Question

Time required for 99.9% completion of a first order reaction is nearly ?

• A. 10 times that required for 25% of the reaction
• B. 10 times that required for 50% of the reaction
• C. 20 times that required for 50% of the reaction
• D. 100 times that required for 90% of the reaction

Answer 1

The correct option is B

10 times that required for 50% of the reaction

For first order reaction, $t=\frac{2.303}{k}log\frac{a}{a-x}$
$t_{99.9}=\frac{2.303}{k}log\frac{100}{100-99.9}=\frac{2.303}{k}\times 3$
$t_{50}=\frac{2.303}{k}log\frac{100}{50}=\frac{2.303}{k}log2=\frac{2.303}{k}\times 0.301$
$t_{\frac{99.9}{t_{50}}}=10approx$.