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Observe the following reaction
$2 A+B \longrightarrow C$
The rate of formation of $C$ is
$2.2 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-13} \mathrm{~min}^{-1}$. What is the value of $-\frac{d[A]}{d t}\left(\right.$ in $\left.\mathrm{mol} \mathrm{L}^{-1} \min ^{-1}\right) ?$
Options:
$2 A+B \longrightarrow C$
The rate of formation of $C$ is
$2.2 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-13} \mathrm{~min}^{-1}$. What is the value of $-\frac{d[A]}{d t}\left(\right.$ in $\left.\mathrm{mol} \mathrm{L}^{-1} \min ^{-1}\right) ?$
Solution:
1549 Upvotes
Verified Answer
The correct answer is:
$4.4 \times 10^{-3}$
$2 A+B \longrightarrow C$
Rate of reaction,
$=-\frac{1}{2} \frac{d[A]}{d t}=-\frac{d[B]}{d t}=\frac{d[C]}{d t}$
$\begin{aligned} \therefore & -\frac{d[A]}{d t}=2 \frac{d[C]}{d t} \\ & =2 \times 2.2 \times 10^{-3}=4.4 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}\end{aligned}$
Rate of reaction,
$=-\frac{1}{2} \frac{d[A]}{d t}=-\frac{d[B]}{d t}=\frac{d[C]}{d t}$
$\begin{aligned} \therefore & -\frac{d[A]}{d t}=2 \frac{d[C]}{d t} \\ & =2 \times 2.2 \times 10^{-3}=4.4 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}\end{aligned}$
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