Second Law Thermodynamics

Answer: B. The second law states that in any spontaneous process, the total entropy of the universe (system + surroundings) increases (NCERT Class 11 Chemistry, Chapter 5, page 161). Option B captures this directly.

Why A is wrong: A describes the first law of thermodynamics (energy conservation), not the second law.

Why C is wrong: C describes the third law of thermodynamics (Nernst statement about entropy at absolute zero).

Why D is wrong: D is a property of enthalpy as a state function — this is a general thermodynamic fact, not the second law.

Answer: B. At the crossover, ΔG = 0, so T = ΔH/ΔS = 50/0.15 = 333.3 K. Above this temperature, TΔS > ΔH, making ΔG < 0 (spontaneous). B is correct.

Why A is wrong: A reverses the direction — below 333 K, TΔS < ΔH, so ΔG > 0 (non-spontaneous). This is the trap of confusing 'above' and 'below' the crossover temperature.

Why C is wrong: C would require ΔH < 0 and ΔS > 0 simultaneously. Here ΔH is positive, so spontaneity is temperature-dependent.

Why D is wrong: D would require ΔH > 0 and ΔS < 0. Here ΔS is positive, so the reaction does become spontaneous at high enough T.

Answer: B. The dissolution is endothermic (ΔH > 0) but produces a large entropy increase (ions dispersing in solvent). At room temperature, TΔS exceeds ΔH, making ΔG = ΔH − TΔS < 0. B is correct.

Why A is wrong: A contradicts the stated fact — the process is endothermic, so ΔH > 0, not negative. This is the trap of judging spontaneity from ΔH sign alone (mistake: predicting spontaneity from ΔH without considering TΔS).

Why C is wrong: C is factually wrong — dissolution of an ionic solid into solution increases entropy (more disorder among dispersed ions and solvent molecules).

Why D is wrong: D would mean the process is non-spontaneous, contradicting the observation that NH₄Cl dissolves spontaneously.

Answer: C. C is correct. An isolated system exchanges neither energy nor matter with its surroundings, so the second law applies to it directly: in any spontaneous change, its entropy increases, and it reaches a maximum at equilibrium. ΔS = 0 applies only to a reversible process.

Why A is wrong: A is wrong because a decrease in the entropy of an isolated system would violate the second law. The entropy of a non-isolated system can fall, but only if the surroundings gain more.

Why B is wrong: B is wrong because constant entropy corresponds to a reversible process, an idealised limit, not to a real spontaneous change.

Why D is wrong: D is wrong because ΔH is irrelevant here: an isolated system cannot exchange heat. The sign of ΔH matters when predicting spontaneity through ΔG for systems at constant T and P, not for an isolated system.

Answer: C. At crossover: T = |ΔH|/|ΔS| = 400/0.50 = 800 K. Below 800 K, ΔG < 0 (spontaneous); above 800 K, the −TΔS term (which is positive here since ΔS < 0) overwhelms ΔH and ΔG > 0. The reaction becomes non-spontaneous above 800 K. C is correct.

Why A is wrong: A results from dividing 400/2.0 or some other arithmetic error. The correct divisor is 0.50, giving 800 K.

Why B is wrong: B results from dividing 400/1.0 — an error in reading ΔS. The actual ΔS magnitude is 0.50 kJ/(mol·K).

Why D is wrong: D is wrong because both ΔH and ΔS are negative — spontaneity is temperature-dependent, not unconditional. At sufficiently high T, the entropy penalty dominates.

Answer: C. C is correct. At the boiling point the phase change is reversible, so ΔS = q_rev/T = ΔH_vap/T_b = 40 700 J mol⁻¹ / 373 K ≈ 109 J K⁻¹ mol⁻¹. The temperature must be in kelvin, and ΔH in joules to give an answer in J K⁻¹ mol⁻¹.

Why A is wrong: A is wrong because 0.109 is ΔH in kJ divided by T, which is 0.109 kJ K⁻¹ mol⁻¹, then labelled J. It is the right value in the wrong unit.

Why B is wrong: B is wrong because 407 divides by the boiling point in degrees Celsius, 100, instead of kelvin. Entropy calculations need absolute temperature.

Why D is wrong: D is wrong because 1.52 × 10⁴ multiplies ΔH by T instead of dividing. Entropy is heat divided by temperature.

Answer: A. ΔG = ΔH − TΔS = −10.0 − (298 × 0.040) = −10.0 − 11.92 = −21.92 ≈ −21.9 kJ/mol. A is correct.

Why B is wrong: B ignores the TΔS term entirely, using ΔG = ΔH alone. This is the classic mistake of predicting spontaneity from enthalpy without considering the entropy contribution.

Why C is wrong: C uses the wrong sign: ΔG = ΔH + TΔS = −10.0 + 11.92 = +1.92. The formula has a minus sign before TΔS, not plus.

Why D is wrong: D uses the temperature in °C instead of kelvin: −10.0 − (25 × 0.040) = −11.0 kJ/mol. T must be 298 K.

Answer: B. At the crossover temperature, ΔG = 0: ΔH = TΔS, so ΔS = ΔH/T = 80/400 = 0.20 kJ/(mol·K). For spontaneity above 400 K, ΔS must be at least 0.20 kJ/(mol·K). B is correct.

Why A is wrong: A gives ΔS = 80/800 = 0.10 — this would place the crossover at 800 K, not 400 K. The reaction would need T > 800 K to be spontaneous, contradicting the stated 400 K threshold.

Why C is wrong: C gives ΔS = 80/200 = 0.40 — this would make the crossover 200 K, meaning the reaction is spontaneous above 200 K. While technically it would also be spontaneous above 400 K, 0.40 is not the minimum ΔS required — 0.20 is.

Why D is wrong: D is negative. An endothermic reaction with negative ΔS has ΔG > 0 at ALL temperatures (never spontaneous), contradicting the premise.