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JEE Main 202430 JanuaryMorning Shift
Exam
JEE Main
Year
2024
Date
30 January
Shift
Morning Shift
Subject
Chemistry
Chapter
Chemical Kinetics

JEE Main 2024 · 30 January · Morning Shift · Chemistry · Chemical Kinetics

The rate of first order reaction is 0 . 04mol L ^ - 1 s ^ - 1 at 10...

The rate of first order reaction is 0.04molL−1s−10 . 04mol{L}^{- 1}{s}^{- 1} at 1010 minutes and 0.03molL−1s−10 . 03mol{L}^{- 1}{s}^{- 1} at 2020 minutes after initiation. Half life of the reaction is ______ minutes. (Given log⁡2=0.3010,log⁡3=0.4771)\log 2 = 0 . 3010, \log 3 = 0 . 4771 )

Round off your answer to the nearest integer.

Correct answer

0

Step-by-step solution

 Rater = k\left[A\right]0 . 04 = k{\left[{A}_{0}\right]}_{10}0 . 03 = k{\left[{A}_{}\right]}_{20}</p><p></p><p>⇒ \dfrac{{\left[{A}_{0}\right]}_{10}}{{\left[A\right]}_{20}} = \dfrac{0 . 04}{0 . 03} = \dfrac{4}{3}</p><p>Fromt=2.303klog[A]10[A]2010=2.303klog4310=2.303k(0.6020−0.4771)k=2.30310×0.1249=2.876×10−2min⁡−1</p><p>Fromt = \dfrac{2 . 303}{k}log\dfrac{{\left[A\right]}_{10}}{{\left[A\right]}_{20}}10 = \dfrac{2 . 303}{k}log\dfrac{4}{3}10 = \dfrac{2 . 303}{k} ( 0 . 6020 - 0 . 4771 ) k = \dfrac{2 . 303}{10} \times 0 . 1249 = 2 . 876 \times {10}^{- 2}\min^{- 1}

t1/2=0.693k=0.6932.876×10−2=0.24×102min=24min.{t}_{1/2} = \dfrac{0 . 693}{k} = \dfrac{0 . 693}{2 . 876 \times {10}^{- 2}} = 0 . 24 \times {10}^{2}min = 24min .

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