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MHT CET · Chemistry · Chemical Kinetics

Calculate the rate constant of the first order reaction if \(80 \%\) of the reactant reacted in 15 minute.

  1. A 0.11 minute \(^{-1}\)
  2. B 0.22 minute \(^{-1}\)
  3. C 0.34 minute \(^{-1}\)
  4. D 0.42 minute \(^{-1}\)
Verified Solution

Answer & Solution

Correct Answer

(A) 0.11 minute \(^{-1}\)

Step-by-step Solution

Detailed explanation

\(80 \%\) of the reactant has reacted.
So, if \([A]_0=100\), then \([A]_{\mathrm{t}}=100-80=20\)
\(\begin{aligned}
\mathrm{k} & =\frac{2.303}{\mathrm{t}} \log _{10} \frac{[\mathrm{A}]_0}{[\mathrm{~A}]_{\mathrm{t}}} \\
& =\frac{2.303}{15} \log _{10} \frac{100}{20} \\
& =\frac{2.303}{15} \log _{10}(5) \\
& =\frac{2.303}{15} \times 0.699 \\
& =0.1073 \\
& \approx 0.11 \text { minute }^{-1}
\end{aligned}\)