Search any question & find its solution
Question:
Answered & Verified by Expert
Calculate the value of the equilibrium constant $\left(\mathrm{K}_{\mathrm{p}}\right)$ for the reaction of oxygen gas oxidising ammonia gas to nitric oxide and water vapour. The pressure of each gas at equilibrium is $0.5 \mathrm{~atm}$.
Options:
Solution:
1234 Upvotes
Verified Answer
The correct answer is:
$0.5 \mathrm{~atm}$
$4 \mathrm{NH}_3 \uparrow+5 \mathrm{O}_2 \uparrow \longrightarrow 4 \mathrm{NO} \uparrow+6 \mathrm{H}_2 \mathrm{O} \uparrow$
$\mathrm{K}_{\mathrm{p}}=\frac{\mathrm{p}_{\mathrm{NO}}^4 \mathrm{p}_{\mathrm{H}_2 \mathrm{O}}^6}{\mathrm{P}_{\mathrm{NH}_3}^4 \mathrm{p}_{\mathrm{O}_2}^5}=\frac{(0.5)^{10}}{(0.5)^9}=0.5 \mathrm{~atm}$
$\mathrm{K}_{\mathrm{p}}=\frac{\mathrm{p}_{\mathrm{NO}}^4 \mathrm{p}_{\mathrm{H}_2 \mathrm{O}}^6}{\mathrm{P}_{\mathrm{NH}_3}^4 \mathrm{p}_{\mathrm{O}_2}^5}=\frac{(0.5)^{10}}{(0.5)^9}=0.5 \mathrm{~atm}$
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.