1. Fundamental Concepts
- Gibbs free energy (ΔG) is a thermodynamic quantity that combines enthalpy (ΔH) and entropy (ΔS) to determine whether a reaction is spontaneous under constant temperature and pressure. (ΔG represents the maximum amount of useful work that can be obtained from a process at constant temperature and pressure.)
- Equation: , where is the change in enthalpy, is the temperature in Kelvin, and is the change in entropy.
2. Key Concepts
Spontaneity Criteria:
ΔG < 0: The process is spontaneous.
ΔG = 0: The system is at equilibrium.
ΔG > 0: The process is non-spontaneous.
Relationship Between ΔH, ΔS, and Temperature:
ΔH < 0, ΔS > 0: spontaneous at all temperatures
ΔH < 0, ΔS < 0: spontaneous at low temperature
ΔH > 0, ΔS > 0: spontaneous at high temperature
ΔH > 0, ΔS < 0: never spontaneous
Interpretation of ΔG Values:
Large negative ΔG → strongly product-favored reaction
Small negative ΔG → reaction occurs but may not go to completion
Large negative ΔG → strongly product-favored reaction
Small negative ΔG → reaction occurs but may not go to completion
3. Examples
Example 1 (Basic)
Problem: Calculate for a reaction with and at .
Step-by-Step Solution:
- Convert to kJ/K:
- Use the Gibbs Free Energy equation:
- Substitute the values:
Example 2 (Medium)
Problem: For the reaction : and .
Determine the temperature above which this reaction becomes spontaneous. (Give your answer to the nearest whole number in Kelvin.)
Determine the temperature above which this reaction becomes spontaneous. (Give your answer to the nearest whole number in Kelvin.)
Step-by-Step Solution:
- Convert to kJ/K:
- To find the temperature above which the reaction becomes spontaneous, use the condition: A reaction is spontaneous when ( ). Therefore,
- Solve for temperature:
- Calculate :
- Final Answer: The reaction becomes spontaneous above 1112 K.
Example 3 (Hard)
Problem: Two liquids, A and B, have the following properties:
Substance A: , boiling point = 350 K
Substance B: , boiling point = 350 K
Substance B: , boiling point = 350 K
((a) Calculate the entropy of vaporization ( ) for each substance. (Give your answer to the nearest whole number.)
(b) At 300 K, determine which substance is more likely to evaporate spontaneously. Assume that and are constant over this temperature range.
Justify your answer using thermodynamic reasoning.
(b) At 300 K, determine which substance is more likely to evaporate spontaneously. Assume that and are constant over this temperature range.
Justify your answer using thermodynamic reasoning.
Step-by-Step Solution:
- At the boiling point, vaporization is at equilibrium, so,
- Calculate
Substance A:
Substance B: - Determine which substance is more likely to evaporate spontaneously at 300 K (compare ΔG):
Substance A:
Substance B:
Both ΔG values are positive → neither evaporates spontaneously at 300 K, but A has the smaller ΔG, so it is closer to spontaneous. - Final answer:
(a)A: , B:
(b)Substance A is more likely to evaporate spontaneously at 300 K.
4. Problem-Solving Techniques
- Unit Conversion: Always ensure that units are consistent (e.g., convert J to kJ, °C to K if necessary).
- Always Check Temperature Units: Temperature must be in Kelvin
- Memorize how ΔH, ΔS, and T together determine spontaneity: Practice predicting spontaneity by analyzing the signs of ΔH and ΔS without doing full calculations.
- Connect ΔG to Equilibrium
ΔG < 0 → reaction proceeds forward
ΔG > 0 → reaction proceeds in reverse
ΔG = 0 → equilibrium