The reaction that occurs in a breath analyser, a device used to determine the alcohol level in a per

Q: Why do we divide by the stoichiometric coefficients?

Dividing by the stoichiometric coefficients normalizes the rates so that the overall rate of reaction is the same regardless of which reactant or product you are measuring.

Q: Why is there a negative sign for the rate of disappearance?

The concentration of reactants decreases over time, making the change in concentration negative. The negative sign ensures the rate of disappearance is a positive quantity.

Q: Can we use any other reactant or product to find the rate?

Yes! As long as you know the rate of appearance or disappearance of one species and the balanced chemical equation, you can find the rate for any other species.

Q: What if the stoichiometric coefficients were 1?

If the coefficients were 1, the rate of disappearance of the reactant would exactly equal the rate of appearance of the product.

Q: Why do we round to the nearest integer?

The question specifically asks for the nearest integer, which is a common format in numerical answer type questions in competitive exams like JEE.

Question

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

Sigma Percentile · Accepted Answer

Have you ever wondered how a breath analyser works? When a person blows into the device, the alcohol (ethanol) in their breath reacts with a chemical mixture. This isn't just a random reaction; it's a highly precise redox reaction that changes color, allowing the device to calculate the exact alcohol concentration in the blood!

In this problem, we are diving deep into the chemical kinetics of this very reaction. We are given the balanced chemical equation and the rate at which one of the products is formed. Our mission? To find out how fast the alcohol is being consumed. Let's break it down step-by-step.

### Analyzing the Setup

The core of our problem is the balanced chemical equation:

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

This equation tells us a story. It says that for every **3 moles of ethanol ($C_2H_6O$)** that are consumed, exactly **2 moles of chromium sulfate ($Cr_2(SO_4)_3$)** are produced.

In chemical kinetics, the rate at which a reaction proceeds must be consistent, regardless of which reactant or product we are looking at. However, because different molecules are consumed and produced in different amounts (dictated by their stoichiometric coefficients), their individual rates of appearance and disappearance will differ.

### The Master Equation

To unify these different rates, we divide the rate of disappearance or appearance of each species by its respective stoichiometric coefficient. This gives us the overall **Rate of Reaction**.

For our specific molecules of interest, ethanol and chromium sulfate, the relationship is:

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

Here, the negative sign indicates that ethanol is disappearing (its concentration is decreasing), while the positive sign indicates that chromium sulfate is appearing.

### Rearranging for the Unknown

We are asked to find the rate of disappearance of ethanol, which is represented by $-\frac{d[C_2H_6O]}{dt}$.

By rearranging our master equation, we can isolate this term:

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

This equation makes intuitive sense: since 3 moles of ethanol are consumed for every 2 moles of chromium sulfate produced, ethanol must be disappearing 1.5 times faster than chromium sulfate is appearing!

### Final Calculation

We are given that the rate of appearance of chromium sulfate is $2.67 	ext{ mol min}^{-1}$. Let's substitute this value into our rearranged equation:

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

Now, it's just a matter of simple arithmetic.

$$-\frac{d[C_2H_6O]}{dt} = 3 	imes 1.335$$

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

The question asks for the answer to the **nearest integer**. Since $4.005$ is extremely close to $4$, we round it down.

**Final Answer: 4**

And there you have it! By understanding the stoichiometry of the reaction and the fundamental principles of chemical kinetics, we've successfully determined the rate at which alcohol is consumed in a breath analyser. Keep practicing, and these concepts will become second nature to you!

Enter Numerical Value:

Visualized Solution

$The Chemical Reaction$

$Rate of Reaction Relation$

$Rate of Disappearance$

$Substitution$

$Calculation$

$Final Answer$

The Sigma Insight: Rate of Chemical Reaction

Analyzing the Setup

The Master Equation

Rearranging for the Unknown

Final Calculation

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The reaction $2 A + B_{2} \to 2 A B$ is an elementary reaction. For a certain quantity of reactants, if the volume of the reaction vessel is reduced by a factor of 3, the rate of the reaction increases by a factor of ........... . (Round off to the nearest integer).

Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Chemical Reaction

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Rate of Reaction Relation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Rate of Disappearance

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substitution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Answer