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MHT CET · Chemistry · Chemical Kinetics

Calculate the rate constant of the first order reaction if \(80 \%\) of the reactant decomposes in 60 minutes.

  1. A \(2.68 \times 10^{-2}\) minute \(^{-1}\)
  2. B \(5.36 \times 10^{-2}\) minute \(^{-1}\)
  3. C \(1.34 \times 10^{-2}\) minute \(^{-1}\)
  4. D \(8.1 \times 10^{-2}\) minute \(^{-1}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(2.68 \times 10^{-2}\) minute \(^{-1}\)

Step-by-step Solution

Detailed explanation

\(\begin{aligned} \mathrm{k} & =\frac{2.303}{\mathrm{t}} \log _{10} \frac{[\mathrm{A}]_0}{[\mathrm{~A}]_{\mathrm{t}}} \\ & =\frac{2.303}{60} \log _{10} \frac{(100)}{(20)}=\frac{2.303}{60} \log _{10} \frac{10}{2} \\ & =\frac{2.303}{60}\left[\log _{10}(10)-\log _{10}(2)\right] \\ & =\frac{2.303}{60} \times 0.699=2.68 \times 10^{-2} \text { minute }^{-1}\end{aligned}\)