Q23 / 160 · AP EAMCET 2023 (18 May Shift 2) · Chemistry

A reaction, $3 \mathrm{X}(\mathrm{g}) \rightarrow 2 \mathrm{Y}(\mathrm{g})+\mathrm{Z}(\mathrm{g})$ takes place in a closed vessel. What is the rate of formation of $Y$ (in $\mathrm{mol} \mathrm{L}^{-1}$ ) if the rate of disappearance of $\mathrm{X}$ is $7.2 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$ ?

  • A
    $3.6 \times 10^{-3}$
  • Correct answer
    B
    $4.8 \times 10^{-3}$
  • C
    $2.4 \times 10^{-3}$
  • D
    $1.2 \times 10^{-3}$

Solution

$\begin{aligned} & \text { }-\frac{1}{3} \frac{\mathrm{d}[\mathrm{X}]}{\mathrm{dt}}=\frac{1}{2} \frac{\mathrm{d}[\mathrm{Y}]}{\mathrm{dt}}=\frac{\mathrm{d}[\mathrm{Z}]}{\mathrm{dt}} \\ & \Rightarrow \frac{\mathrm{d}[\mathrm{Y}]}{\mathrm{dt}}=\frac{2}{3} \frac{\mathrm{d}[\mathrm{X}]}{\mathrm{dt}}=\frac{2}{3} \times 7.2 \times 10^{-3} \\ & =4.8 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~S}^{-1}\end{aligned}$

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

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