A) 0
B) 1.5
C) 1
D) 2
Correct Answer: D
Solution :
[d] \[a\xrightarrow{{}}B\] \[\begin{align} & \begin{matrix} \text{Initial concentration} & \text{Rate of reaction} \\ \end{matrix} \\ & \begin{matrix} 2\times {{10}^{-3}}M & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2.40\times {{10}^{-4}}M{{s}^{-1}} \\ \end{matrix} \\ & \begin{matrix} 1\times {{10}^{-3}}M & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0.60\times {{10}^{-4}}M{{s}^{-1}} \\ \end{matrix} \\ \end{align}\] rate of reaction \[r=k{{[A]}^{x}}\] where x = order of reaction hence \[2.40\times {{10}^{-4}}=k{{\left[ 2\times {{10}^{-3}} \right]}^{x}}\] ......(i) \[0.60\times {{10}^{-4}}=k{{[1\times {{10}^{-3}}]}^{x}}\] ......(ii) On dividing eqn.(i) from eqn. (ii) we get \[4={{(2)}^{x}}\] \[\therefore ~~x=2\] i.e. order of reaction = 2
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