MHT CET · Chemistry · Chemical Kinetics
Calculate rate constant of first order reaction if concentration of reactant decreases by \(90 \%\) in 30 minute?
- A \(2.16 \times 10^{-2} \mathrm{~min}^{-1}\)
- B \(3.52 \times 10^{-2} \mathrm{~min}^{-1}\)
- C \(4.81 \times 10^{-2} \mathrm{~min}^{-1}\)
- D \(7.67 \times 10^{-2} \cdot \mathrm{~min}^{-1}\)
Answer & Solution
Correct Answer
(D) \(7.67 \times 10^{-2} \cdot \mathrm{~min}^{-1}\)
Step-by-step Solution
Detailed explanation
Concentration decreases by \(90 \%\). Hence, \(10 \%\) reactant is left after 30 minutes.
\(\begin{aligned}
\mathrm{k} & =\frac{2.303}{\mathrm{t}} \log _{10} \frac{[\mathrm{~A}]_0}{[\mathrm{~A}]_{\mathrm{t}}} \\
\mathrm{k} & =\frac{2.303}{30} \log _{10} \frac{100}{10} \\
& =\frac{2.303}{30 \mathrm{~min}}=7.67 \times 10^{-2} \mathrm{~min}^{-1}
\end{aligned}\)
\(\begin{aligned}
\mathrm{k} & =\frac{2.303}{\mathrm{t}} \log _{10} \frac{[\mathrm{~A}]_0}{[\mathrm{~A}]_{\mathrm{t}}} \\
\mathrm{k} & =\frac{2.303}{30} \log _{10} \frac{100}{10} \\
& =\frac{2.303}{30 \mathrm{~min}}=7.67 \times 10^{-2} \mathrm{~min}^{-1}
\end{aligned}\)
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