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Question Paper from: IIT JEE Main 2021 TestSeries Paper 1

In the first-order reaction t_99.9%is 60 minutes, t_50%of the reaction would be nearly

36 minutes

6 minutes

10 minutes

30 minutes

Solution:

Chapter Name: Chemical Kinetics

Difficulty Level: Moderate

First Order Reaction

$K = \frac{2.303}{t} l o g \frac{a}{( a - x )}$

$K = \frac{2.303}{60} l o g \frac{100}{( 100 - 99.9 )}$

$K = \frac{2.303}{60} l o g \frac{100}{0.1}$

$K = \frac{2.303}{60} \times 3$

$K = 0.115$

for t_50%

$0.115 = \frac{2.303}{t} l o g \frac{100}{50}$

$0.115 = \frac{2.303}{t} l o g 2$

$0.115 = \frac{2.303 \times 0.3010}{t}$

$t = \frac{0.693}{0.115} = 6 min$

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IIT JEE Main 2021 TestSeries Paper 1

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