1. Tardigrade
  2. Question
  3. Chemistry
  4. The rate of a first order reaction is 0.04 mol L-1 s-1 at 30 min and 0.03 mol L-1 s-1 at 40 min. Thus, half-life of the reaction is

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Q. The rate of a first order reaction is $0.04$ mol $L^{-1} s^{-1}$ at $30$ min and $0.03$ mol $L^{-1} s^{-1}$ at $40$ min. Thus, half-life of the reaction is

Chemical Kinetics

Solution:

$k =\frac{2.303}{t}$ log $\left(\frac{a}{a -x}\right)$
$\frac{kt}{2.303}$ log $a -$ log $\left(a -x\right)$
$\frac{kt_{1}}{2.303} =$ log $a -$ log $\left(a -x_{1}\right)$ at time $t_{1} ....\left(i\right)$
$\frac{kt_{2}}{2.303} =$ log $a -$ log $\left(a -x_{2}\right)$ at time $t_{2} ....\left(ii\right)$
Subtracting (i) from (ii)
$\therefore \frac{k}{2.303}\left(t_{2} - t_{1}\right) =$ log $\left(\frac{a -x_{1}}{a -x_{2}}\right)$
Rate, $r_{1} = k\left(a -x_{1}\right);r_{2} = k\left(a -x_{2}\right)$
$\therefore \frac{\left(a -x_{1}\right)}{\left(a -x_{2}\right)} =\frac{r_{1}}{r_{2}} \Rightarrow \frac{0.04}{0.03} =\frac{4}{3}$
$\frac{k\left(40 -30\right)}{2.303} =$ log $\frac{4}{3}$
$\therefore k = \frac{2.303}{10}$ log $\frac{4}{3} = \frac{2.303}{10}$ log $1.33$
$ = \frac{2.303 \times0.125}{10} = 0.02878$ min$^{-1}$
$T_{50} = \frac{0.693}{0.02878} = 24.1$ min