Equilibrium Constant - Temperature Dependence
For an endothermic reaction at 400 K, K_c = 2.0 \times 10^{-3}. When the temperature is raised to 600 K, which of the following best describes the change in K_c?
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About This Question
- Subject
- chemistry
- Chapter
- equilibrium
- Topic
- equilibrium constant - temperature dependence
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
Kcincreasesbecausehighertemperaturefavourstheendothermicforwardreaction
The temperature dependence of the equilibrium constant is given by the van't Hoff equation: d(ln K)/dT = \Delta H° / RT^2. For an endothermic reaction, \Delta H° > 0, so d(ln K)/dT > 0, meaning K increases as temperature increases. Raising temperature for an endothermic reaction provides the energy needed to drive the forward reaction further, shifting equilibrium to the right and increasing the ratio of products to reactants at the new equilibrium, which means K_c increases. Option A is incorrect reasoning: while Le Chatelier's principle says the system opposes the temperature rise by absorbing heat, this means the endothermic forward reaction is favoured, which increases K_c — not decreases it. Option C is fundamentally wrong; K_c is temperature-dependent and changes with temperature, unlike rate constants which change differently. Option D is the opposite of reality for endothermic reactions; K_c increases with temperature for endothermic reactions and decreases for exothermic ones. This van't Hoff relationship is a critical NCERT thermodynamic concept linking \Delta G° = -RT ln K to temperature. Plausibility check: at higher temperature, the new K_c > 2.0 \times 10^{-3} for an endothermic process, consistent with our conclusion.
This medium difficulty chemistry question is from the chapter equilibrium, covering the topic of equilibrium constant - temperature dependence. It appeared in the 2025 exam.
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