JEE Mains · Chemistry · STD 12 - 3. Chemical kinetics
\(A \rightarrow B\) The above reaction is of zero order. Half life of this reaction is \(50\,min\). The time taken for the concentration of \(A\) to reduce to one-fourth of its initial value is \(...........\min\)(Nearest integer)
- A \(74\)
- B \(75\)
- C \(72\)
- D \(73\)
Answer & Solution
Correct Answer
(B) \(75\)
Step-by-step Solution
Detailed explanation
Assume reaction starts with \(1\) mole \(A\) \(\left( t _{1 / 2}=\frac{ a }{2 k }, K =\frac{1}{2 \times 50}\right.\) For \(75 \%\) completion \(a -\frac{ a }{4}= kt\) \(t =\frac{3}{4} \frac{ a }{ k }=\frac{3}{4} \times \frac{100}{ a }=75\)
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