CUET UG Chemistry Physical Chemistry Thermodynamics Hesss Law Gibbs Energy Spontaneity

Must-Know

• Hess’s Law states that the total enthalpy change for a reaction is the same regardless of the pathway, provided initial and final conditions are identical; for example, ΔH for C(s) → CO₂(g) is –393.5 kJ/mol whether it occurs directly or via CO(g) intermediate.
• The standard enthalpy of formation (Δ_fH°) of an element in its most stable form at 298 K and 1 atm is zero; e.g., Δ_fH° of O₂(g), C(graphite), and H₂(g) is 0 kJ/mol.
• For the reaction C(graphite) + ½O₂(g) → CO(g), ΔH = –110.5 kJ/mol; this cannot be measured directly but is calculated using Hess’s Law from combustion data.
• Gibbs energy change (ΔG) determines spontaneity: if ΔG < 0, the process is spontaneous; if ΔG > 0, it is non-spontaneous; if ΔG = 0, the system is at equilibrium.
• ΔG = ΔH – TΔS is the Gibbs equation; used to predict spontaneity from enthalpy and entropy changes.
• A reaction with ΔH < 0 and ΔS > 0 is spontaneous at all temperatures; e.g., combustion of glucose.
• A reaction with ΔH > 0 and ΔS < 0 is non-spontaneous at all temperatures; e.g., formation of ozone from O₂ at ground level.
• A reaction with ΔH < 0 and ΔS < 0 is spontaneous only at low temperatures; e.g., synthesis of ammonia: N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = –92.4 kJ/mol.
• A reaction with ΔH > 0 and ΔS > 0 is spontaneous only at high temperatures; e.g., decomposition of CaCO₃(s) → CaO(s) + CO₂(g), which occurs above 1100 K.
• Standard Gibbs energy of formation (Δ_fG°) of an element in its standard state is zero; e.g., Δ_fG° of Cl₂(g), Fe(s), and O₂(g) is 0 kJ/mol.
• ΔG° = –RT ln K, where K is the equilibrium constant; at equilibrium, ΔG = 0 and Q = K.
• For a reaction with K > 1, ΔG° < 0; for K < 1, ΔG° > 0; e.g., if K = 10, ΔG° = –RT ln(10) ≈ –5.7 kJ/mol at 298 K.
• The combustion of methane: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l), has ΔG° = –818 kJ/mol, confirming spontaneity.
• Entropy (S) increases with molecular complexity and physical state change from solid to liquid to gas; e.g., S° for H₂O(s) < H₂O(l) < H₂O(g).
• Standard molar entropy (S°) of a substance is always positive; e.g., S° of H₂(g) is 130.7 J/mol·K at 298 K.
• ΔS° for a reaction is calculated as ΣS°(products) – ΣS°(reactants); e.g., for N₂(g) + 3H₂(g) → 2NH₃(g), ΔS° = –198.3 J/mol·K.
• For exothermic reactions, ΔH is negative; for endothermic, ΔH is positive; e.g., photosynthesis has ΔH > 0.
• Bond breaking is endothermic (absorbs energy), bond formation is exothermic (releases energy); net ΔH = Σ(bond energies broken) – Σ(bond energies formed).
• The standard enthalpy change for the reaction: 2Al(s) + Fe₂O₃(s) → Al₂O₃(s) + 2Fe(s) is –851.5 kJ/mol, calculated via Hess’s Law using formation data.
• ΔG is a state function, so it obeys Hess’s Law; ΔG°_reaction = ΣΔ_fG°(products) – ΣΔ_fG°(reactants).

Difficulty Level

Intermediate — requires understanding of sign conventions, temperature dependence, and integration of ΔH, ΔS, and ΔG concepts, but avoids complex derivations.

Common CUET Traps

• Trap: Assuming all exothermic reactions are spontaneous.
  Avoid: Check ΔG = ΔH – TΔS; if ΔS is highly negative and T is high, ΔG may be positive even if ΔH is negative.
• Trap: Confusing ΔG and ΔG° — thinking ΔG° < 0 means reaction proceeds to completion.
  Avoid: ΔG° < 0 means K > 1, but equilibrium may still have significant reactants; spontaneity does not imply completeness.
• Trap: Using Δ_fH° values for compounds not in standard states or misapplying Hess’s Law with unbalanced equations.
  Avoid: Ensure all substances are in standard states (1 atm, 298 K) and equations are stoichiometrically balanced before summing ΔH values.

Practice MCQs

1. For a reaction with ΔH = –40 kJ/mol and ΔS = –100 J/mol·K, at what temperature is the reaction at equilibrium?
   
   A) 200 K
   
   B) 300 K
   
   C) 400 K
   
   D) 500 K
   Answer: C
   Explanation: At equilibrium, ΔG = 0 ⇒ T = ΔH/ΔS = (–40,000 J/mol)/(–100 J/mol·K) = 400 K.
   Why others fail: Option B (300 K) is a common miscalculation if signs are ignored or units not converted.

2. Which of the following reactions has ΔG° = 0?
   
   A) H₂(g) → 2H(g)
   
   B) O₂(g) → 2O(g)
   
   C) C(graphite) → C(diamond)
   
   D) Fe(s) at 298 K and 1 atm
   Answer: D
   Explanation: ΔG° = 0 for elements in their standard states; Fe(s) is the standard state of iron.
   Why others fail: Option C is tempting because both are carbon forms, but C(diamond) is not the standard state, so ΔG° ≠ 0.

3. The standard Gibbs energy change for a reaction is –5.7 kJ/mol at 298 K. What is the equilibrium constant (K)?
   
   A) 0.1
   
   B) 1
   
   C) 10
   
   D) 100
   Answer: C
   Explanation: ΔG° = –RT ln K ⇒ –5700 = –(8.314)(298) ln K ⇒ ln K ≈ 2.3 ⇒ K ≈ 10.
   Why others fail: Option B (K = 1) corresponds to ΔG° = 0, a common trap when formula is misremembered.

4. For which reaction is ΔH equal to Δ_fH° of the product?
   
   A) CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)
   
   B) C(graphite) + 2H₂(g) → CH₄(g)
   
   C) 2C(graphite) + 3H₂(g) → C₂H₆(g)
   
   D) H₂(g) + ½O₂(g) → H₂O(g)
   Answer: B
   Explanation: Δ_fH° is defined for formation of one mole of compound from elements in standard states; B forms one mole of CH₄.
   Why others fail: Option D forms H₂O(g), but multiple products or incorrect stoichiometry invalidate others.

5. A reaction is spontaneous only at high temperatures. Which combination of ΔH and ΔS is correct?
   
   A) ΔH > 0, ΔS > 0
   
   B) ΔH < 0, ΔS < 0
   
   C) ΔH > 0, ΔS < 0
   
   D) ΔH < 0, ΔS > 0
   Answer: A
   Explanation: When ΔH > 0 and ΔS > 0, ΔG = ΔH – TΔS becomes negative only at high T.
   Why others fail: Option D (ΔH < 0, ΔS > 0) is always spontaneous, a tempting choice due to positive entropy change.

Last‑Minute Revision

• ⚠️ ΔG < 0 ⇒ spontaneous; ΔG > 0 ⇒ non-spontaneous; ΔG = 0 ⇒ equilibrium.
• ⚠️ Δ_fH° and Δ_fG° of elements in standard states are zero — verify from NCERT.
• ⚠️ Hess’s Law: ΔH is path-independent — use with formation or combustion data.
• ⚠️ ΔG° = –RT ln K — key for linking thermodynamics and equilibrium.
• ⚠️ For spontaneity, ΔG (not ΔH) is the deciding factor.
• ⚠️ S° > 0 for all substances; increases with disorder.
• ⚠️ ΔS° = ΣS°(products) – ΣS°(reactants) — not Δ_fS°.
• ⚠️ Units: ΔH in kJ/mol, ΔS in J/mol·K — convert before using in ΔG = ΔH – TΔS.
• ⚠️ Combustion reactions are exothermic: ΔH < 0.
• ⚠️ Bond breaking: +ΔH; bond forming: –ΔH.
• ⚠️ C(graphite) is standard state of carbon; Δ_fH° of diamond ≠ 0.
• ⚠️ ΔG determines spontaneity, not speed — kinetics vs thermodynamics.
• ⚠️ If ΔH < 0 and ΔS > 0, reaction is spontaneous at all T — mnemonic: "Both favorable, always go".
• ⚠️ If ΔH > 0 and ΔS < 0, never spontaneous — "Both unfavorable, never go".
• ⚠️ For ΔH < 0, ΔS < 0: spontaneous at low T — "Enthalpy-driven".
• ⚠️ For ΔH > 0, ΔS > 0: spontaneous at high T — "Entropy-driven".
• ⚠️ Standard conditions: 298 K, 1 bar (now used instead of 1 atm — verify from NCERT).
• ⚠️ ΔG = ΔG° + RT ln Q — used for non-equilibrium conditions.