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  • For a reaction when concentration of A doubled, r

For a reaction \mathrm{A+B \rightarrow P} when concentration of A doubled, rate becomes 4 times and when concentration of A and B both doubled rate becomes doubled. Find the order of the reaction

Option: 1

\mathrm{2^{\text {nd }}}


Option: 2

\mathrm{3^{\text {nd }}}


Option: 3

\mathrm{1^{\text {nd }}}


Option: 4

0


Answers (1)

best_answer

To find order of rom first we will find rate law
\mathrm{ \begin{aligned} \frac{r_2}{r_1} & =\left(\frac{C_2}{C_1}\right)^n \\ 4 & =(2)^n \\\\ n & =2 \\\\ \frac{r_2}{r_1} & =\left(\frac{C_2}{C_1}\right)_A^x\left(\frac{C_2}{c_1}\right)^y \\\\ 2 & =(2)_A^x(2)_B^y \\\\ \frac{2}{4} & =(2)_B^y \\\\ \frac{1}{2} & =(2)_B^y \\\\ y & =-1 \\\\ \text { rate } & =K[A]^x[B]^y \end{aligned} }

\mathrm{ =k[A]^2[B]^{-1} }

\mathrm{n=1 \, \, first \, \, order}

Posted by

Nehul

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