Question 73. Time required for completion of 99.9 \% of a First order reaction is times of half life \left(t_{1 / 2}\right) of the reaction.
Answer (10)
Explanation:
\frac{t_{99.9\%}}{t_{1/2}}=\frac{\frac{2.303}k\left(\frac a{a-x}\right)}{\frac{2.303}k\log2} =\frac{\log\left(\frac{100}{100-99.9}\right)}{\log2} =\frac{\log10^3}{\log2}=\frac3{0.3}=10