1. Tardigrade
  2. Question
  3. Chemistry
  4. Rate constant k of a reaction varies with temperature according to the equation log k=constant-(Ea/2.303 RT)⋅ When a graph is plotted for logk versus 1/T a straight line with a slope -5632 is obtained. The energy of activation for this reaction is

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Q. Rate constant k of a reaction varies with temperature according to the equation
$log k=constant-\frac{E_{a}}{2.303 RT}\cdot$
When a graph is plotted for logk versus 1/T a straight line with a slope -5632 is obtained. The energy of activation for this reaction is

Chemical Kinetics

Solution:

log k=constant $-\frac{E_{a}}{2.303 RT}$
$\frac{E_{a}}{2.303R}=-5632$
$E_{a}=5632\times2.303\times8.314=107836$
$E_{a} \approx107.836 KJ \, mol^{-1}.$

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