Question

The rate of a first order reaction is $0.04{\text{mol L}}^{-1}{\text{min}}^{-1}$ at $10\text{minutes}$ and $0.02{\text{mol L}}^{-1}{\text{min}}^{-1}$ at $20\text{minutes}$. Find the half-life of the reaction (in $\text{min}$).
$(\text{Given}:\ln 2=0.693)$

Answer 1

Let the concentrations of the reactant after $10\text{minutes}$ and $20\text{minutes}$ be $C_1$ and $C_2$ respectively.
$\therefore \text{Rate after}10\text{minutes}=k.C_1=0.04{\text{mol L}}^{-1}{\text{s}}^{-1}\text{Rate after}20\text{minutes}=k.C_2=0.02{\text{mol L}}^{-1}{\text{s}}^{-1}$
$\therefore \frac{C_1}{C_2}=\frac{0.04}{0.02}=2$
Let $\frac{C_1}{C_2}$ be the ratio of initial and final concentrations.
$\therefore k=\frac{1}{10}ln\frac{C_1}{C_2}=\frac{1}{10}ln2$
$\Rightarrow k=\frac{1}{10}\times 0.693=0.0693{\text{min}}^{-1}$
We know, $t_{\frac{1}{2}}$ for a first order reaction is,
$t_{\frac{1}{2}}=\frac{0.693}{k}=\frac{0.693}{0.0693}=10\text{min}$