ExamBro
ExamBro
MHT CET · Chemistry · Chemical Kinetics

A first order reaction is \(25 \%\) completed in 40 minutes. What is the rate constant \(\mathrm{k}\) for the reaction?

  1. A \(\frac{2.303 \times \log 1 \cdot 33}{40}\)
  2. B \(2 \cdot 303 \times \log \frac{4}{3}\)
  3. C \(\frac{2 \cdot 303}{40} \times \log \frac{1}{4}\)
  4. D \(\frac{2 \cdot 303 \times \log 4}{40 \times 3}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{2.303 \times \log 1 \cdot 33}{40}\)

Step-by-step Solution

Detailed explanation

(B)
\(\begin{aligned}[\mathrm{A}]_{0} &=100,[\mathrm{~A}]_{\mathrm{t}}=100-25=75, \mathrm{t}=40 \min \\ \mathrm{k} &=\frac{2.303}{\mathrm{t}} \log _{10} \frac{[\mathrm{A}]_{0}}{[\mathrm{~A}]_{\mathrm{t}}} \\ \therefore \mathrm{k} &=\frac{2.303}{40} \log _{10} \frac{100}{75}=\frac{2.303 \times \log 1.33}{40} \end{aligned}\)