Question

$t_{1/4}$ can be taken as the time taken for the concentration of a reactant to drop to $\frac{3}{4}$ of its initial value. If the rate constant for a first order reaction is $k$, the $t_{1/4}$ can be written as

• A. 0.75/k
• B. 0.69/k
• C. 0.29/k
• D. 0.10/k

Answer 1

Answer: (C)

For first-order kinetics
$k=\frac{2.303}{t}\log \left(\frac{a}{a-x}\right)$
$\therefore k=\frac{2.303}{t_{1/4}}\log \frac{a}{\frac{3a}{4}}$
$t_{1/4}=\frac{2.303\log \frac{4}{3}}{k}$
$=\frac{2.303(\log 4-\log 3)}{k}$
$=\frac{2.303(0.6020-0.4771)}{k}$
$=\frac{2.303\times 0.1249}{k}$
$=\frac{0.2876}{k}$
$=\frac{0.29}{k}$