For the reaction $2A + B \to A_2B$,the rate of reaction with respect to $A$ is $3.9 \times 10^{-9} \ mol \ L^{-1} \ s^{-1}$. Calculate the rate of consumption of $B$ and the rate of formation of $A_2B$.
For the reaction $2A + B \to A_2B$,the rate expression is given by:
$Rate = -\frac{1}{2} \frac{d[A]}{dt} = -\frac{d[B]}{dt} = \frac{d[A_2B]}{dt}$
Given that the rate of reaction with respect to $A$ is $-\frac{d[A]}{dt} = 3.9 \times 10^{-9} \ mol \ L^{-1} \ s^{-1}$.
$1$. Rate of consumption of $B$ $(-\frac{d[B]}{dt})$:
$-\frac{d[B]}{dt} = \frac{1}{2} \times (-\frac{d[A]}{dt}) = \frac{1}{2} \times 3.9 \times 10^{-9} = 1.95 \times 10^{-9} \ mol \ L^{-1} \ s^{-1}$.
$2$. Rate of formation of $A_2B$ $(\frac{d[A_2B]}{dt})$:
$\frac{d[A_2B]}{dt} = \frac{1}{2} \times (-\frac{d[A]}{dt}) = 1.95 \times 10^{-9} \ mol \ L^{-1} \ s^{-1}$.
$Rate = -\frac{1}{2} \frac{d[A]}{dt} = -\frac{d[B]}{dt} = \frac{d[A_2B]}{dt}$
Given that the rate of reaction with respect to $A$ is $-\frac{d[A]}{dt} = 3.9 \times 10^{-9} \ mol \ L^{-1} \ s^{-1}$.
$1$. Rate of consumption of $B$ $(-\frac{d[B]}{dt})$:
$-\frac{d[B]}{dt} = \frac{1}{2} \times (-\frac{d[A]}{dt}) = \frac{1}{2} \times 3.9 \times 10^{-9} = 1.95 \times 10^{-9} \ mol \ L^{-1} \ s^{-1}$.
$2$. Rate of formation of $A_2B$ $(\frac{d[A_2B]}{dt})$:
$\frac{d[A_2B]}{dt} = \frac{1}{2} \times (-\frac{d[A]}{dt}) = 1.95 \times 10^{-9} \ mol \ L^{-1} \ s^{-1}$.
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