Question

The decomposition of $H_2{O}_2$ can be followed by titration with $KMn{O}_4$ and is found to be a first-order reaction. The rate constant is $4.5\times {10}^{-2}$. In an experiment, the initial titre value was 25 mL. The titre value will be 5 mL after a lapse of:

• A. $4.5\times {10}^{-2}\times 5min$
• B. $\frac{lo{g}_{e}5}{4.5\times {10}^{-2}}min$
• C. $\frac{lo{g}_{e}5/4}{4.5\times {10}^{-2}}min$
• D. None of the above

Answer 1

Answer: (B)

$\frac{lo{g}_{e}5}{4.5\times {10}^{-2}}min$

As we know,
$t=\frac{2.303}{k}log\frac{V_0}{V_1}$
$=\frac{1}{k}ln\frac{V_0)}{V_1}$
$=\frac{1}{4.5\times {10}^{-2}mi{n}^{-1}}In\frac{25mL}{5mL}$
$=\frac{lo{g}_{e}5}{4.5\times {10}^{-2}}min$