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(B, D) The relationship between the equilibrium constant $K$ and temperature is given by the van't Hoff equation: $\ln K = -\frac{\Delta H^\circ}{

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$[D]$ With increase in temperature,the value of $K$ for exothermic reaction decreases because favourable change in entropy of the surrounding decreases

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(B, D) The relationship between the equilibrium constant $K$ and temperature is given by the van't Hoff equation: $\ln K = -\frac{\Delta H^\circ}{RT} + \frac{\Delta S^\circ}{R}$.
For an endothermic reaction $(\Delta H > 0)$,increasing temperature increases $K$ because the entropy of the surroundings $(\Delta S_{surr} = -\frac{\Delta H}{T})$ becomes less negative (less unfavourable).
For an exothermic reaction $(\Delta H < 0)$,increasing temperature decreases $K$ because the entropy of the surroundings $(\Delta S_{surr} = -\frac{\Delta H}{T})$ becomes less positive (less favourable).
Thus,statements $[B]$ and $[D]$ correctly describe the effect of temperature on $K$ in terms of entropy changes of the surroundings.

For a reaction taking place in a container in equilibrium with its surroundings,the effect of temperature on its equilibrium constant $K$ in terms of change in entropy is described by:

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