Q28 / 160 · AP EAMCET 2005 · Chemistry

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 $\left.\mathrm{L}^{-1} \min ^{-1}\right) ?$

  • A
    $2.2 \times 10^{-3}$
  • B
    $1.1 \times 10^{-3}$
  • Correct answer
    C
    $4.4 \times 10^{-3}$
  • D
    $5.5 \times 10^{-3}$

Solution

$2 A+B \longrightarrow C$ rate of reaction $\begin{aligned} & =-\frac{1}{2} \frac{d[A]}{d t}=-\frac{d[B]}{d t}=\frac{d[C]}{d t} \\ \therefore \quad-\frac{d[A]}{d t} & =2 \frac{d[C]}{d t}=2 \times 2.2 \times 10^{-3} \\ & =4.4 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1} \end{aligned}$

Hence, the correct answer is option: 'C'.

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