If the rate constant of a first order reaction is $4.606 imes 10^{-3} \mathrm{~s}^{-1}$, then find the time required for $400 \mathrm{~g}$ of the reactant to reduce to $50 \mathrm{~g}$.

Question

If the rate constant of a first order reaction is $4.606 	imes 10^{-3} \mathrm{~s}^{-1}$, then find the time required for $400 \mathrm{~g}$ of the reactant to reduce to $50 \mathrm{~g}$.

Aicharya · Accepted Answer

Given, Rate constant, $k=4.606 	imes 10^{-3} \mathrm{~s}^{-1}$ Concentration at, time $t=50 \mathrm{~g}$ Initial concentration $=400 \mathrm{~g}$ Now, formula, $ \begin{aligned} k & =\frac{2.303}{t} \log \frac{A_0}{A_t} \ t & =\frac{2.303}{4.606 	imes 10^{-3}} \log \frac{400}{50} \ & =\frac{2.303}{0.00460} \log 8 \ & =\frac{2.303 	imes 0.9030}{0.00460} \ & =452.08 \mathrm{~s}=\frac{452.08}{60} \mathrm{~min} \ t & =7.52 \mathrm{~min} \end{aligned} $

If the rate constant of a first order reaction is $4.606 \times 10^{-3} \mathrm{~s}^{-1}$, then find the…

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