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NEET CHEMISTRYThermodynamicsMedium

Question

Match List-I (Equations/Conditions) with List-II (Type of processes) and select the correct option.

List-I (Equations) (a) Kp>QK_p > Q (b) T<ΔHΔST < \frac{\Delta H}{\Delta S} (for ΔH>0\Delta H > 0) (c) Kp=QK_p = Q (d) T>ΔHΔST > \frac{\Delta H}{\Delta S} (for ΔH>0\Delta H > 0)

List-II (Type of processes) (i) Non-spontaneous (ii) Equilibrium (iii) Spontaneous and endothermic (iv) Spontaneous

A

(a)-(iii), (b)-(iv), (c)-(ii), (d)-(i)

B

(a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)

C

(a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)

D

(a)-(i), (b)-(ii), (c)-(iii), (d)-(iv)

Step-by-Step Solution

  1. (a) Kp>QK_p > Q: When the reaction quotient (QQ) is less than the equilibrium constant (KpK_p), the reaction proceeds in the forward direction to reach equilibrium. Therefore, the forward process is Spontaneous (iv). [NCERT Chem 11, Equilibrium, Sec 6.3]
  2. (c) Kp=QK_p = Q: When QQ equals KpK_p, the system is at Equilibrium (ii). [NCERT Chem 11, Equilibrium, Sec 6.3]
  3. (d) T>ΔHΔST > \frac{\Delta H}{\Delta S}: For an endothermic reaction (ΔH>0\Delta H > 0), spontaneity requires ΔG=ΔH−TΔS<0\Delta G = \Delta H - T\Delta S < 0. This occurs when TΔS>ΔHT\Delta S > \Delta H, or T>ΔHΔST > \frac{\Delta H}{\Delta S}. Thus, the process is Spontaneous and endothermic (iii). [NCERT Chem 11, Thermodynamics, Sec 5.6]
  4. (b) T<ΔHΔST < \frac{\Delta H}{\Delta S}: Conversely, if the temperature is low (TΔS<ΔHT\Delta S < \Delta H) for an endothermic reaction, ΔG\Delta G will be positive. Thus, the process is Non-spontaneous (i).

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Thermodynamics. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYThermodynamicsequationsconditionslistiiprocessesselectcorrect

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