For a chemical reaction $A \to B$ , the rate of the reaction is $2.0 \times 10^{- 3} m o l d m^{- 3} \sec^{- 1},$ when the Initial concentration is $0.05 m o l d m^{- 3}$ , the rate of the same reaction is $1.6 \times 10^{- 2} m o l d m^{- 3} \sec^{- 1}$ . When the Initial concentration is $0.1 m o l d m^{- 3}$ , find the order of the reaction.

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Detailed Solution

Let the rate equation for the reaction be

$\begin{array}{l} r a t e = K {[A]}^{n} \\ r_{1} = 2.0 \times 10^{- 3} m o l d m^{- 3} \sec^{- 1}, \\ [A_{0}] = 0.05 m o l d m^{- 3} \\ 2.0 \times 10^{- 3} = K {[0.05]}^{n} \to (1) \\ a n d r_{2} = 1.6 \times 10^{- 2} m o l e d m^{- 3} \sec^{- 1}, \\ [A_{0}] = 0.1 m o l d m^{- 3} \\ ∴ 1.6 \times 10^{- 2} = K {[0.1]}^{n} \to (2) \\ (1) d i v i d e d b y (2) \\ \frac{2.0 \times 10^{- 3}}{1.6 \times 10^{- 2}} = \frac{{[0.05]}^{n}}{{[0.1]}^{n}} = \frac{1}{8} = {(\frac{1}{2})}^{n} \end{array}$