JEE MAIN - Chemistry (2023 - 25th January Evening Shift - No. 16)
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.
Answer
1
Explanation
(A) $\underset{1-\alpha}{\mathrm{A}} \longrightarrow$ Products
$\mathrm{k}=4.6 \times 10^{-3} \mathrm{~s}^{-1}$
$\mathrm{kt}=\ln \frac{1}{1-\alpha}$
$\alpha=1-\mathrm{e}^{-\mathrm{kt}}$
Reaction completes at infinite time.
(B) For first order reaction,
Half-life $=\frac{0.693}{4.6 \times 10^{-3}}=150.65 \mathrm{~s}$
(C) For $10 \%$ completion, $t=t_1$
$$ \begin{aligned} & a=100, a-x=100-10=90 \\\\ & k=\frac{2.303}{t_1} \log \frac{100}{90}=4.6 \times 10^{-3} \\\\ & t_1=\frac{2.303}{4.6 \times 10^{-3}} \times 0.04575 \end{aligned} $$
For $90 \%$ completion $t=t_2, a=100, a-x=100-90=10$
$$ \begin{aligned} & k=\frac{2.303}{t_2} \log \frac{100}{10}=4.6 \times 10^{-3} \\\\ & \frac{2.303}{t_2} \times 1=4.6 \times 10^{-3} \Rightarrow t_2=\frac{2.303}{4.6 \times 10^{-3}} \\\\ & \frac{t_2}{t_1}=\frac{1}{0.04575}=21.85 \quad \text { (Incorrect) } \end{aligned} $$
(E) Unit of rate = mol Lā1sā1
Unit of rate constant = sā1
Both have different units.
$\therefore$ Number of correct statements $=1$
$\mathrm{k}=4.6 \times 10^{-3} \mathrm{~s}^{-1}$
$\mathrm{kt}=\ln \frac{1}{1-\alpha}$
$\alpha=1-\mathrm{e}^{-\mathrm{kt}}$
Reaction completes at infinite time.
(B) For first order reaction,
Half-life $=\frac{0.693}{4.6 \times 10^{-3}}=150.65 \mathrm{~s}$
(C) For $10 \%$ completion, $t=t_1$
$$ \begin{aligned} & a=100, a-x=100-10=90 \\\\ & k=\frac{2.303}{t_1} \log \frac{100}{90}=4.6 \times 10^{-3} \\\\ & t_1=\frac{2.303}{4.6 \times 10^{-3}} \times 0.04575 \end{aligned} $$
For $90 \%$ completion $t=t_2, a=100, a-x=100-90=10$
$$ \begin{aligned} & k=\frac{2.303}{t_2} \log \frac{100}{10}=4.6 \times 10^{-3} \\\\ & \frac{2.303}{t_2} \times 1=4.6 \times 10^{-3} \Rightarrow t_2=\frac{2.303}{4.6 \times 10^{-3}} \\\\ & \frac{t_2}{t_1}=\frac{1}{0.04575}=21.85 \quad \text { (Incorrect) } \end{aligned} $$
(E) Unit of rate = mol Lā1sā1
Unit of rate constant = sā1
Both have different units.
$\therefore$ Number of correct statements $=1$
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