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KCET · Chemistry · Chemical Kinetics

For a chemical reaction \(\mathrm{A} \rightarrow \mathrm{B}\), the rate of the reaction is \(2 \times 10^{-3} \mathrm{~mol} \mathrm{} \mathrm{dm}^{-3} \mathrm{~s}^{-1}\), when the initial concentration is \(0.05 \mathrm{~mol} \mathrm{dm}^{-3}\). The rate of the same reaction is \(1.6 \times 10^{-2} \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}\) when the initial concentration is \(0.1 \mathrm{~mol} \mathrm{} \mathrm{dm}^{-3}\). The order of the reaction is

  1. A 2
  2. B 0
  3. C 3
  4. D 1
Verified Solution

Answer & Solution

Correct Answer

(C) 3

Step-by-step Solution

Detailed explanation

\(\mathrm{A} \longrightarrow \mathrm{B}\)
Rate \(=k[A]^{n}\)
\((\text { Rate })_{1}=\mathrm{k}(0.05)^{\mathrm{n}}=2 \times 10^{-3} \quad \text{...(i)}\)
\((\text { Rate })_{2}=\mathrm{k}(0.1)^{\mathrm{n}}=1.6 \times 10^{-2} \quad \text{...(ii)}\)
Dividing the Eq. (ii) by Eq. (i).
\[
\begin{gathered}
\frac{(\text { Rate })_{2}}{(\text { Rate })_{1}}=\frac{\mathrm{k}(0.1)^{\mathrm{n}}}{\mathrm{k}(0.05)^{\mathrm{n}}}=\frac{1.6 \times 10^{-2}}{2 \times 10^{-3}} \\
\quad(2)^{\mathrm{n}}=8 \quad \text { or } \quad(2)^{\mathrm{n}}=2^{3} \\
\therefore \quad \mathrm{n}=3
\end{gathered}
\]