JEE Mains · Chemistry · STD 12 - 3. Chemical kinetics
A flask is filled with equal moles of \(A\) and \(B\). The half lives of \(A\) and \(B\) are \(100\, s\) and \(50 \,s\) respectively and are independent of the initial concentration. The time required for the concentration of \(A\) to be four times that of \(B\) is \(....\,s.\) (Given : \(\ln 2=0.693\) )
- A \(855\)
- B \(400\)
- C \(200\)
- D \(300\)
Answer & Solution
Correct Answer
(C) \(200\)
Step-by-step Solution
Detailed explanation
\(k _{ A }=\frac{\ln 2}{100} ; k _{ B }=\frac{\ln 2}{50}\) \(A _{ t }= A _{0} \times e ^{- k _{ A } t }\) \(A _{ t }= A _{0} \times e ^{\left(\frac{-\ln 2}{100} \times t \right)}\) \(B _{ t }= B _{0} \times e ^{\left(\frac{-\ln 2}{50} \times t \right)}\) \(A _{0}= B _{0}\)…
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