JEE Mains · Chemistry · STD 12 - 3. Chemical kinetics
The rate constant for a first order reaction is given by the following equation: \(\ln k=33.24-\frac{2.0 \times 10^{4} \,K }{ T }\) The Activation energy for the reaction is given by \(...\,k\,\,J mol ^{-1}\). (In Nearest integer) (Given: \(R =8.3\, J\, K ^{-1} \,mol ^{-1}\) )
- A \(15\)
- B \(166\)
- C \(961\)
- D \(247\)
Answer & Solution
Correct Answer
(B) \(166\)
Step-by-step Solution
Detailed explanation
\(\ln k =\ln A -\frac{ E _{ A }}{ RT }\) Given: \(\ln k =33.24-\frac{2.0 \times 10^{4}}{ T }\) \(\therefore \text { on comparing } \frac{ E _{ A }}{ R }=2.0 \times 10^{4}\) \(\therefore E _{ A }=2.0 \times 10^{4} \times R\)…
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