JEE Main 2026 Apr 4 Shift 2, Chemistry Q24: Reactions of Various Orders

For a first-order reaction: $k=\frac{1}{t}\ln \frac{[A{]}_0}{[A{]}_t}$.

From t = 0 to t = 20 min:

$k=\frac{1}{20}\ln \frac{0.6500}{0.00065}=\frac{1}{20}\ln (1000)=\frac{3\ln 10}{20}$ min${}^{-1}$.

From t = 0 to t = x:

$x=\frac{1}{k}\ln \frac{0.6500}{0.0650}=\frac{20}{3\ln 10}\ln (10)=\frac{20}{3}\approx 6.67$ min.

Nearest integer: 7.