For a first order reaction, the time required for completion of 90% reaction is 'x' times the half life of the reaction. The value of 'x' is (Given : ln 10 = 2.303 and log 2 = 0.3010)

Chemical Kinetics and Nuclear Chemistry PYQ — JEE Main 2022 · 24 Jun · Shift 2 · Q5 with Solution

View written solutionFree

Correct answer: C

For a first-order reaction,

t = \frac{2.303}{k} \log \frac{a}{a-x}

where $a$ is initial amount and $x$ is amount reacted.

If 90% reaction is completed, then 10% remains. So,

\frac{a}{a-x} = \frac{100}{10} = 10

Hence,

t_{90\%} = \frac{2.303}{k} \log 10 = \frac{2.303}{k}

For a first-order reaction,

t_{1/2} = \frac{0.693}{k}

Also using the given value $\log 2 = 0.3010$ ,

t_{1/2} = \frac{2.303}{k} \log 2 = \frac{2.303}{k}(0.3010)

t_{90\%} = x\, t_{1/2}

Therefore,

x = \frac{t_{90\%}}{t_{1/2}} = \frac{\frac{2.303}{k}}{\frac{2.303}{k}(0.3010)} = \frac{1}{0.3010}

x \approx 3.32

\boxed{\text{C: } 3.32}

Stored correct answer is C, which matches the derived answer.