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NEET CHEMISTRYEquilibriumMedium

Question

For a given exothermic reaction, KpK_p and KpK_p' are the equilibrium constants at temperatures T1T_1 and T2T_2 respectively. Assuming that the heat of reaction is constant in the temperature range between T1T_1 and T2T_2, it is readily observed that: (Assume T2>T1T_2 > T_1)

A

Kp>KpK_p > K_p'

B

Kp<KpK_p < K_p'

C

Kp=KpK_p = K_p'

D

Kp=1KpK_p = \frac{1}{K_p'}

Step-by-Step Solution

According to Le Chatelier's principle, for an exothermic reaction, an increase in temperature shifts the equilibrium in the backward direction, which decreases the value of the equilibrium constant. Since T2>T1T_2 > T_1, the equilibrium constant at the higher temperature (KpK_p') will be less than the equilibrium constant at the lower temperature (KpK_p). Therefore, Kp>KpK_p > K_p'. Alternatively, using the van't Hoff equation: logKpKp=ΔH2.303R(T2T1T1T2)\log \frac{K_p'}{K_p} = \frac{\Delta H}{2.303 R} \left( \frac{T_2 - T_1}{T_1 T_2} \right) Since T2>T1T_2 > T_1, the term (T2T1)(T_2 - T_1) is positive. For an exothermic reaction, ΔH\Delta H is negative. Thus, the entire RHS is negative, making logKpKp<0\log \frac{K_p'}{K_p} < 0. This implies KpKp<1\frac{K_p'}{K_p} < 1, or Kp>KpK_p > K_p'.

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Equilibrium. 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.

CHEMISTRYEquilibriumexothermicreactionequilibriumconstantstemperatures

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