JEE Mains · Chemistry · STD 12 - 3. Chemical kinetics
\(A\) and \(B\) decompose via first order kinetics with half-lives \(54.0\, min\) and \(18.0\, min\) respectively. Starting from an equimolar non reactive mixture of \(A\) and \(B\), the time taken for the concentration of \(A\) to become \(16\) times that of \(B\) is ...... \(min.\) (Round off to the Nearest Integer).
- A \(110\)
- B \(108\)
- C \(208\)
- D \(136\)
Answer & Solution
Correct Answer
(B) \(108\)
Step-by-step Solution
Detailed explanation
Given \(t _{2}=54\, min \quad T _{1 / 2}=18 \,min\) \(A\) \(\quad\) \(B\) \(\Rightarrow \quad\) To calculate \(:\left[ A _{t}\right]=16 \times\left[ B _{ t }\right] \ldots .(1)\) time \(=?\) \(\Rightarrow \quad\) For \(I\) order kinetic :…
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