Q6 / 150 · MHT CET 2023 (13 May Shift 1) · Chemistry

Calculate the rate constant of the first order reaction if $20 \%$ of the reactant decomposes in 15 minutes.

  • Correct answer
    A
    $1.488 \times 10^{-2}$ minute $^{-1}$
  • B
    $1.881 \times 10^{-2}$ minute $^{-1}$
  • C
    $1.984 \times 10^{-2}$ minute $^{-1}$
  • D
    $1.18 \times 10^{-2}$ minute $^{-1}$

Solution

$20 \%$ of the reactant has decomposed. So, if $[A]_0=100$, then $[A]_{\mathrm{t}}=100-20=80$ $$ \begin{aligned} k & =\frac{2.303}{t} \log _{10} \frac{[\mathrm{A}]_0}{[\mathrm{~A}]_{\mathrm{t}}} \\ & =\frac{2.303}{15} \log _{10} \frac{100}{80} \\ & =\frac{2.303}{15} \log _{10} \frac{5}{4} \\ & =\frac{2.303}{15} \times\left(\log _{10} 5-\log _{10} 4\right) \\ & =\frac{2.303}{15} \times(0.699-0.602) \\ & =0.01488=1.488 \times 10^{-2} \text { minute }^{-1} \end{aligned} $$

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

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