Q26 / 160 · TS EAMCET 2018 (05 May Shift 1) · Chemistry

A particular reaction has a rate constant $1.15 \times 10^{-3} \mathrm{~s}^{-1}$. How long does it take for $6 \mathrm{~g}$ of the reactant of reduce to $3 \mathrm{~g}$ ? $(\log 2=0.301)$

  • A
    $301 \mathrm{~s}$
  • Correct answer
    B
    $603 \mathrm{~s}$
  • C
    $840 \mathrm{~s}$
  • D
    $15 \mathrm{~s}$

Solution

Given, rate constant $(k)=1.15 \times 10^{-3} \mathrm{~s}^{-1}$ Reaction time $=$ ? Initial amount of reactant $\left(A_o\right)=6 \mathrm{~g}$ Final amount of reactant $\left(A_f\right)=3 \mathrm{~g}$ From rate constant equation, $$ \begin{aligned} k & =\frac{2.303}{t} \log \frac{A_o}{A_f} \\ 1.15 \times 10^{-3} & =\frac{2.303}{t} \log \frac{6}{3} \\ t & =\frac{2.303}{1.15 \times 10^{-3}} \log \frac{6}{3}=\frac{2.303}{1.15 \times 10^{-3}} \times .3010 \\ t & =.6027 \times 10^{+3} \mathrm{~s} \\ \text { or } \quad t & =603 \mathrm{~s} \end{aligned} $$

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

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