COMEDK · Chemistry · 7. Chemical Equilibrium
Consider the gaseous equilibrium
\(2\text{AB}_{2(g)} \rightleftharpoons 2\text{AB}_{(g)} + \text{B}_{2(g)}\)
The expression relating the degree of dissociation (\(\alpha\)) and equilibrium constant (\(K_p\)) and total pressure P is:
- A \(\left(\dfrac{K_p}{2P}\right)^{1/3}\)
- B \(\left(\dfrac{2K_p}{P}\right)^{1/3}\)
- C \(\left(\dfrac{2K_p}{P}\right)^{2}\)
- D \(\left(\dfrac{K_p}{2P}\right)^{1/2}\)
Answer & Solution
Correct Answer
(B) \(\left(\dfrac{2K_p}{P}\right)^{1/3}\)
Step-by-step Solution
Detailed explanation
The given reaction is \(2\text{AB}_{2(g)} \rightleftharpoons 2\text{AB}_{(g)} + \text{B}_{2(g)}\) Let the initial moles of \(\text{AB}_2\) be \(2\). Moles at equilibrium: \(\text{AB}_2 = 2(1-\alpha)\) \(\text{AB} = 2\alpha\) \(\text{B}_2 = \alpha\) Total moles at equilibrium…
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