In a first order reaction concentration of reactant decreases from $20 \mathrm{~m} \mathrm{~mol}$ to $10…
In a first order reaction concentration of reactant decreases from $20 \mathrm{~m} \mathrm{~mol}$ to $10 \mathrm{~m} \mathrm{~mol}$ in $1.151 \mathrm{~min}$. What is rate constant?
$1.15 \mathrm{~min}^{-1}$
$3.0 \mathrm{~min}^{-1}$
$5.50 \mathrm{~min}^{-1}$
$0.60 \mathrm{~min}^{-1}$
Solution
for 1st order reaction
$\begin{aligned}
& \text {rate constant }(\mathrm{k})=\frac{2.303}{t} \log \frac{\mathrm{a}_{\mathrm{o}}}{\mathrm{a}_{\mathrm{t}}} \\
& \mathrm{a}_{\mathrm{o}}=\text { Initial amount }=20 \mathrm{~m} \mathrm{~mol} \\
& \text { at }=\text { final amount }=10 \mathrm{~m} \mathrm{~mol} \\
& \mathrm{k}=\frac{2.303}{1.151} \cdot \log \frac{20}{10} \\
& =\frac{2.303}{1.151} \times \log 2 \\
& =\frac{2.303 \times .0 .3010}{1.151} \\
& \mathrm{k}=0.60 \mathrm{~min}^{-1}
\end{aligned}$
II $^{\text {nd }}$ Method
As concentration decreases from $20 \mathrm{~m} \mathrm{~mol}$ to $10 \mathrm{~m} \mathrm{~mol}$ in $1.151 \mathrm{~min}$ $\ell_{1 / 2}=1.151 \mathrm{~min}=\frac{\ell \mathrm{n} 2}{\mathrm{~K}}$
$\mathrm{k}=\frac{\ln 2}{1.151}=\frac{0.693}{1.151}$ $\mathrm{k}=0.60 \mathrm{~min}^{-1}$