The variation of equilibrium constant K with temperature is represented by :

Chemistry · JEE Advanced · NTA Exams — Equilibrium

\(\log _{\mathrm{e}} \mathrm{~K}_{2}-\log _{\mathrm{e}} \mathrm{~K}_{\mathrm{l}}=-\frac{\Delta \mathrm{H}}{\mathrm{R}} \int_{\mathrm{T}_{1}}^{\mathrm{T}_{2}}\left(\frac{1}{\mathrm{~T}}\right) \mathrm{dT}\)
\(\left\{\frac{\mathrm{d} \log _{\mathrm{e}} \mathrm{~K}}{\mathrm{dT}}\right\}_{\mathrm{P}}=-\frac{\Delta \mathrm{H}}{\mathrm{RT}}\)
log_e K = Constant \(-\frac{\mathrm{RT}}{\Delta \mathrm{H}}\)
log_e K = Constant \(-\frac{\Delta \mathrm{H}}{\mathrm{RT}}\)

Answer

(A) _ e ~K _ 2 - _ e ~K _ l =- H R _ T _ 1 ^ T _ 2 ( 1 ~T ) dT, (D) log e K = Constant - H RT

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