Search any question & find its solution
Question:
Answered & Verified by Expert
At $500 \mathrm{~K}$, for the reaction
$\mathrm{N}_2(g)+3 \mathrm{H}_2(g) \rightleftharpoons 2 \mathrm{NH}_3(g)$, the $K_p$ is $0.036 \mathrm{~atm}^{-2}$. What is its $K_C$ in $\mathrm{L}^2 \mathrm{~mol}^{-1}$ ?
$\left(R=0.082 \mathrm{~L}\right.$ atom $\left.\mathrm{mol}^{-1} \mathrm{~K}^{-1}\right)$.
Options:
$\mathrm{N}_2(g)+3 \mathrm{H}_2(g) \rightleftharpoons 2 \mathrm{NH}_3(g)$, the $K_p$ is $0.036 \mathrm{~atm}^{-2}$. What is its $K_C$ in $\mathrm{L}^2 \mathrm{~mol}^{-1}$ ?
$\left(R=0.082 \mathrm{~L}\right.$ atom $\left.\mathrm{mol}^{-1} \mathrm{~K}^{-1}\right)$.
Solution:
1907 Upvotes
Verified Answer
The correct answer is:
$60.5$
$K_p=K_C(R T)^{Δn}$
$\mathrm{N}_2(g)+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(g)$
$\begin{aligned} \Delta n & =2-(3+1)=2-4=-2 \\ K_p & =0.036 \quad \text { (given) }\end{aligned}$
$0.036=K_C(0.082 \times 500)^{-2}$
$\Rightarrow \quad K_C=\frac{0.036}{(0.082 \times 500)^{-2}}=60.516 \mathrm{~L}^2 \mathrm{~mol}^{-2}$
$\mathrm{N}_2(g)+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(g)$
$\begin{aligned} \Delta n & =2-(3+1)=2-4=-2 \\ K_p & =0.036 \quad \text { (given) }\end{aligned}$
$0.036=K_C(0.082 \times 500)^{-2}$
$\Rightarrow \quad K_C=\frac{0.036}{(0.082 \times 500)^{-2}}=60.516 \mathrm{~L}^2 \mathrm{~mol}^{-2}$
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.