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 \(\ln \left(\frac{\mathrm{x}}{\mathrm{y}}\right)=\ln \mathrm{x}-\ln \mathrm{y}\)
- A \(450\)
- B \(200\)
- C \(100\)
- D \(700\)
Answer & Solution
Correct Answer
(D) \(700\)
Step-by-step Solution
Detailed explanation
\(\quad \quad \quad \quad\quad \quad \mathrm{A} \longrightarrow \quad \mathrm{B}\) \(\mathrm{t}=0\quad \quad \quad \quad 1 \text { mole } \quad 0\) \(\mathrm{t}=100 \mathrm{~min}\quad 1-\mathrm{x} \quad \quad 2 \mathrm{x}\)…
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