Question

1 mole of He and 1 mole A are taken in a 10 L rigid container at 400 K, and equilibrium \( A \rightleftharpoons B \) is established. Calculate the partial pressure of He and B at equilibrium if \( K_c = 4 \).

• A. 3.28 atm, 2.62 atm
• B. 2.6 atm, 3.28 atm
• C. 2.6 atm, 2.6 atm
• D. 3.28 atm, 3.28 atm

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The problem involves a gaseous equilibrium and an inert gas (Helium). We must determine the moles at equilibrium to find the individual partial pressures in the container.
Step 2: Key Formula or Approach:
1. Equilibrium constant \(K_c = \frac{[B]}{[A]}\).
2. Ideal Gas Law: \(PV = nRT\).
Step 3: Detailed Explanation:
Let \(x\) moles of A react to reach equilibrium:
Reaction: \(A \rightleftharpoons B\)
At \(t=0\): \(1 \text{ mol}\), \(0 \text{ mol}\).
At equilibrium: \(1-x \text{ mol}\), \(x \text{ mol}\).
Given \(V = 10 \text{ L}\). Concentrations: \([A] = \frac{1-x}{10}\), \([B] = \frac{x}{10}\).
\[ K_c = \frac{x/10}{(1-x)/10} = \frac{x}{1-x} = 4 \]
\[ x = 4 - 4x \implies 5x = 4 \implies x = 0.8 \text{ moles of B at equilibrium}. \]
Partial pressure of B (\(P_B\)):
\[ P_B = \frac{n_B RT}{V} = \frac{0.8 \times 0.0821 \times 400}{10} = 2.6272 \text{ atm} \approx 2.62 \text{ atm}. \]
Partial pressure of He (\(P_{He}\)):
He is inert and its moles remain 1.
\[ P_{He} = \frac{n_{He} RT}{V} = \frac{1 \times 0.0821 \times 400}{10} = 3.284 \text{ atm} \approx 3.28 \text{ atm}. \]
Step 4: Final Answer:
The partial pressure of Helium is 3.28 atm and B is 2.62 atm.