For the reaction: 2A + B to A_2B; the rate = | vedclass

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Question

(A) The reaction is $2A + B 	o A_2B$. Given the stoichiometry,the change in concentration is $\Delta[A] = [A]_0 - [A]_t = 0.2 - 0.12 = 0.08 \ mol/L$.

Vedclass · Accepted Answer

$3.11 	imes 10^{-8}$

Vedclass · Answer

(A) The reaction is $2A + B 	o A_2B$. Given the stoichiometry,the change in concentration is $\Delta[A] = [A]_0 - [A]_t = 0.2 - 0.12 = 0.08 \ mol/L$. Since $2 \ mol$ of $A$ react with $1 \ mol$ of $B$,the amount of $B$ consumed is $\frac{1}{2} 	imes 0.08 = 0.04 \ mol/L$. Therefore,the concentration of $B$ at this time is $[B]_t = [B]_0 - 0.04 = 0.4 - 0.04 = 0.36 \ mol/L$. The rate law is given by $	ext{Rate} = K[A][B]^2$. Substituting the values: $	ext{Rate} = (2.0 	imes 10^{-6}) 	imes (0.12) 	imes (0.36)^2$. $	ext{Rate} = 2.0 	imes 10^{-6} 	imes 0.12 	imes 0.1296 = 3.1104 	imes 10^{-8} \ mol \ L^{-1} \ s^{-1}$.

$Experiment$	$[X] / (mol \ L^{-1})$	$[Y] / (mol \ L^{-1})$	$\text{Initial rate} / (mol \ L^{-1} \ min^{-1})$
$I$	$0.1$	$0.1$	$2 \times 10^{-3}$
$II$	$0.2$	$0.2$	$4 \times 10^{-3}$
$III$	$0.4$	$0.4$	$M \times 10^{-3}$
$IV$	$0.1$	$0.2$	$2 \times 10^{-3}$

For the reaction: $2A + B \to A_2B$; the rate $= K[A][B]^2$ with $K = 2.0 \times 10^{-6} \ L^2 \ mol^{-2} \ s^{-1}$. Initial concentrations of $A$ and $B$ are $[A]_0 = 0.2 \ mol/L$ and $[B]_0 = 0.4 \ mol/L$ respectively. Calculate the rate of reaction after $[A]$ is reduced to $0.12 \ mol/L$.

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