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If the rate of disappearance of $\mathrm{N}_2 \mathrm{O}_5$ in the following reaction is $1.2 \times 10^{-5}$ in $\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}$, the rate of production of $\mathrm{NO}_2 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$ is
$2 \mathrm{~N}_2 \mathrm{O}_5(g) \longrightarrow 4 \mathrm{NO}_2(g)+\mathrm{O}_2(g)$
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$2 \mathrm{~N}_2 \mathrm{O}_5(g) \longrightarrow 4 \mathrm{NO}_2(g)+\mathrm{O}_2(g)$
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Verified Answer
The correct answer is:
$2.4 \times 10^{-5}$
Rate of disapperance of reactant $\mathrm{N}_2 \mathrm{O}_5=$ Rate of apperance of products $\mathrm{NO}_2$ and $\mathrm{O}_2$.
For the reaction,
$\begin{aligned} & 2 \mathrm{~N}_2 \mathrm{O}_5(g) \longrightarrow 4 \mathrm{NO}_2(g)+\mathrm{O}_2(g) \\ & \quad-\frac{1}{2} \frac{d\left[\mathrm{~N}_2 \mathrm{O}_5\right]}{d t}=\frac{1}{4} \frac{d\left[\mathrm{NO}_2\right]}{d t}=\frac{d\left[\mathrm{O}_2\right]}{d t}\end{aligned}$
$\therefore \quad \frac{d\left[\mathrm{NO}_2\right]}{d t}=\frac{4}{2} \frac{d\left[\mathrm{~N}_2 O_5\right]}{d t}$
$=2 \times 1.2 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$
$\left(\frac{d\left[\mathrm{~N}_2 \mathrm{O}_5\right]}{d t}\right)=2.4 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$
For the reaction,
$\begin{aligned} & 2 \mathrm{~N}_2 \mathrm{O}_5(g) \longrightarrow 4 \mathrm{NO}_2(g)+\mathrm{O}_2(g) \\ & \quad-\frac{1}{2} \frac{d\left[\mathrm{~N}_2 \mathrm{O}_5\right]}{d t}=\frac{1}{4} \frac{d\left[\mathrm{NO}_2\right]}{d t}=\frac{d\left[\mathrm{O}_2\right]}{d t}\end{aligned}$
$\therefore \quad \frac{d\left[\mathrm{NO}_2\right]}{d t}=\frac{4}{2} \frac{d\left[\mathrm{~N}_2 O_5\right]}{d t}$
$=2 \times 1.2 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$
$\left(\frac{d\left[\mathrm{~N}_2 \mathrm{O}_5\right]}{d t}\right)=2.4 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$
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