The half life period of a first order chemical reaction is 6.93 minutes. The time required for the completion of 99% of the chemical reaction will be (log 2=0.301) :

Chemical Kinetics and Nuclear Chemistry PYQ — JEE Main 2009 · Shift 0 · Q19 with Solution

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Correct answer: C

For a first-order reaction, $t_{1/2} = \frac{0.693}{k}$

Given: $t_{1/2} = 6.93\ \text{min}$

So, $6.93 = \frac{0.693}{k}$

Hence, $k = \frac{0.693}{6.93} = 0.1\ \text{min}^{-1}$

If the reaction is 99% complete, then 1% of the reactant remains.

So, $\frac{[A]_t}{[A]_0} = 0.01$

For a first-order reaction, $t = \frac{2.303}{k} \log \frac{[A]_0}{[A]_t}$

Thus, $t = \frac{2.303}{0.1} \log \frac{1}{0.01}$

Since, $\frac{1}{0.01} = 100$

and $\log 100 = 2$

therefore, $t = 23.03 \times 2 = 46.06\ \text{minutes}$

$46.06\ \text{minutes}$ corresponds to Option C.

Stored correct answer: C

Derived answer: C

So, the derived answer agrees with the stored correct answer.