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A first order reaction is $25 \%$ completed in 40 minutes. What is the rate constant $\mathrm{k}$ for the reaction?
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Verified Answer
The correct answer is:
$\frac{2.303 \times \log 1 \cdot 33}{40}$
(B)
$\begin{aligned}[\mathrm{A}]_{0} &=100,[\mathrm{~A}]_{\mathrm{t}}=100-25=75, \mathrm{t}=40 \min \\ \mathrm{k} &=\frac{2.303}{\mathrm{t}} \log _{10} \frac{[\mathrm{A}]_{0}}{[\mathrm{~A}]_{\mathrm{t}}} \\ \therefore \mathrm{k} &=\frac{2.303}{40} \log _{10} \frac{100}{75}=\frac{2.303 \times \log 1.33}{40} \end{aligned}$
$\begin{aligned}[\mathrm{A}]_{0} &=100,[\mathrm{~A}]_{\mathrm{t}}=100-25=75, \mathrm{t}=40 \min \\ \mathrm{k} &=\frac{2.303}{\mathrm{t}} \log _{10} \frac{[\mathrm{A}]_{0}}{[\mathrm{~A}]_{\mathrm{t}}} \\ \therefore \mathrm{k} &=\frac{2.303}{40} \log _{10} \frac{100}{75}=\frac{2.303 \times \log 1.33}{40} \end{aligned}$
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