A B is a first order reaction. The concentration of A is decreased from x mol L^{-1} to y\, {mol}\, {L}^{-1} in 100 min . What is the average velocity of the reaction in mol {L}^{-1}\, {min}^{-1}\, ? [AP EAPCET 23 May 2025 S1]

Correct Answer is : ∣x−y∣100\;\; ≤ft|x-y|/100100∣x−y∣

Correct Answer is : ∣x−y∣100\;\; ≤ft|x-y|/100100∣x−y∣

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Chemical Kinetics

Section: Chemistry

Question : 3 of 18

Marks: +1, -0

A \longrightarrow B

x

L^{-1}

y\, \mathrm{mol}\, \mathrm{L}^{-1}

\mathrm{L}^{-1}\, \mathrm{min}^{-1}\, ?

Solution:

We want to find the average velocity (rate) for the reaction as

A

goes to

B

when the amount of

A

drops from

x

y\, \mathrm{mol/L}

in 100 minutes.

The change in the concentration of

A

is:

\Delta[A]=y-x\, \mathrm{mol/L}

This tells us how much

A

has changed in value.

The average velocity means how much

A

changes per minute. We can find it by dividing the total change in

A

by the time taken: Average velocity

=\; \frac{\Delta[A]}{\Delta t}

where

\Delta t = 100

minutes.

So, Average velocity

=\; \frac{y-x}{100}\, \mathrm{mol}\, \mathrm{L}^{-1}\, \mathrm{min}^{-1}

If you need the answer as a positive value (since rates are usually positive), write it as

\; \frac{\left|x-y\right|}{100}\, \mathrm{mol}\, \mathrm{L}^{-1}\, \mathrm{min}^{-1}

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