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JEE Mains · Chemistry · STD 12 - 3. Chemical kinetics

For the first order reaction \(A \rightarrow 2 B, 1\) \(mole\) of reactant \(A\) gives \(0.2\) moles of \(B\) after \(100\, minutes.\) The half life of the reaction is \(.....\,min\) [Use \(: \ln 2=0.69, \ln 10=2.3\) Properties of logar \(\ln \left(\frac{\mathrm{x}}{\mathrm{y}}\right)=\ln \mathrm{x}-\ln \mathrm{y}\)

  1. A \(450\)
  2. B \(200\)
  3. C \(100\)
  4. D \(700\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(700\)

Step-by-step Solution

Detailed explanation

\(\quad \quad \quad \quad\quad \quad \mathrm{A} \longrightarrow \quad \mathrm{B}\) \(\mathrm{t}=0\quad \quad \quad \quad 1 \text { mole } \quad 0\) \(\mathrm{t}=100 \mathrm{~min}\quad 1-\mathrm{x} \quad \quad 2 \mathrm{x}\)…