JEE Mains · Chemistry · STD 12 - 3. Chemical kinetics
\(A \rightarrow B\) (first reaction)
\(C \rightarrow D\) (second reaction)
Consider the above two first-order reactions. The rate constant for first reaction at 500 K is double of the same at 300 K . At \(500 K, 50 \%\) of the reaction becomes complete in 2 hour. The activation energy of the second reaction is half of that of first reaction. If the rate constant at 500 K of the second reaction becomes double of the rate constant of first reaction at the same temperature; then rate constant for the second reaction at 300 K is __________ \(\times 10^{-1}\) hour \(^{-1}\) (nearest integer).
- A 4.5
- B 4.9
- C 5
- D 5.5
Answer & Solution
Correct Answer
(C) 5
Step-by-step Solution
Detailed explanation
For \(A \xrightarrow{ K _1} B\) \(\ln (2)=\frac{E_{a_1}}{R}\left[\frac{1}{300}-\frac{1}{500}\right]\) \(E _{ a _1}=\frac{\ln 2 \times R \times 1500}{2}\) \(E _{ a _2}=\frac{ E _{ a _1}}{2}=\frac{\ln 2 \times R \times 1500}{4}\) \(\left( K _1\right)_{ at~500 K}=\frac{\ln 2}{2}\)…
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