MHT CET · Chemistry · Chemical Kinetics
Calculate the rate constant of first order reaction if the concentration of the reactant decreases by \(90 \%\) in 30 minutes.
- A \(7.7 \times 10^{-2}\) minute \(^{-1}\)
- B \(4.2 \times 10^{-2}\) minute \(^{-1}\)
- C \(2.1 \times 10^{-2}\) minute \(^{-1}\)
- D \(3.5 \times 10^{-2}\) minute \(^{-1}\)
Answer & Solution
Correct Answer
(A) \(7.7 \times 10^{-2}\) minute \(^{-1}\)
Step-by-step Solution
Detailed explanation
Concentration decreases by \(90 \%\). Hence, \(10 \%\) reactant is left after 30 minutes.
\(\begin{aligned}
\mathrm{k} & =\frac{2.303}{\mathrm{t}} \log _{10} \frac{[\mathrm{A}]_0}{[\mathrm{~A}]_{\mathrm{t}}} \\
\mathrm{k} & =\frac{2.303}{30} \log _{10} \frac{100}{10} \\
& =\frac{2.303}{30 \mathrm{~min}} \\
& =7.7 \times 10^{-2} \mathrm{~min}^{-1}
\end{aligned}\)
\(\begin{aligned}
\mathrm{k} & =\frac{2.303}{\mathrm{t}} \log _{10} \frac{[\mathrm{A}]_0}{[\mathrm{~A}]_{\mathrm{t}}} \\
\mathrm{k} & =\frac{2.303}{30} \log _{10} \frac{100}{10} \\
& =\frac{2.303}{30 \mathrm{~min}} \\
& =7.7 \times 10^{-2} \mathrm{~min}^{-1}
\end{aligned}\)
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