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COMEDK · Chemistry · 19. Chemical Kinetics

The concentration of a reactant \(X\) decreases from 0.1 \(\mathrm{M}\) to \(0.025 \mathrm{M}\) in \(40 \mathrm{~min}\). If the reaction follows I order kinetics, the rate of the reaction when the concentration of \(X\) is \(0.01 \mathrm{M}\) will be

  1. A \(1.73 \times 10^{-4} \mathrm{M} \mathrm{min}^{-1}\)
  2. B \(3.47 \times 10^{-4} \mathrm{M} \mathrm{min}^{-1}\)
  3. C \(3.47 \times 10^{-5} \mathrm{Mmin}^{-1}\)
  4. D \(1.73 \times 10^{-5} \mathrm{M} \mathrm{min}^{-1}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(3.47 \times 10^{-4} \mathrm{M} \mathrm{min}^{-1}\)

Step-by-step Solution

Detailed explanation

For lst order reaction \[ \begin{aligned} k &=\frac{2303}{t} \log \frac{\left[R_{0}\right]}{[R]}=\frac{2303}{40} \log \left[\frac{0.1}{0.025}\right] \\ &=\frac{2303}{40} \times \log 4=0.0347 \mathrm{~min}^{-1} \end{aligned} \] Rate expression,…