JEE Main 2025 Apr 2 Shift 2, Chemistry Q21: Reactions of Various Orders

Reading from the curve, the concentration drops from 40 g L${}^{-1}$ to 20 g L${}^{-1}$ over a time interval of 10 s (one half-life). This is the signature of first-order kinetics.

Rate constant from this half-life:

$k=\frac{1}{t}\ln \frac{[A{]}_0}{[A{]}_t}=\frac{1}{10}\ln \frac{40}{20}=\frac{\ln 2}{10}{\text{s}}^{-1}$.

Time required for the concentration to fall from 50 g L${}^{-1}$ to 2.5 g L${}^{-1}$ (a factor of 20):

$t=\frac{1}{k}\ln \frac{50}{2.5}=\frac{10}{\ln 2}\ln 20$.

Using $\log 2=0.3010$, so $\ln 2=2.303\times 0.3010$ and $\ln 20=2.303\times 1.3010$:

$t=10\times \frac{1.3010}{0.3010}\approx 43.2\text{s}$.

$\therefore$ Nearest integer $=43$ s.