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 : \(R =2\,cal\,K ^{-1}\,mol ^{-1}\) )

- A \(8\)
- B \(5\)
- C \(4\)
- D \(3\)
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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