Solveeit Logo

Question

Chemistry Question on Chemical Kinetics

For a first order reaction, the time required for 99.9%99.9\% of the reaction to take place is nearly

A

10 times that required for half of the reaction

B

100 times that required for two-third of the reaction

C

10 times that required for one-fourth of the reaction

D

20 times that required for half of the reaction

Answer

10 times that required for half of the reaction

Explanation

Solution

Time required to complete 99.9%99.9 \% of reaction
t99.9=2.303klog10010099.9t_{99.9}=\frac{2.303}{k} \log \frac{100}{100-99.9}
=2.303klog1000=6.909k=\frac{2.303}{k} \log 1000=\frac{6.909}{k}
and t1/2=2.303klog10010050 t_{1 / 2}=\frac{2.303}{k} \log \frac{100}{100-50}
=2.303klog2=0.693k=\frac{2.303}{k} \log 2=\frac{0.693}{k}
So, t99.9t_{99.9} is 1010 times that of required for half of the reaction.