Question

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

• A. $32.4$ min
• B. $51.2$ min
• C. $46.8$ min
• D. $24.1$ min

Answer 1

Answer: (D)

$k=\frac{2.303}{t}$ log $\left(\frac{a}{a-x}\right)$
$\frac{kt}{2.303}$ log $a-$ log $\left(a-x\right)$
$\frac{k{t}_1}{2.303}=$ log $a-$ log $\left(a-x_1\right)$ at time $t_1....\left(i\right)$
$\frac{k{t}_2}{2.303}=$ 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 $1.33$
$=\frac{2.303\times 0.125}{10}=0.02878$ min${}^{-1}$
$T_{50}=\frac{0.693}{0.02878}=24.1$ min