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Q. Which of following lines correctly show the temperature dependence of equilibrium constant $K$, for an exothermic reaction ?Chemistry Question Image

JEE MainJEE Main 2018Thermodynamics

Solution:

Equilibrium constant $K =\left(\frac{ A _{ f }}{ A _{ b }}\right) e ^{-\frac{\Delta H ^{\circ}}{ RT }}$

ln $K =$ ln $\left(\frac{ A _{ f }}{ A _{ b }}\right)-\frac{\Delta H ^{\circ}}{ R }\left(\frac{1}{ T }\right)$

$y = C + m x$

Comparing with equation of straight line,

Slope $=\frac{-\Delta H^{\circ}}{R}$

Since, reaction is exothermic, $\Delta H ^{\circ}=- ve$, therefore, slope $= +ve $

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