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The rate constant, k of a zero order reaction $2 \mathrm{NH}_3(\mathrm{~g})$ $\underset{1130 \mathrm{~K}}{\stackrel{\mathrm{Pt}}{\longrightarrow}} \mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g})$ is $\mathrm{y} \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$. The rate of formation of hydrogen (in $\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}$ ) is
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Verified Answer
The correct answer is:
$3 y \times 10^{-4}$
$\mathrm{r}=\mathrm{K}\left[\mathrm{NH}_3\right]^{\circ}=\mathrm{K}=\mathrm{y} \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$
Now,
$\begin{aligned}
& \frac{1}{3} \frac{\mathrm{d}\left[\mathrm{H}_2\right]}{\mathrm{dt}}=\mathrm{r} \\
& \Rightarrow \frac{\mathrm{d}\left[\mathrm{H}_2\right]}{\mathrm{dt}}=3 \mathrm{r}=3 \mathrm{y} \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}
\end{aligned}$
Now,
$\begin{aligned}
& \frac{1}{3} \frac{\mathrm{d}\left[\mathrm{H}_2\right]}{\mathrm{dt}}=\mathrm{r} \\
& \Rightarrow \frac{\mathrm{d}\left[\mathrm{H}_2\right]}{\mathrm{dt}}=3 \mathrm{r}=3 \mathrm{y} \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}
\end{aligned}$
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