MHT CET · Chemistry · Chemical Kinetics
A first order reaction takes 40 minute 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 \(7.2 \times 10^{-3}\) minute \({ }^{-1}\)
Answer & Solution
Correct Answer
(A) \(5.6 \times 10^{-3}\) minute \({ }^{-1}\)
Step-by-step Solution
Detailed explanation
If \([A]_0=100\), then \([A]_t=100-20=80\)
For a first order reaction,
\(\begin{aligned}
\mathrm{k} & =\frac{2.303}{\mathrm{t}} \log _{10} \frac{[\mathrm{~A}]_0}{[\mathrm{~A}]_{\mathrm{t}}}=\frac{2.303}{40 \mathrm{~min}} \log _{10} \frac{100}{80} \\
& =\frac{2.303}{40 \mathrm{~min}} \log _{10} 1.25=\frac{2.303}{40 \mathrm{~min}} \times 0.097 \\
& =5.57 \times 10^{-3} \mathrm{minute}^{-1} \approx 5.6 \times 10^{-3} \mathrm{minute}^{-1}
\end{aligned}\)
For a first order reaction,
\(\begin{aligned}
\mathrm{k} & =\frac{2.303}{\mathrm{t}} \log _{10} \frac{[\mathrm{~A}]_0}{[\mathrm{~A}]_{\mathrm{t}}}=\frac{2.303}{40 \mathrm{~min}} \log _{10} \frac{100}{80} \\
& =\frac{2.303}{40 \mathrm{~min}} \log _{10} 1.25=\frac{2.303}{40 \mathrm{~min}} \times 0.097 \\
& =5.57 \times 10^{-3} \mathrm{minute}^{-1} \approx 5.6 \times 10^{-3} \mathrm{minute}^{-1}
\end{aligned}\)
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