A first order reaction has the rate constant, $\mathrm{k=4.6\times10^{-3}~s^{-1}}$. The number of correct statement/s from the following is/are __________

Given : $\mathrm{\log3=0.48}$

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 ($\mathrm{1-e^{-kt}}$)

E. The rate and the rate constant have the same unit.

(A) $\underset{1-\alpha}{\mathrm{A}} \longrightarrow$ Products <br/><br/> $\mathrm{k}=4.6 \times 10^{-3} \mathrm{~s}^{-1}$ <br/><br/> $\mathrm{kt}=\ln \frac{1}{1-\alpha}$ <br/><br/> $\alpha=1-\mathrm{e}^{-\mathrm{kt}}$ <br/><br/> Reaction completes at infinite time. <br/><br/>(B) For first order reaction, <br/><br/> Half-life $=\frac{0.693}{4.6 \times 10^{-3}}=150.65 \mathrm{~s}$ <br/><br/> (C) For $10 \%$ completion, $t=t_1$ <br/><br/>$$ \begin{aligned} & a=100, a-x=100-10=90 \\\\ & k=\frac{2.303}{t_1} \log \frac{100}{90}=4.6 \times 10^{-3} \\\\ & t_1=\frac{2.303}{4.6 \times 10^{-3}} \times 0.04575 \end{aligned} $$ <br/><br/>For $90 \%$ completion $t=t_2, a=100, a-x=100-90=10$ <br/><br/>$$ \begin{aligned} & k=\frac{2.303}{t_2} \log \frac{100}{10}=4.6 \times 10^{-3} \\\\ & \frac{2.303}{t_2} \times 1=4.6 \times 10^{-3} \Rightarrow t_2=\frac{2.303}{4.6 \times 10^{-3}} \\\\ & \frac{t_2}{t_1}=\frac{1}{0.04575}=21.85 \quad \text { (Incorrect) } \end{aligned} $$ <br/><br/> (E) Unit of rate = mol L^–1s^–1 <br/><br/>Unit of rate constant = s^–1 <br/><br/>Both have different units. <br/><br/> $\therefore$ Number of correct statements $=1$