For the reaction \(\mathrm{A} \rightarrow \mathrm{B}\) the following graph was obtained. The time required (in seconds) for the concentration of A to reduce to \(2.5 \mathrm{~g} \mathrm{~L}^{-1}\) (if the initial concentration of A was \(50 \mathrm{~g} \mathrm{~L}^{-1}\) ) is \_\_\_\_. (Nearest integer) Given : \(\log 2=0.3010\)

JEE Main · 2025 · 02 Apr 2025 · Shift 2 · Chemical Kinetics

For the reaction

A \to B

the following graph was obtained. The time required (in seconds) for the concentration of A to reduce to

2.5 g L^{- 1}

(if the initial concentration of A was

50 g L^{- 1}

) is ____. (Nearest integer)

Given :

lo g 2 = 0.3010

Answer: 43

The smooth exponential fall of concentration with time marks first-order kinetics: the concentration halves in equal intervals, from 40 to 20 g L

^{- 1}

between 5 s and 15 s and from 20 to 10 g L

^{- 1}

between 15 s and 25 s, so the half-life is 10 s and

k = \frac{ln 2}{10} = 0.0693 s^{- 1}

. For the concentration to fall from 50 g L

^{- 1}

to 2.5 g L

^{- 1}

t = \frac{2.303}{k} lo g \frac{50}{2.5} = \frac{2.303}{0.0693} lo g 20 = \frac{2.303 \times 1.301}{0.0693} = 43.2 s

. To the nearest integer the time required is 43 s.

JEE Main · 2025 · 02 Apr 2025 · Shift 2 · Integer answer