Question

For a first-order reaction, the time required for 99.0% completion is twice for the time required for the completion of 90% of the reaction.

If true enter 1, else enter 0.

Answer 1

For 99% completion,
$t_{99}$%$=\frac{2.303}{K}log\frac{100}{100-99}$ .....(i)

For 90% completion,
$t_{90}$%$=\frac{2.303}{K}log\frac{100}{100-90}$ .....(ii)

$\frac{t_{99}}{t_{90}}=\frac{log100}{log10}=2$

$\therefore t_{99}$% $=2\times t_{90}$%

Hence, it is a true statement.