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
- A \(1.73 \times 10^{-4} \mathrm{M} \mathrm{min}^{-1}\)
- B \(3.47 \times 10^{-4} \mathrm{M} \mathrm{min}^{-1}\)
- C \(3.47 \times 10^{-5} \mathrm{Mmin}^{-1}\)
- D \(1.73 \times 10^{-5} \mathrm{M} \mathrm{min}^{-1}\)
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,…
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