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JEE Mains · Chemistry · STD 12 - 3. Chemical kinetics
For a first order reaction, \(A\to P\), \(t_{1/2}\) (half-life) is\(10\,days\). The time required for \(\frac{{{1^{th}}}}{4}\) conversion of \(A\) (in days) is: \((\ln\, 2\, = 0.693,\, \ln\, 3\, = 1.1)\).
- A \(3.2\)
- B \(2.5\)
- C \(4.1\)
- D \(5\)
Answer & Solution
Correct Answer
(C) \(4.1\)
Step-by-step Solution
Detailed explanation
The half life \(t_{1/2}\,=\,10\,days\) The decay constant \(k\, = \,\frac{{0.693}}{{{t_{1/2}}}}\, = \,\frac{{0.693}}{{10\,\,days}}\, = \,0.0693\,\,day{s^{ - 1}}\) The time required for one fourth conversion \(t\, = \,\frac{{2.303}}{k}\,{\log _{10}}\frac{a}{{a - x}}\)…
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