Q24 / 160 · TS EAMCET 2024 (10 May Shift 2) · Chemistry

For a first order reaction, the graph between $\log \frac{a}{(a-x)}$ (on $y$-axis) and time (in min, on $x$-axis) gave a straight line passing through origin. The slope is $2 \times 10^{-3} \mathrm{~min}^{-1}$. What is the rate constant (in $\min ^{-1}$ )?

  • A
    $2 \times 10^{-3}$
  • B
    $\frac{2 \times 10^{-3}}{2.303}$
  • Correct answer
    C
    $4.606 \times 10^{-3}$
  • D
    $0.5 \times 10^{-5}$

Solution

From first order reaction, $\begin{aligned} & \text { log } \frac{a}{a-x}=\frac{k t}{2.303} \\ & y=\mathrm{mx}+\mathrm{c} \\ & \text { slope }(\mathrm{m})=\frac{\mathrm{k}}{2.303} \\ & \text { given, } \mathrm{m}=2 \times 10^{-3} \mathrm{~min}^{-1} \\ & \frac{\mathrm{k}}{2.303}=2 \times 10^{-3} \mathrm{~min}^{-1} \\ & \therefore \mathrm{k}=2.303 \times 2 \times 10^{-3} \mathrm{~min}^{-1} \\ & =4.606 \times 10^{-3} \mathrm{~min}^{-1}\end{aligned}$

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

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