Q29 / 160 · AP EAMCET 2021 (23 Aug Shift 1) · Chemistry

A reaction has rate constant $k=2.4 \times 10^{-4} \mathrm{~s}^{-1}$. Then, find the ratio of $t_{99.9}$ to $t_{50}$.

  • A
    1
  • B
    5
  • Correct answer
    C
    10
  • D
    15

Solution

Unit of $k$ is $\mathrm{s}^{-1}$. $\therefore$ It is first order reaction $$ \begin{aligned} t_{99.9 \%} & =\frac{2.303}{k} \log \frac{100}{0.1} \\ & =\frac{2.303}{k} \log 1000=\frac{2.303}{k} \log 10^3 \\ & =\frac{2.303}{k} \times 3 \log 10=\frac{2.303}{k} \times 3 \quad[\because \log 10=1] \\ t_{50 \%} & =\frac{0.693}{k} \end{aligned} $$ Ratio of $t_{99.9 \%}$ and $t_{50 \%}$ is $\frac{t_{99.9 \%}}{t_{50 \%}}=\frac{3 \times 2.303 / k}{0.693 / k}=10$.

Hence, the correct answer is option: 'C'.

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