CUET UG Chemistry: Physical Chemistry - Thermodynamics, Hess's 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, ?H for C(s)-CO?(g) is –393.5 kJ/mol whether it occurs directly or via CO(g) intermediate.
• The standard enthalpy of formation (?_fH°) of an element in its most stable form at 298 K and 1 atm is zero; e.g., ?_fH° of O?(g), C(graphite), and H?(g) is 0 kJ/mol.
• For the reaction C(graphite) + ½O?(g)-CO(g), ?H = –110.5 kJ/mol; this cannot be measured directly but is calculated using Hess’s Law from combustion data.
• Gibbs energy change (?G) determines spontaneity: if ?G < 0, the process is spontaneous; if ?G > 0, it is non-spontaneous; if ?G = 0, the system is at equilibrium.
• ?G = ?H – T?S is the Gibbs equation; used to predict spontaneity from enthalpy and entropy changes.
• A reaction with ?H < 0 and ?S > 0 is spontaneous at all temperatures; e.g., combustion of glucose.
• A reaction with ?H > 0 and ?S < 0 is non-spontaneous at all temperatures; e.g., formation of ozone from O? at ground level.
• A reaction with ?H < 0 and ?S < 0 is spontaneous only at low temperatures; e.g., synthesis of ammonia: N?(g) + 3H?(g)-2NH?(g), ?H = –92.4 kJ/mol.
• A reaction with ?H > 0 and ?S > 0 is spontaneous only at high temperatures; e.g., decomposition of CaCO?(s)-CaO(s) + CO?(g), which occurs above 1100 K.
• Standard Gibbs energy of formation (?_fG°) of an element in its standard state is zero; e.g., ?_fG° of Cl?(g), Fe(s), and O?(g) is 0 kJ/mol.
• ?G° = –RT ln K, where K is the equilibrium constant; at equilibrium, ?G = 0 and Q = K.
• For a reaction with K > 1, ?G° < 0; for K < 1, ?G° > 0; e.g., if K = 10, ?G° = –RT ln(10)-–5.7 kJ/mol at 298 K.
• The combustion of methane: CH?(g) + 2O?(g)-CO?(g) + 2H?O(l), has ?G° = –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 H?O(s) < H?O(l) < H?O(g).
• Standard molar entropy (S°) of a substance is always positive; e.g., S° of H?(g) is 130.7 J/mol·K at 298 K.
• ?S° for a reaction is calculated as ?S°(products) – ?S°(reactants); e.g., for N?(g) + 3H?(g)-2NH?(g), ?S° = –198.3 J/mol·K.
• For exothermic reactions, ?H is negative; for endothermic, ?H is positive; e.g., photosynthesis has ?H > 0.
• Bond breaking is endothermic (absorbs energy), bond formation is exothermic (releases energy); net ?H = ?(bond energies broken) – ?(bond energies formed).
• The standard enthalpy change for the reaction: 2Al(s) + Fe?O?(s)-Al?O?(s) + 2Fe(s) is –851.5 kJ/mol, calculated via Hess’s Law using formation data.
• ?G is a state function, so it obeys Hess’s Law; ?G°_reaction = _fG°(products) – _fG°(reactants).

Difficulty Level

Intermediate — requires understanding of sign conventions, temperature dependence, and integration of ?H, ?S, and ?G concepts, but avoids complex derivations.

Common CUET Traps

• Trap: Assuming all exothermic reactions are spontaneous.
  Avoid: Check ?G = ?H – T?S; if ?S is highly negative and T is high, ?G may be positive even if ?H is negative.
• Trap: Confusing ?G and ?G° — thinking ?G° < 0 means reaction proceeds to completion.
  Avoid: ?G° < 0 means K > 1, but equilibrium may still have significant reactants; spontaneity does not imply completeness.
• Trap: Using ?_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 ?H values.

Practice MCQs

1. For a reaction with ?H = –40 kJ/mol and ?S = –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, ?G = 0-T = ?H/?S = (–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 ?G° = 0?
   A) H?(g)-2H(g)
   B) O?(g)-2O(g)
   C) C(graphite)-C(diamond)
   D) Fe(s) at 298 K and 1 atm
   Answer: D
   Explanation: ?G° = 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 ?G°-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: ?G° = –RT ln K-–5700 = –(8.314)(298) ln K-ln K-2.3-K-10.
   Why others fail: Option B (K = 1) corresponds to ?G° = 0, a common trap when formula is misremembered.

4. For which reaction is ?H equal to ?_fH° of the product?
   A) CH?(g) + 2O?(g)-CO?(g) + 2H?O(l)
   B) C(graphite) + 2H?(g)-CH?(g)
   C) 2C(graphite) + 3H?(g)-C?H?(g)
   D) H?(g) + ½O?(g)-H?O(g)
   Answer: B
   Explanation: ?_fH° is defined for formation of one mole of compound from elements in standard states; B forms one mole of CH?.
   Why others fail: Option D forms H?O(g), but multiple products or incorrect stoichiometry invalidate others.

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

Last?Minute Revision

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