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