sample 5difficulty 4/5
The van't Hoff plot has a NEGATIVE slope. What does this indicate?
- A
ΔS° = 0
- B
K is independent of T
- C
The reaction is exothermic
- Dcheck_circle
The reaction is endothermic (ΔH° > 0)
Why
ln K = -ΔH°/R · (1/T) + ΔS°/R. A negative slope means -ΔH°/R < 0, i.e. ΔH° > 0 (endothermic).
- A