JEE Mains · Chemistry · STD 12 - 3. Chemical kinetics
For a first order reaction \(A \rightarrow B\), the rate constant, \(k =5.5 \times 10^{-14} \,s ^{-1}\). The time required for \(67\, \%\) completion of reaction is \(x \,\times 10^{-1}\) times the half life of reaction. The value of \(x\) is \(....\) (Nearest integer)
- A \(160\)
- B \(16\)
- C \(35\)
- D \(90\)
Answer & Solution
Correct Answer
(B) \(16\)
Step-by-step Solution
Detailed explanation
\(t _{67 \,\%}=\frac{1}{ k } \ln \left(\frac{1}{1-0.67}\right)=\frac{ t _{1 / 2}}{\ln 2} \times \ln \left(\frac{1}{1-\frac{2}{3}}\right)\) \(t _{67 \%}=\frac{ t _{1 / 2}}{\log 2} \times \log 3=\frac{ t _{1 / 2} \times 0.4771}{0.301}\)…
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