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Consider a gas phase reaction which occurs in a closed vessel
$2 \mathrm{~A} \rightarrow 4 \mathrm{~B}+\mathrm{C}$
The concentration of $\mathrm{B}$ is found to be increased by $5 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1}$ in 10 seconds.
The rate of disappearance of $\mathrm{A}$ (in $\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}$ ) is
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$2 \mathrm{~A} \rightarrow 4 \mathrm{~B}+\mathrm{C}$
The concentration of $\mathrm{B}$ is found to be increased by $5 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1}$ in 10 seconds.
The rate of disappearance of $\mathrm{A}$ (in $\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}$ ) is
Solution:
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Verified Answer
The correct answer is:
$2.5 \times 10^{-4}$
$\begin{aligned} & \frac{1}{2} \frac{\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}=\frac{1}{4} \frac{\mathrm{d}[\mathrm{B}]}{\mathrm{dt}} \Rightarrow \frac{\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}=\frac{1}{2} \times \frac{\mathrm{d}[\mathrm{B}]}{\mathrm{dt}} \\ & =\frac{1}{2} \times \frac{\Delta \mathrm{B}}{\Delta \mathrm{t}} \\ & =\frac{1}{2} \times \frac{5 \times 10^{-3}}{10} \\ & =2.5 \times 10^{-4} \mathrm{Ms}^{-1} .\end{aligned}$
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