AP EAMCET · Chemistry · Chemical Equilibrium
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 \(\left.\mathrm{mol}^{-1} \mathrm{~K}^{-1}\right)\).
- A \(2.1 \times 10^{-4}\)
- B \(2.1 \times 10^{-5}\)
- C \(60.5\)
- D \(605\)
Answer & Solution
Correct Answer
(C) \(60.5\)
Step-by-step Solution
Detailed explanation
\(K_p=K_C(R T)^{Δn}\) \(\mathrm{N}_2(g)+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(g)\) \(\begin{aligned} \Delta n & =2-(3+1)=2-4=-2 \\ K_p & =0.036 \quad \text { (given) }\end{aligned}\) \(0.036=K_C(0.082 \times 500)^{-2}\)…
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