Q67 / 180 · NEET 2019 (Odisha) · Chemistry
A first order reaction has a rate constant of $2.303 \times 10^{-3} \mathrm{~s}^{-1}$. The time required for $40 \mathrm{~g}$ of this reactant to reduce to $10 \mathrm{~g}$ will be
[Given that $\log _{10} 2=0.3010$ ]
- A$230.3 \mathrm{~s}$
- B$301 \mathrm{~s}$
- C$2000 \mathrm{~s}$
- Correct answerD$602 \mathrm{~s}$
Solution
For first order reaction,
$$
\mathrm{t}=\frac{2.303}{\mathrm{k}} \log \frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}}
$$
Given : $\mathrm{k}=2.303 \times 10^{-3} \mathrm{~s}^{-1}, \mathrm{a}=40 \mathrm{~g}, \mathrm{a}-\mathrm{x}=10 \mathrm{~g}$
On substituting the given values in Eq. (i), we get
$$
\begin{aligned}
t & =\frac{2.303}{2.303 \times 10^{-3}} \log \frac{40}{10} \\
& =10^3 \log 2^2=2 \times 10^3 \times \log 2 \\
& =2 \times 10^3 \times 0.3010=602 \mathrm{~s}
\end{aligned}
$$
Alternative method
For first order reaction, $\mathrm{t}_{1 / 2}=\frac{0.693}{\mathrm{k}}$
$$
\mathrm{t}_{1 / 2}\left(\mathrm{t}_{50 \%}\right)=\frac{0.693}{2.303 \times 10^{-3}}=301 \mathrm{~s}
$$
Also, $\quad \mathrm{t}_{75 \%}=2 \mathrm{t}_{50 \%}$
$$
\therefore \quad \mathrm{t}_{75 \%}=2 \times 301=602 \mathrm{~s}
$$
Hence, the correct answer is option: 'D'.