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In the reaction of formation of sulphur trioxide by contact process $2 \mathrm{SO}_2+\mathrm{O}_2 \rightleftharpoons 2 \mathrm{SO}_3$ the rate of reaction was measured as
$\frac{\mathrm{d}\left[\mathrm{O}_2\right]}{\mathrm{dt}}=-2.5 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$. The rate of reaction is terms of $\left[\mathrm{SO}_2\right]$ in $\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}$ will be:
Options:
$\frac{\mathrm{d}\left[\mathrm{O}_2\right]}{\mathrm{dt}}=-2.5 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$. The rate of reaction is terms of $\left[\mathrm{SO}_2\right]$ in $\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}$ will be:
Solution:
2420 Upvotes
Verified Answer
The correct answer is:
$-5.00 \times 10^{-4}$
$-5.00 \times 10^{-4}$
From rate law
$$
\begin{aligned}
&-\frac{1}{2} \frac{\mathrm{dSO}_2}{\mathrm{dt}}=-\frac{\mathrm{dO}_2}{\mathrm{dt}}=\frac{1}{2} \frac{\mathrm{dSO}_3}{\mathrm{dt}} \\
&\therefore-\frac{\mathrm{dSO}_2}{\mathrm{dt}}=-2 \times \frac{\mathrm{dO}_2}{\mathrm{dt}} \\
&=-2 \times 2.5 \times 10^{-4} \\
&=-5 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}
\end{aligned}
$$
$$
\begin{aligned}
&-\frac{1}{2} \frac{\mathrm{dSO}_2}{\mathrm{dt}}=-\frac{\mathrm{dO}_2}{\mathrm{dt}}=\frac{1}{2} \frac{\mathrm{dSO}_3}{\mathrm{dt}} \\
&\therefore-\frac{\mathrm{dSO}_2}{\mathrm{dt}}=-2 \times \frac{\mathrm{dO}_2}{\mathrm{dt}} \\
&=-2 \times 2.5 \times 10^{-4} \\
&=-5 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}
\end{aligned}
$$
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