MHT CET · Chemistry · Chemical Kinetics
What is rate constant of a first order reaction if time required to decrease concentration of reactant from \(1.6 \mathrm{M}\) to \(0.4 \mathrm{M}\) is 12 hours?
- A 0.116 hour \(^{-1}\)
- B 0.167 hour \(^{-1}\)
- C 0.4 hour \(^{-1}\)
- D 0.78 hour \(^{-1}\)
Answer & Solution
Correct Answer
(A) 0.116 hour \(^{-1}\)
Step-by-step Solution
Detailed explanation
We know
\(\begin{aligned} \mathrm{K} & =\frac{2.303}{\mathrm{t}} \log \frac{\mathrm{a}_0}{\mathrm{a}_{\mathrm{t}}} \\ & =\frac{2.303}{12} \log \frac{1.6}{0.4} \\ & =\frac{2.303}{12} \log ^4 \\ & =\frac{2.303}{12} \times 2 \log ^2 \\ & =0.115\end{aligned}\)
\(\begin{aligned} \mathrm{K} & =\frac{2.303}{\mathrm{t}} \log \frac{\mathrm{a}_0}{\mathrm{a}_{\mathrm{t}}} \\ & =\frac{2.303}{12} \log \frac{1.6}{0.4} \\ & =\frac{2.303}{12} \log ^4 \\ & =\frac{2.303}{12} \times 2 \log ^2 \\ & =0.115\end{aligned}\)
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