JEE Mains · Chemistry · STD 12 - 3. Chemical kinetics
\(A \rightarrow B\) The rate constants of the above reaction at \(200 \,K\) and \(300 \,K\) are \(0.03 \,min ^{-1}\) and \(0.05 \,min ^{-1}\) respectively. The activation energy for the reaction is \(....J\) (Nearest integer) (Given : In \(10=2.3\) \(R =8.3\,J\,K ^{-1}\, mol ^{-1}\) \(\log 5=0.70\) \(\log 3=0.48\) \(\log 2=0.30\)
- A \(2510\)
- B \(2530\)
- C \(2540\)
- D \(2520\)
Answer & Solution
Correct Answer
(D) \(2520\)
Step-by-step Solution
Detailed explanation
\(\log \frac{ K _{300}}{ K _{200}}=\frac{ E _{ a }}{2.3 \times 8.314}\left(\frac{1}{ T _1}-\frac{1}{ T _2}\right)\) \(\log \frac{0.05}{0.03}=\frac{ Ea }{2.305 \times 8.314} \times\left[\frac{1}{200}-\frac{1}{300}\right]\) \(E _{ a }=2519.88\,J \Rightarrow E _{ a }=2520\,J\)
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