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  • For a reaction   the observed rate law : <im

For a reaction \mathrm{A+B\rightarrow P}  the observed rate law :

\mathrm{rate=\frac{K _1K^{2}_2}{k-1}\left [ A \right ]\left [ B \right ]}

then find out the expression of  K observed ,A observed and E observed ?

 

Option: 1

\mathrm{ \begin{aligned} & \text { k observed }=\frac{k_1 k_2^2}{k-1} \\ & \text { A observed }=\frac{A_1 A_2^2}{A-1} \\ & \text { E observed }=E_1+E_2^2-E-1 \end{aligned} }


Option: 2

\mathrm{\begin{aligned} & \text { k%20observed }=\frac{k_1 k_2^2}{k-1} \\ & \text { A%20observed }=\frac{2A_1 A_2}{A-1} \\ & \text { E%20observed }=E_1+2E_2-E-1 \end{aligned} }


Option: 3

\mathrm{\begin{aligned} & \text { k observed }=\frac{k_1 k_2^2}{k-1} \\ & \text { A observed }=\frac{A_1 A_2}{2A-1} \\ & \text { E observed }=E_1+E_2^2-E-1 \end{aligned} }


Option: 4

\mathrm{\begin{aligned} & \text { k%20observed }=\frac{k_1 k_2^2}{k-1} \\ & \text { A%20observed }=\frac{A_1 A_2^2}{A-1} \\ & \text { E%20observed }=E_1+2E_2-E-1 \end{aligned} }


Answers (1)

best_answer

Given:-\mathrm{ \text { rate }=\frac{k_1 k_2^2}{k-1}[A][B] }

\mathrm{ \text { kobs. }=\frac{k_1 k_2^2}{k-1} }

\mathrm{ \text { Aobs. } e^{\frac{(-E a)0 bs}{R T}} \Rightarrow \frac{A_1 e^{\frac{-E 1}{R T}} \times A_2^2 e^{\frac{-2 E_2}{R T}}}{A-1 e^{-\frac{E-1}{R T}}} }

\mathrm{ \text { Aobs. } e^{\frac{-(E a)0 bs}{R T}}=\frac{A_1 A_2^2}{A-1} e^{-\frac{E 1}{R T}-\frac{2 E_2}{R T}+\frac{E-1}{R T}} }

\mathrm{ \text { So A%20observed } =\frac{A_1 A_2^2}{A-1} }

\mathrm{ \text { (Ea){observed} } =E_1+2E_2-E-1 }

Hence option 4 is correct.

Posted by

manish painkra

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