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JEE Mains · Chemistry · STD 12 - 3. Chemical kinetics

For a reaction, given below is the graph of \(\ln k\,v s \frac{1}{ T }\). The activation energy for the reaction is equal to \(......cal\) \(mol ^{-1}\). (Nearest integer). (Given : \(R =2\,cal\,K ^{-1}\,mol ^{-1}\) )

  1. A \(8\)
  2. B \(5\)
  3. C \(4\)
  4. D \(3\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(8\)

Step-by-step Solution

Detailed explanation

\(K = Ae ^{- Ea / RT }\) \(\ln k =\frac{- Ea }{ RT }+\ln A\) Slope \(=\frac{ Ea }{ R }=\frac{20}{5}\) \(E _{ a }=4\,R =8\,Cal / mol\)
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