If the rate constant of a reaction is $0.03 \mathrm{~s}^{-1}$, how much time does it take for $7.2 \

Question

If the rate constant of a reaction is $0.03 \mathrm{~s}^{-1}$, how much time does it take for $7.2 \mathrm{~mol} \mathrm{~L}^{-1}$ concentration of the reactant to get reduced to $0.9 \mathrm{~mol} \mathrm{~L}^{-1}$ ? (Given $: \log 2=0.301$ )

Professor PC Thomas Classes · Answer

(c) : Rate constant, $k=0.03 \mathrm{~s}^{-1}$
For $1^{	ext {st }}$ order reaction, $k=\frac{2.303}{t} \log \frac{a}{a-x}$
So, $t=\frac{2.303}{0.03} \log \frac{7.2}{0.9}=\frac{2.303}{0.03} \log 8=\frac{2.303}{0.03} 	imes 3 \log 2$
$
=\frac{2.303}{0.03} 	imes 3 	imes 0.301 \
=69.32 \mathrm{~s}
$
[Given, $\log 2=0.301$ ]