MHT CET · Chemistry · Chemical Kinetics
A first order reaction takes 23.03 minutes for \(20 \%\) decomposition. Calculate its rate constant.
- A \(5.6 \times 10^{-3}\) minute \(^{-1}\)
- B \(4.5 \times 10^{-3}\) minute \(^{-1}\)
- C \(6.5 \times 10^{-3}\) minute \(^{-1}\)
- D \(9.69 \times 10^{-3}\) minute \(^{-1}\)
Answer & Solution
Correct Answer
(D) \(9.69 \times 10^{-3}\) minute \(^{-1}\)
Step-by-step Solution
Detailed explanation
\(
[A]_0=100 \%,[A]_t=100-20=80 \%
\)
Substitution of these in above
\(
\begin{aligned}
\mathrm{k} & =\frac{2.303}{\mathrm{t}} \log _{10} \frac{[\mathrm{A}]_0}{[\mathrm{~A}]_{\mathrm{t}}} \\
\mathrm{k} & =\frac{2.303}{\mathrm{t}} \log _{10} \frac{100}{80} \\
& =\frac{2.303}{23.03 \mathrm{~min}} \log _{10}(1.25) \\
& =\frac{2.303}{23.03 \mathrm{~min}} \times 0.0969 \\
& =9.69 \times 10^{-3} \text { minute }^{-1}
\end{aligned}
\)
[A]_0=100 \%,[A]_t=100-20=80 \%
\)
Substitution of these in above
\(
\begin{aligned}
\mathrm{k} & =\frac{2.303}{\mathrm{t}} \log _{10} \frac{[\mathrm{A}]_0}{[\mathrm{~A}]_{\mathrm{t}}} \\
\mathrm{k} & =\frac{2.303}{\mathrm{t}} \log _{10} \frac{100}{80} \\
& =\frac{2.303}{23.03 \mathrm{~min}} \log _{10}(1.25) \\
& =\frac{2.303}{23.03 \mathrm{~min}} \times 0.0969 \\
& =9.69 \times 10^{-3} \text { minute }^{-1}
\end{aligned}
\)
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