Thermodynamics — ΔG ΔH ΔS + Hess's Law
Thermodynamics & Thermochemistry
45
JEE Qs
12%
Hard
75
min
Master the application of ΔG = ΔH - TΔS and Hess's Law with careful attention to signs, units, and standard state conditions to accurately predict spontaneity and calculate reaction energies.
🧮 Key Formulas
✅ Key Points for JEE
- 1Hess's Law enables calculation of enthalpy (and similarly, entropy and Gibbs energy) changes for complex reactions by algebraically combining known reaction equations and their respective thermodynamic values.
- 2The sign of ΔG (Gibbs Free Energy change) is the definitive criterion for spontaneity: ΔG < 0 (spontaneous), ΔG > 0 (non-spontaneous, reverse is spontaneous), ΔG = 0 (equilibrium).
- 3Predicting spontaneity requires careful consideration of the signs of ΔH and ΔS, and the temperature (T in Kelvin) in the equation ΔG = ΔH - TΔS, as spontaneity can be temperature-dependent.
- 4Standard state conditions (1 bar pressure, 298 K, 1 M concentration for solutions) are essential for determining standard thermodynamic values (ΔH°, ΔS°, ΔG°).
- 5The equilibrium constant (K_eq) is quantitatively linked to the standard Gibbs free energy change (ΔG° = -RT ln K_eq), providing insight into the extent of a reaction at equilibrium.
⚠️ Common Mistakes
- ✕Failing to convert temperature to Kelvin (T) when using the equation ΔG = ΔH - TΔS, leading to incorrect calculations of spontaneity.
- ✕Incorrectly applying Hess's Law by forgetting to multiply the thermodynamic values (ΔH, ΔS, ΔG) when multiplying reaction coefficients, or not changing the sign when reversing a reaction.
- ✕Confusing ΔS_system, ΔS_surroundings, and ΔS_total; only ΔS_total (or ΔG_system) determines the overall spontaneity of a process.
- ✕Mismatched units (e.g., Joules for ΔS and kilojoules for ΔH) in calculations without proper conversion, which results in significant errors.
- ✕Incorrectly interpreting the meaning of ΔG values, especially confusing spontaneity under standard conditions (ΔG°) with spontaneity under non-standard conditions (ΔG).
📝 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