Time required for 99.9 \% completion of a first order reaction is \underline\hspace2cm times the time required for completion of 90 \% reaction.(nearest integer)

Chemical Kinetics and Nuclear Chemistry PYQ — JEE Main 2024 · 6 Apr · Shift 1 · Q22 with Solution

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Correct answer: 3

For a first-order reaction,

kt = \ln\left(\frac{[A]_0}{[A]}\right)

If fraction reacted is $x$ , then fraction remaining is $(1-x)$ . So,

t = \frac{1}{k}\ln\left(\frac{1}{1-x}\right)

For $90\%$ completion,

x = 0.90 \Rightarrow 1-x = 0.10

Thus,

t_{90} = \frac{1}{k}\ln\left(\frac{1}{0.10}\right)=\frac{1}{k}\ln(10)

For $99.9\%$ completion,

x = 0.999 \Rightarrow 1-x = 0.001

Thus,

t_{99.9} = \frac{1}{k}\ln\left(\frac{1}{0.001}\right)=\frac{1}{k}\ln(1000)

\frac{t_{99.9}}{t_{90}} = \frac{\ln(1000)}{\ln(10)}

Since,

\ln(1000)=\ln(10^3)=3\ln(10)

So,

\frac{t_{99.9}}{t_{90}} = 3

\boxed{3}

The time required for $99.9\%$ completion is 3 times the time required for $90\%$ completion.