For a first order reaction, $(A) \rightarrow$ products, the concentration of $A$ changes from $0.1…

For a first order reaction, $(A) \rightarrow$ products, the concentration of $A$ changes from $0.1 \mathrm{~M}$ to $0.025 \mathrm{~M}$ in $40$ minutes. The rate of reaction when the concentration of $A$ is $0.01 \mathrm{~M}$ is :
  1. $1.73 \times 10^{-5} \mathrm{M} / \mathrm{min}$
  2. $3.47 \times 10^{-4} \mathrm{M} / \mathrm{min}$
  3. $3.47 \times 10^{-5} \mathrm{M} / \mathrm{min}$
  4. $1.73 \times 10^{-4} \mathrm{M} / \mathrm{min}$

Solution

$\begin{aligned} & \mathrm{k}=\frac{2.303}{40} \log \frac{0.1}{0.025} \\ & \mathrm{k}=\frac{0.693}{20} \end{aligned}$ For a F.O.R., rate $=k[\mathrm{~A}]$; rate $=\frac{0.693}{20} \times 10^{-2}=3.47 \times 10^{-4} \mathrm{~M} / \mathrm{min}$.

Asked in: JEE Main 2012 (Offline)

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