JEE MAIN - Chemistry (2022 - 28th June Morning Shift - No. 21)
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)
Answer
16
Explanation
$$
\begin{aligned}
&\because \mathrm{kt}=\ln \frac{\mathrm{A}_{0}}{\mathrm{~A}} \\\\
&\frac{\ln 2}{\mathrm{t}_{\frac{1}{2}}} \mathrm{t}_{67 \%}=\ln \frac{\mathrm{A}_{0}}{0.33 \mathrm{~A}_{0}} \\\\
&\frac{\log 2}{\mathrm{t}_{\frac{1}{2}}} \mathrm{t}_{67 \%}=\log \frac{1}{0.33} \\\\
&\mathrm{t}_{67 \%}=1.566 \mathrm{t}_{1 / 2} \\\\
&\mathrm{x}=15.66
\end{aligned}
$$
Nearest integer $=16$
Nearest integer $=16$
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