Select the rate law for the reaction A + B lo | vedclass

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(A) The rate law is given by $R = K[A]^{\alpha}[B]^{\beta}$.From the given data:$0.10 = K[0.012]^{\alpha}[0.035]^{\beta}$ $(i)$$0.80 = K[0.024]^{\alph

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$Rate = K[B]^3$

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(A) The rate law is given by $R = K[A]^{\alpha}[B]^{\beta}$.From the given data:$0.10 = K[0.012]^{\alpha}[0.035]^{\beta}$ $(i)$$0.80 = K[0.024]^{\alpha}[0.070]^{\beta}$ $(ii)$$0.10 = K[0.024]^{\alpha}[0.035]^{\beta}$ $(iii)$$0.80 = K[0.012]^{\alpha}[0.070]^{\beta}$ $(iv)$Dividing $(iii)$ by $(i)$:$\frac{0.10}{0.10} = \frac{K[0.024]^{\alpha}[0.035]^{\beta}}{K[0.012]^{\alpha}[0.035]^{\beta}}$$1 = 2^{\alpha} \Rightarrow \alpha = 0$.Dividing $(iv)$ by $(i)$:$\frac{0.80}{0.10} = \frac{K[0.012]^{\alpha}[0.070]^{\beta}}{K[0.012]^{\alpha}[0.035]^{\beta}}$$8 = 2^{\beta} \Rightarrow \beta = 3$.Thus,the rate law is $R = K[A]^0[B]^3$ or $R = K[B]^3$.

$Exp$	$[A]$	$[B]$	$Rate$
$1$	$0.012$	$0.035$	$0.10$
$2$	$0.024$	$0.070$	$0.80$
$3$	$0.024$	$0.035$	$0.10$
$4$	$0.012$	$0.070$	$0.80$

$Expt.$	$[X]_0 / mol \ L^{-1}$	$[Y]_0 / mol \ L^{-1}$	$Rate / mol \ L^{-1} s^{-1}$
$1$	$0.25$	$0.25$	$1.0 \times 10^{-6}$
$2$	$0.50$	$0.25$	$4.0 \times 10^{-6}$
$3$	$0.25$	$0.50$	$8.0 \times 10^{-6}$

Select the rate law for the reaction $A + B \longrightarrow C$ based on the following data:

$Exp$ $[A]$ $[B]$ $Rate$

$1$ $0.012$ $0.035$ $0.10$

$2$ $0.024$ $0.070$ $0.80$

$3$ $0.024$ $0.035$ $0.10$

$4$ $0.012$ $0.070$ $0.80$

Similar Questions

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Select the rate law for the reaction $A + B \longrightarrow C$ based on the following data: $Exp$ $[A]$ $[B]$ $Rate$ $1$ $0.012$ $0.035$ $0.10$ $2$ $0.024$ $0.070$ $0.80$ $3$ $0.024$ $0.035$ $0.10$ $4$ $0.012$ $0.070$ $0.80$