MHT CET · Chemistry · Chemical Kinetics
Calculate the rate constant of the first order reaction if \(80 \%\) of the reactant reacted in 15 minute.
- A 0.11 minute \(^{-1}\)
- B 0.22 minute \(^{-1}\)
- C 0.34 minute \(^{-1}\)
- D 0.42 minute \(^{-1}\)
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 \([A]_{\mathrm{t}}=100-80=20\)
\(\begin{aligned}
\mathrm{k} & =\frac{2.303}{\mathrm{t}} \log _{10} \frac{[\mathrm{A}]_0}{[\mathrm{~A}]_{\mathrm{t}}} \\
& =\frac{2.303}{15} \log _{10} \frac{100}{20} \\
& =\frac{2.303}{15} \log _{10}(5) \\
& =\frac{2.303}{15} \times 0.699 \\
& =0.1073 \\
& \approx 0.11 \text { minute }^{-1}
\end{aligned}\)
So, if \([A]_0=100\), then \([A]_{\mathrm{t}}=100-80=20\)
\(\begin{aligned}
\mathrm{k} & =\frac{2.303}{\mathrm{t}} \log _{10} \frac{[\mathrm{A}]_0}{[\mathrm{~A}]_{\mathrm{t}}} \\
& =\frac{2.303}{15} \log _{10} \frac{100}{20} \\
& =\frac{2.303}{15} \log _{10}(5) \\
& =\frac{2.303}{15} \times 0.699 \\
& =0.1073 \\
& \approx 0.11 \text { minute }^{-1}
\end{aligned}\)
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