Gibbs Free Energy — ΔG = ΔH - TΔS
Thermodynamics & Thermochemistry
9
JEE Qs
8%
Hard
75
min
Master the spontaneity conditions based on ΔH, ΔS, and T, and deeply understand the relationship between ΔG° and K_eq for predicting reaction feasibility and equilibrium position.
🧮 Key Formulas
✅ Key Points for JEE
- 1Gibbs Free Energy (ΔG) is the ultimate criterion for spontaneity: ΔG < 0 for spontaneous, ΔG > 0 for non-spontaneous, ΔG = 0 for equilibrium.
- 2The sign of ΔG depends on the signs of ΔH and ΔS, and importantly, on temperature (T) as TΔS term dominates at higher temperatures.
- 3For standard conditions, ΔG° relates directly to the equilibrium constant (K_eq), providing a quantitative measure of reaction extent.
- 4ΔG represents the maximum useful (non-PV) work that can be extracted from a system at constant temperature and pressure.
- 5Remember to always use temperature (T) in Kelvin and ensure consistent units for ΔH and ΔS (e.g., kJ/mol and J/mol.K respectively, converting one before calculation).
⚠️ Common Mistakes
- ✕Confusing standard Gibbs free energy (ΔG°) with non-standard Gibbs free energy (ΔG); ΔG° is for standard conditions, while ΔG applies to any conditions.
- ✕Incorrectly using units, especially temperature in Celsius instead of Kelvin, or mixing kJ and J without proper conversion for ΔH and TΔS terms.
- ✕Assuming ΔH and ΔS are always constant over a wide temperature range; while often approximated in JEE, their temperature dependence can be a subtle point.
📝 Practice Questions
See allQ32.Given are statements for certain thermodynamic variables, (A) Internal energy, volume (V) and mass (M) are extensive variables. (B) Pressure (P), temperature (T) and density ( ρ ) are intensive variables. (C) Volume (V), temperature (T) and density ( ρ ) are intensive variables. (D) Mass (M), temperature (T) and internal energy are extensive variables. Choose the correct answer from the options given below : (1) (B) and (C) Only (2) (C) and (D) Only (3) (D) and (A) Only (4) (A) and (B) Only
Q64.Ice at −5∘C is heated to become vapor with temperature of 110∘C at atmospheric pressure. The entropy change associated with this process can be obtained from 2025 (23 Jan Shift 1) JEE Main Previous Year Paper (1) (2) (3) (4)
Q53.A liquid when kept inside a thermally insulated closed vessel at 25∘C was mechanically stirred from outside. What will be the correct option for the following thermodynamic parameters ? (1) ΔU < 0, q = 0, w > 0 (2) ΔU = 0, q = 0, w = 0 (3) ΔU > 0, q = 0, w > 0 (4) ΔU = 0, q < 0, w > 0
Q31.Water of mass m gram is slowly heated to increase the temperature from T1 to Tz The change in entropy of the water, given specific heat of water is 1Jkg−1 K−1 , is : (1) m ln ( T2T1 ) (2) zero (3) m ln ( T1T2 ) (4) m (T2 −T1)
Q74.Niobium ( Nb ) and ruthenium (Ru) have " x " and " y " number of electrons in their respective 4 d orbitals. The value of x + y is ______ -.
Q70.The correct stability order of the following species/molecules is: (1) q > r > p (2) r > q > p (3) q > p > r (4) p > q > r Q71. 1 The standard enthalpy and standard entropy of decomposition of N2O4 to NO2 are 55.0 kJ mol−1 and 175.0 J/K/mol respectively. The standard free energy change for this reaction at 25∘C in J mol−1 is ______ (Nearest integer)
NCERT Chapters
- Class 11 Chemistry Ch 6: Thermodynamics