The reaction that occurs in a breath analyser, a device used to determine the alcohol level in a person's blood stream is 2K_2Cr_2O_7 + 8H_2SO_4 + 3C_2H_6O \to 2Cr_2(SO_4)_3 + 3C_2H_4O_2 + 2K_2SO_4 + 11H_2O If the rate of appearance of Cr_2(SO_4)_3 is 2.67 mol min - 1 at a particular time, the rate of disappearance of C_2H_6O at the same time is \underline\hspace2cm mol min - 1. (Nearest integer)

Chemical Kinetics and Nuclear Chemistry PYQ — JEE Main 2021 · 27 Aug · Shift 1 · Q14 with Solution

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Correct answer: 4

$2K_2Cr_2O_7 + 8H_2SO_4 + 3C_2H_6O \to 2Cr_2(SO_4)_3 + 3C_2H_4O_2 + 2K_2SO_4 + 11H_2O$

For a reaction

$aA + bB \to cC + dD$

the rate relation is

$-\frac{1}{a}\frac{d[A]}{dt} = \frac{1}{c}\frac{d[C]}{dt}$

Here, comparing ethanol $C_2H_6O$ and $Cr_2(SO_4)_3$ :

So,

$-\frac{1}{3}\frac{d[C_2H_6O]}{dt} = \frac{1}{2}\frac{d[Cr_2(SO_4)_3]}{dt}$

Given rate of appearance of $Cr_2(SO_4)_3$ is

$\frac{d[Cr_2(SO_4)_3]}{dt} = 2.67\ \text{mol min}^{-1}$

Therefore,

$-\frac{1}{3}\frac{d[C_2H_6O]}{dt} = \frac{1}{2}(2.67)$

$-\frac{d[C_2H_6O]}{dt} = 3 \times \frac{2.67}{2}$

$-\frac{d[C_2H_6O]}{dt} = 4.005\ \text{mol min}^{-1}$

$4.005 \approx 4$

So, the rate of disappearance of $C_2H_6O$ is

$\boxed{4\ \text{mol min}^{-1}}$