A schematic plot of K_eq versus inverse of temperature ( 1/T ) for a reaction is shown below. The reaction must be: ![image](https://pub-7faa453bd1ec4 - Chemistry MCQ AIEEE 2026 | ExamTest.live

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A schematic plot of ln K_{e q} versus inverse of temperature (1/ T) for a reaction is shown below. The reaction must be:

Solution & Step-by-step Explanation

The van't Hoff equation is: ln K_{e q} = - \frac{Δ H ^{\circ}}{R} (\frac{1}{T}) + \frac{Δ S ^{\circ}}{R} In the plot of ln K_{e q} vs 1/ T :Slope = - Δ H^{\circ} / R .From the schematic (described as increasing ln K with 1/ T), the slope is positive. - \frac{Δ H ^{\circ}}{R} > 0 ⟹ Δ H^{\circ} < 0 A negative enthalpy change (Δ H^{\circ} < 0) indicates an exothermic reaction.

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A schematic plot of ln K_{e q} versus inverse of temperature (1/ T) for a reaction is shown below. The reaction must be:

exothermic

endothermic

one with negligible enthalpy change

highly spontaneous at ordinary temperature

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