MHT CET · Chemistry · Chemical Kinetics
In a first order reaction \(87.5 \%\) of reactant is converted in to product in 15 minutes. The rate constant for the reaction is given by
- A \(\frac{0.693}{5} \mathrm{~min}^{-1}\)
- B \(\frac{0.693}{15} \mathrm{~min}^{-1}\)
- C \(\frac{5}{0.693} \min ^{-1}\)
- D \(0.693 \times 5 \mathrm{~min}^{-1}\)
Answer & Solution
Correct Answer
(A) \(\frac{0.693}{5} \mathrm{~min}^{-1}\)
Step-by-step Solution
Detailed explanation
\(87.5 \%\) of reactant is converted into product in 15 minutes, means \(12.5 \%\) remains unreacted
\([\mathrm{A}]_{0}=100, \quad[\mathrm{~A}]_{t}=100-87.5=12.5\)
\(k=\frac{2.303}{15} \log _{10} \frac{[A]_{0}}{[A]_{t}}\)
\(k=\frac{2.303}{15} \log _{10} \frac{100}{12.5}\)
\(k=\frac{2.303}{15} \log _{10} 8=\frac{0.693}{5} \min ^{-1}\)
\([\mathrm{A}]_{0}=100, \quad[\mathrm{~A}]_{t}=100-87.5=12.5\)
\(k=\frac{2.303}{15} \log _{10} \frac{[A]_{0}}{[A]_{t}}\)
\(k=\frac{2.303}{15} \log _{10} \frac{100}{12.5}\)
\(k=\frac{2.303}{15} \log _{10} 8=\frac{0.693}{5} \min ^{-1}\)
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