JEE Main 202127 AugustMorning Shift

Exam: JEE Main
Year: 2021
Date: 27 August
Shift: Morning Shift
Subject: Chemistry
Chapter: Some Basic Concepts of Chemistry

JEE Main 2021 · 27 August · Morning Shift · Chemistry · Some Basic Concepts of Chemistry

The reaction that occurs in a breath analyser, a device used to...

The reaction that occurs in a breath analyser, a device used to determine the alcohol level in a person's blood stream is

$2{K}_{2}{Cr}_{2}{O}_{7} + 8{H}_{2}{SO}_{4} + 3{C}_{2}{H}_{6}O \to 2{Cr}_{2}{\left({SO}_{4}\right)}_{3} + 3{C}_{2}{H}_{4}{O}_{2} + 2{K}_{2}{SO}_{4} + 11{H}_{2}O$

If the rate of appearance of ${Cr}_{2}{\left({SO}_{4}\right)}_{3}$ is $2 . 67mol{min}^{- 1}$ at a particular time, the rate of disappearance of ${C}_{2}{H}_{6}O$ at the same time is mol $\min^{- 1} .$ (Nearest integer)

Correct answer

Step-by-step solution

$\dfrac{1}{3}\dfrac{- d\left[{C}_{2}{H}_{6}O\right]}{dt} = \dfrac{1}{2}\dfrac{d\left[{Cr}_{2}{\left({SO}_{4}\right)}_{3}\right]}{dt}$

$\dfrac{1}{3}$ (Rate of disappearance of {C}_{2}{H}_{6}O) = \dfrac{1}{2} $(Rate of disappearance of {Cr}_{2}{\left({SO}_{4}\right)}_{3})$

Rate of disappearance of ${C}_{2}{H}_{6}O = \left(\dfrac{2 . 67mol / min \times 3}{2}\right) = 4 . 005mol / min$

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