MHT CET · Chemistry · Chemical Kinetics
What is the value of rate constant of first order reaction, if it takes 15 minutes for consumption of \(20 \%\) of reactants?
- A \(1 \cdot 84 \times 10^{-2} \min ^{-1}\)
- B \(1 \cdot 38 \times 10^{-2} \min ^{-1}\)
- C \(1 \cdot 07 \times 10^{-2} \min ^{-1}\)
- D \(1 \cdot 48 \times 10^{-2} \min ^{-1}\)
Answer & Solution
Correct Answer
(D) \(1 \cdot 48 \times 10^{-2} \min ^{-1}\)
Step-by-step Solution
Detailed explanation
(B)
\([\mathrm{A}]_{0}=\) Original amount of reactant \(=100\)
\([\mathrm{A}]_{t}=\) Reactant remaining unreacted \(=100-20=80\)
For first order reaction,
\(\begin{aligned}
\mathrm{k} &=\frac{2.303}{\mathrm{t}} \log _{10} \frac{[\mathrm{A}]_{0}}{[\mathrm{~A}]_{t}} \\
\therefore \mathrm{k} &=\frac{2.303}{15 \mathrm{~min}} \log _{10} \frac{100}{80}=0.0148 \mathrm{~min}^{-1} \\
\therefore \mathrm{k} &=1.48 \times 10^{-2} \mathrm{~min}^{-1}
\end{aligned}\)
\([\mathrm{A}]_{0}=\) Original amount of reactant \(=100\)
\([\mathrm{A}]_{t}=\) Reactant remaining unreacted \(=100-20=80\)
For first order reaction,
\(\begin{aligned}
\mathrm{k} &=\frac{2.303}{\mathrm{t}} \log _{10} \frac{[\mathrm{A}]_{0}}{[\mathrm{~A}]_{t}} \\
\therefore \mathrm{k} &=\frac{2.303}{15 \mathrm{~min}} \log _{10} \frac{100}{80}=0.0148 \mathrm{~min}^{-1} \\
\therefore \mathrm{k} &=1.48 \times 10^{-2} \mathrm{~min}^{-1}
\end{aligned}\)
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