Verified Solution • 2025–26 Session moderate

QJEEM22Ch163e: JEE Main 2022 Chemistry - Class 12 JEEM22Ch163e Solution: For a first order reaction A B, the rate constant, k = 5.5 10…

For a first order reaction A →\to B, the rate constant, k = 5.5 ×\times 10 −- 14 s −- 1. The time required for 67% completion of reaction is x ×\times 10 −- 1 times the half life of reaction. The value of x is _____________ (Nearest integer)

(Given : log 3 = 0.4771)

Step-by-Step Verified Solution

amp;∵kt=ln⁡A0 Aamp;ln⁡2t12t67%=ln⁡A00.33 A0amp;log⁡2t12t67%=log⁡10.33amp;t67%=1.566t1/2amp;x=15.66\begin{aligned}&\because \mathrm{kt}=\ln \frac{\mathrm{A}_{0}}{\mathrm{~A}} \\\\&\frac{\ln 2}{\mathrm{t}_{\frac{1}{2}}} \mathrm{t}_{67 \%}=\ln \frac{\mathrm{A}_{0}}{0.33 \mathrm{~A}_{0}} \\\\&\frac{\log 2}{\mathrm{t}_{\frac{1}{2}}} \mathrm{t}_{67 \%}=\log \frac{1}{0.33} \\\\&\mathrm{t}_{67 \%}=1.566 \mathrm{t}_{1 / 2} \\\\&\mathrm{x}=15.66\end{aligned}

Nearest integer =16=16

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