Search any question & find its solution
Question:
Answered & Verified by Expert
The $\mathrm{K}_{\mathrm{p}}$ value at equilibrium of $\mathrm{SO}_3$ formation reaction from $\mathrm{SO}_2(\mathrm{~g})$ and $\mathrm{O}_2(\mathrm{~g})$ is $5 \mathrm{~atm}^{-1}$. What is the equilibrium partial pressure of $\mathrm{O}_2$ if the equilibrium pressure of $\mathrm{SO}_2$ and $\mathrm{SO}_3$ are equal?
Options:
Solution:
2337 Upvotes
Verified Answer
The correct answer is:
$0.2 \mathrm{~atm}$
The equilibrium reaction is:
$\begin{aligned}
& 2 \mathrm{SO}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_3(\mathrm{~g}) \\
& \mathrm{K}_{\mathrm{P}}=\frac{\left(\mathrm{P}_{\mathrm{SO}_3}\right)^2}{\left(\mathrm{P}_{\mathrm{SO}_2}\right)^2\left(\mathrm{P}_{\mathrm{O} 2}\right)}
\end{aligned}$
At equilibrium; $\mathrm{P}_{\mathrm{SO}_2}=\mathrm{P}_{\mathrm{SO}_3}$
$\begin{aligned}
& 5=\frac{\left(\mathrm{P}_{\mathrm{SO}_2}\right)^2}{\left(\mathrm{P}_{\mathrm{SO}_2}\right)^2\left(\mathrm{P}_{\mathrm{O}_2}\right)} \\
& \mathrm{P}_{\mathrm{O}_2}=0.2 \mathrm{~atm} .
\end{aligned}$
$\begin{aligned}
& 2 \mathrm{SO}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_3(\mathrm{~g}) \\
& \mathrm{K}_{\mathrm{P}}=\frac{\left(\mathrm{P}_{\mathrm{SO}_3}\right)^2}{\left(\mathrm{P}_{\mathrm{SO}_2}\right)^2\left(\mathrm{P}_{\mathrm{O} 2}\right)}
\end{aligned}$
At equilibrium; $\mathrm{P}_{\mathrm{SO}_2}=\mathrm{P}_{\mathrm{SO}_3}$
$\begin{aligned}
& 5=\frac{\left(\mathrm{P}_{\mathrm{SO}_2}\right)^2}{\left(\mathrm{P}_{\mathrm{SO}_2}\right)^2\left(\mathrm{P}_{\mathrm{O}_2}\right)} \\
& \mathrm{P}_{\mathrm{O}_2}=0.2 \mathrm{~atm} .
\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.