A first order reaction correct statement

JEE MAIN 2023

25-1-23 S2

Question

A first order reaction has the rate constant, $\mathrm{k}=4.6 \times 10^{-3} \mathrm{~s}^{-1}$. The number of correct statement/s from the following is/are $\_\_\_\_$ .
Given : $\log 3=0.48$
A. Reaction completes in 1000 s .
B. The reaction has a half-life of 500 s .
C. The time required for $10 \%$ completion is 25 times the time required for $90 \%$ completion.
D. The degree of dissociation is equal to $\left(1-e^{-k t}\right)$.
E. The rate and the rate constant have the same unit.

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Solution

Sol. $$ \begin{aligned} & t_{10 \%}=\frac{1}{K} \ln \left(\frac{a}{a-x}\right)=\frac{1}{K} \ln \left(\frac{100}{90}\right) \\ & t_{10 \%}=\frac{2.303}{K}(\log 10-\log 9) \\ & t_{10 \%}=\frac{2.303}{K} \times(0.04) \end{aligned} $$ Similarly $\begin{aligned} & t_{90 \%}=\frac{1}{K} \ln \left(\frac{100}{10}\right) \\ & t_{90 \%}=\frac{2.303}{K} \\ & \frac{t_{90 \%}}{t_{10 \%}}=\frac{1}{0.04}=25 \\ & e^{k t}=\frac{a}{a-x} \\ & \frac{a-x}{a}=e^{-k t} \\ & 1-\frac{x}{a}=e^{-k t} \\ & x=a\left(1-e^{-k t}\right) \\ & \alpha=\frac{x}{a}=\left(1-e^{-k t}\right)\end{aligned}$

Question Tags

JEE Main

Chemistry

Medium

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